1. Relations

Exercise 1.1

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1 A. Question

Let A be the set of all human beings in a town at a particular time. Determine whether each of the following relations are reflexive, symmetric and transitive:

R = {(x, y) : x and y work at the same place}

Answer

We have been given that,

A is the set of all human beings in a town at a particular time.

Here, R is the binary relation on set A.

So, recall that

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

Using these criteria, we can solve these.

We have,

R = {(x, y): x and y work at the same place}

Check for Reflexivity:

Since x & x are the same people then, x & x works at the same place.

Take yourself, for example, if you work at Bloomingdale then you work at Bloomingdale. Since you can’t work in two places at a particular time,

So, ∀ x ∈ A, then (x, x) ∈ R.

∴ R is Reflexive.

Check for Symmetry:

If x & y works at the same place,then, y and x also work at the same place.

If you & your friend, Chris was working in Bloomindale, then Chris and you are working in Bloomingdale only.

The only difference is in the way of writing, either you write your name and your friend’s name or your friend’s name and your name, it’s the same.

So, if (x, y) ∈ R, then (y, x) ∈ R

∀ x, y ∈ A

∴ R is Symmetric.

Check for Transitivity:

If x & y works at the same place and y & z works at the same place.

Then, x & z also works at the same place.

Say, if she & I was working in Bloomingdale and she & you were also working in Bloomingdale. Then, you and I are working in the same company.

So, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R.

∀ x, y, z ∈ A

∴ R is Transitive.

Hence, R is reflexive, symmetric and transitive.

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1 B. Question

Let A be the set of all human beings in a town at a particular time. Determine whether each of the following relations are reflexive, symmetric and transitive:

R = {(x, y) : x and y live in the same locality}

Answer

We have been given that,

A is the set of all human beings in a town at a particular time.

Here, R is the binary relation on set A.

So, recall that

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

Using these criteria, we can solve these.

We have,

R = {(x, y): x and y live in the same locality}

Check for Reflexivity:

Since x & x are the same people,then, x & x live in the same locality.

Take yourself, for example, if you lived in colony x then you live in colony x.

Since you can’t live in two places at a particular time.

So, ∀ x ∈ A, then (x, x) ∈ R.

∴ R is Reflexive.

Check for Symmetry:

If x & y live in the same locality, then, y & x also lives in the the same locality.

If you & your friend, Chris are neighbors, then you and Chris are neighbors only.

The only difference is in the way of writing, either you write your name and your friend’s name or your friend’s name and your name, it’s the same.

So, if (x, y) ∈ R, then (y, x) ∈ R.

∀ x, y ∈ R

∴ R is Symmetric.

Check for Transitivity:

If x & y lives in the same locality and y & z lives in the same locality.

Then, x & z also lives in the same locality.

Say, if she and I were living in colony x and she & you were also working in colony x. Then, you and I are living in the same colony.

So, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R.

∀ x, y, z ∈ A

∴ R is Transitive.

Hence, R is reflexive, symmetric and transitive.

1 C. Question

Let A be the set of all human beings in a town at a particular time. Determine whether each of the following relations are reflexive, symmetric and transitive:

R = {(x, y) : x is wife of y}

Answer

We have been given that,

A is the set of all human beings in a town at a particular time.

Here, R is the binary relation on set A.

So, recall that

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

Using these criteria, we can solve these.

We have,

R = {(x, y): x is wife of y}

Check for Reflexivity:

Since x and x are the same people.

Then, x cannot be the wife of itself.

A person cannot be a wife of itself.

Wendy is the wife of Sam; Wendy can’t be the wife of herself.

So, ∀ x ∈ A, then (x, x) ∉ R.

∴ R is not reflexive.

Check for Symmetry:

If x is the wife of y.Then, y cannot be the wife of x.

If Wendy is the wife of Sam, then Sam is the husband of Wendy.

Sam cannot be the wife of Wendy.

So, if (x, y) ∈ R, then (y, x) ∉ R.

∀ x, y ∈ A

∴ R is not symmetric.

Check for Transitivity:

If x is the wife of y and y is the wife of z, which is not logically possible.

Then, x is not the wife of z.

It’s easy, take Wendy, Sam, and Mac.

If Wendy is the wife of Sam, Sam can’t be the wife of Mac.

Thus, the possibility of Wendy being the wife of Mac also eliminates.

So, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∉ R.

∀ x, y, z ∈ A

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

1 D. Question

Let A be the set of all human beings in a town at a particular time. Determine whether each of the following relations are reflexive, symmetric and transitive:

R = {(x, y) : x is father of y}

Answer

We have been given that,

A is the set of all human beings in a town at a particular time.

Here, R is the binary relation on set A.

So, recall that

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

Using these criteria, we can solve these.

We have,

R = {(x, y): x is father of y}

Check for Reflexivity:

Since x and x are the same people.

Then, x cannot be the father of itself.

A person cannot be a father of itself.

Leo is the father of Thiago

So, ∀ x ∈ A, then (x, x) ∉ R.

∴ R is not reflexive.

Check for Symmetry:

If x is the father of y.

Then, y cannot be the father of x.

If Sam is the father of Mac, then Mac is the son of Sam.

Mac cannot be the father of Sam.

So, if (x, y) ∈ R, then (y, x) ∉ R.

∀ x, y ∈ A

∴ R is not symmetric.

Check for Transitivity:

If x is the father of y and y is the father of z,then, x is not the father of z.

Take Mickey, Sam, and Mac.

If Mickey is the father of Sam, and Sam is the father of Mac.

Thus, Mickey is not the father of Mac, but the grandfather of Mac.

So, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∉ R.

∀ x, y, z ∈ A

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

2. Question

Relations R₁, R₂, R₃and R₄are defined on a set A = {a, b, c} as follows :

R₁= {(a, a) (a, b) (a, c) (b, b) (b, c), (c, a) (c, b) (c, c)}

R₂= {(a, a)}

R₃= {(b, a)}

R₄= {(a, b) (b, c) (c, a)}

Find whether or not each of the relations R₁, R₂, R_3,R₄on A is (i) reflexive (iii) symmetric (iii) transitive. Answer

We have set,

A = {a, b, c}

Here, R₁, R₂, R_3,and R₄are the binary relations on set A.

So, recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

So, using these results let us start determining given relations.

We have

R₁= {(a, a) (a, b) (a, c) (b, b) (b, c) (c, a) (c, b) (c, c)}

(i). Reflexive:

For all a, b, c ∈ A. [∵ A = {a, b, c}]

Then, (a, a) ∈ R₁

(b, b) ∈ A

(c, c) ∈ A

[∵ R₁= {(a, a) (a, b) (a, c) (b, b) (b, c) (c, a) (c, b) (c, c)}]

So, ∀ a, b, c ∈ A, then (a, a), (b, b), (c, c) ∈ R.

∴ R₁is reflexive.

(ii). Symmetric:

If (a, a), (b, b), (c, c), (a, c), (b, c)∈ R₁

Then, clearly (a, a), (b, b), (c, c), (c, a), (c, b)∈ R₁

∀ a, b, c ∈ A

[∵ R₁= {(a, a) (a, b) (a, c) (b, b) (b, c) (c, a) (c, b) (c, c) }]

But, we need to try to show a contradiction to be able to determine the symmetry. So, we know (a, b) ∈ R₁

But, (b, a) ∉ R₁

So, if (a, b) ∈ R₁, then (b, a) ∉ R₁.

∀ a, b ∈ A

∴ R₁is not symmetric.

(iii). Transitive:

If (b, c) ∈ R₁and (c, a) ∈ R₁

But, (b, a) ∉ R₁[Check the Relation R₁that does not contain (b, a)] ∀ a, b ∈ A

[∵ R₁= {(a, a) (a, b) (a, c) (b, b) (b, c) (c, a) (c, b) (c, c)}] So, if (b, c) ∈ R₁and (c, a) ∈ R₁, then (b, a) ∉ R₁.

∀ a, b, c ∈ A

∴ R₁is not transitive.

Now, we have

R₂= {(a, a)}

(i). Reflexive:

Here, only (a, a) ∈ R₂

for a ∈ A. [∵ A = {a, b, c}]

[∵ R₂= {(a, a)}]

So, for a ∈ A, then (a, a) ∈ R₂.

∴ R₂is reflexive.

(ii). Symmetric:

For symmetry,

If (x, y) ∈ R, then (y, x) ∈ R

∀ x, y ∈ A.

Notice, in R₂we have

R₂= {(a, a)}

So, if (a, a) ∈ R₂, then (a, a) ∈ R₂.

Where a ∈ A.

∴ R₂is symmetric.

(iii). Transitive:

Here,

(a, a) ∈ R₂and (a, a) ∈ R₂

Then, obviously (a, a) ∈ R₂

Where a ∈ A.

[∵ R₂= {(a, a)}]

So, if (a, a) ∈ R₂and (a, a) ∈ R₂, then (a, a) ∈ R₂, where a ∈ A.

∴ R₂is transitive.

Now, we have

R₃= {(b, a)}

(i). Reflexive:

∀ a, b ∈ A [∵ A = {a, b, c}]

But, (a, a) ∉ R₃

Also, (b, b) ∉ R₃

[∵ R₃= {(b, a)}]

So, ∀ a, b ∈ A, then (a, a), (b, b) ∉ R₃

∴ R₃is not reflexive.

(ii). Symmetric:

If (b, a) ∈ R₃

Then, (a, b) should belong to R₃.

∀ a, b ∈ A. [∵ A = {a, b, c}]

But, (a, b) ∉ R₃

[∵ R₃= {(b, a)}]

So, if (a, b) ∈ R₃, then (b, a) ∉ R₃

∀ a, b ∈ A

∴ R₃is not symmetric.

(iii). Transitive:

We have (b, a) ∈ R₃but do not contain any other element in R₃. Transitivity can’t be proved in R₃.

[∵ R₃= {(b, a)}]

So, if (b, a) ∈ R₃but since there is no other element. ∴ R₃is not transitive.

Now, we have

R₄= {(a, b) (b, c) (c, a)}

(i). Reflexive:

∀ a, b, c ∈ A [∵ A = {a, b, c}]

But, (a, a) ∉ R₄

Also, (b, b) ∉ R₄and (c, c) ∉ R₄

[∵ R₄= {(a, b) (b, c) (c, a)}]

So, ∀ a, b, c ∈ A, then (a, a), (b, b), (c, c) ∉ R₄

∴ R₄is not reflexive.

(ii). Symmetric:

If (a, b) ∈ R₄, then (b, a) ∈ R₄

But (b, a) ∉ R₄

[∵ R₄= {(a, b) (b, c) (c, a)}]

So, ∀ a, b ∈ A, if (a, b) ∈ R₄, then (b, a) ∉ R₄.

⇒ R₄is not symmetric.

It is sufficient to show only one case of ordered pairs violating the definition. ∴ R₄is not symmetric.

(iii). Transitivity:

We have,

(a, b) ∈ R₄and (b, c) ∈ R₄

⇒ (a, c) ∈ R₄

But, is it so?

No, (a, c) ∉ R₄

So, it is enough to determine that R₄is not transitive.

∀ a, b, c ∈ A, if (a, b) ∈ R₄and (b, c) ∈ R₄, then (a, c) ∉ R₄.

∴ R₄is not transitive.

3 A. Question

Test whether the following relations R₁, R₂and R₃are (i) reflexive (ii) symmetric and (iii) transitive : R₁on Q₀defined by (a, b) ϵ R₁⇔ a = 1/b

Answer

Here, R₁, R₂, R_3,and R₄are the binary relations.

So, recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

So, using these results let us start determining given relations.

We have

R₁on Q₀defined by (a, b) ∈ R₁⇔

Check for Reflexivity:

∀ a, b ∈ Q₀,

(a, a), (b, b) ∈ R₁needs to be proved for reflexivity.

If (a, b) ∈ R₁

Then, …(1)

So, for (a, a) ∈ R₁

Replace b by a in equation (1), we get

But, we know

⇒ (a, a) ∉ R₁

So, ∀ a ∈ Q₀, then (a, a) ∉ R₁

∴ R₁is not reflexive.

Check for Symmetry:

If (a, b) ∈ R₁

Then, (b, a) ∈ R₁

∀ a, b ∈ Q₀

If (a, b) ∈ R₁

We have, …(2)

Now, for (b, a) ∈ R₁

Replace a by b & b by a in equation (2), we get

⇒ (b, a) ∈ R₂

So, if (a, b) ∈ R₁, then (b, a) ∈ R₁

∀ a, b ∈ Q₀

∴ R₁is symmetric.

Check for Transitivity:

If (a, b) ∈ R₁and (b, c) ∈ R₁

We need to eliminate b.

We have

Putting in , we get

But,

⇒ (a, c) ∉ R₁

So, if (a, b) ∈ R₁and (b, c) ∈ R₁, then (a, c) ∉ R₁

∀ a, b, c ∈ Q₀

∴ R₁is not transitive.

3 B. Question

Test whether the following relations R₁, R₂and R₃are (i) reflexive (ii) symmetric and (iii) transitive : R₂on Z defined by (a, b) ϵ R₂⇔ |a – b| ≤ 5

Answer

Here, R₁, R₂, R_3,and R₄are the binary relations.

So, recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

So, using these results let us start determining given relations.

We have

R₂on Z defined by (a, b) ∈ R₂⇔ |a – b| ≤ 5

Check for Reflexivity:

∀ a ∈ Z,

(a, a) ∈ R₂needs to be proved for reflexivity.

If (a, b) ∈ R₂

Then, |a – b| ≤ 5 …(1)

So, for (a, a) ∈ R₁

Replace b by a in equation (1), we get

|a – a| ≤ 5

⇒ 0 ≤ 5

⇒ (a, a) ∈ R₂

So, ∀ a ∈ Z, then (a, a) ∈ R₂

∴ R₂is reflexive.

Check for Symmetry:

∀ a, b ∈ Z

If (a, b) ∈ R₂

We have, |a – b| ≤ 5 …(2)

Replace a by b & b by a in equation (2), we get

|b – a| ≤ 5

Since, the value is in mod, |b – a| = |a – b|

⇒ The statement |b – a| ≤ 5 is true.

⇒ (b, a) ∈ R₂

So, if (a, b) ∈ R₂, then (b, a) ∈ R₂

∀ a, b ∈ Q₀

∴ R₁is symmetric.

Check for Transitivity:

∀ a, b, c ∈ Z

If (a, b) ∈ R₂and (b, c) ∈ R₂

⇒ |a – b| ≤ 5 and |b – c| ≤ 5

Since, inequalities cannot be added or subtract. We need to take example to check for, |a – c| ≤ 5

Take values a = 18, b = 14 and c = 10

Check: |a – b| ≤ 5

⇒ |18 – 14| ≤ 5

⇒ |4| ≤ 5 is true.

Check: |b – c| ≤ 5

⇒ |14 – 10| ≤ 5

⇒ |4| ≤ 5

Check: |a – c| ≤ 5

⇒ |18 – 10| ≤ 5

⇒ |8| ≤ 5 is not true.

⇒ (a, c) ∉ R₂

So, if (a, b) ∈ R₂and (b, c) ∈ R₂, then (a, c) ∉ R₁

∀ a, b, c ∈ Z

∴ R₂is not transitive.

3 C. Question

Test whether the following relations R₁, R₂and R₃are (i) reflexive (ii) symmetric and (iii) transitive : R₃on R defined by (a, b) ϵ R₃⇔ a²– 4 ab + 3b²= 0.

Answer

Here, R₁, R₂, R_3,and R₄are the binary relations.

So, recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

So, using these results let us start determining given relations.

We have

R₃on R defined by (a, b) ∈ R₃⇔ a²– 4ab + 3b²= 0

Check for Reflexivity:

∀ a ∈ R,

(a, a) ∈ R₃needs to be proved for reflexivity. If (a, b) ∈ R₃, then we have

a²– 4ab + 3b²= 0

Replace b by a, we get

a²– 4aa + 3a²= 0

⇒ a²– 4a²+ 3a²= 0

⇒ –3a²+ 3a²= 0

⇒ 0 = 0, which is true.

⇒ (a, a) ∈ R₃

So, ∀ a ∈ R, (a, a) ∈ R₃

∴ R₃is reflexive.

Check for Symmetry:

∀ a, b ∈ R

If (a, b) ∈ R₃, then we have

a²– 4ab + 3b²= 0

⇒ a²– 3ab – ab + 3b²= 0

⇒ a (a – 3b) – b (a – 3b) = 0

⇒ (a – b) (a – 3b) = 0

⇒ (a – b) = 0 or (a – 3b) = 0

⇒ a = b or a = 3b …(1)

Replace a by b and b by a in equation (1), we get ⇒ b = a or b = 3a …(2)

Results (1) and (2) does not match.

⇒ (b, a) ∉ R₃

∴ R₃is not symmetric.

Check for Transitivity:

∀ a, b, c ∈ R

If (a, b) ∈ R₃and (b, c) ∈ R₃

⇒ a²– 4ab + 3b²= 0 and b²– 4bc + 3c²= 0 ⇒ a²– 3ab – ab + 3b²= 0 and b²– 3bc – bc + 3c² ⇒ a (a – 3b) – b (a – 3b) = 0 and b (b – 3c) – c (b – 3c) = 0 ⇒ (a – b) (a – 3b) = 0 and (b – c) (b – 3c) = 0 ⇒ (a – b) = 0 or (a – 3b) = 0

And (b – c) = 0 or (b – 3c) = 0

⇒ a = b or a = 3b And b = c or b = 3c

What we need to prove here is that, a = c or a = 3c

Take a = b and b = c

Clearly implies that a = c.

[∵ if a = b, just substitute a in place of b in b = c. We get, a = c]

Now, take a = 3b and b = 3c

If a = 3b

Substitute in b = 3c. We get

⇒ a = 9c, which is not the desired result.

⇒ (a, c) ∉ R₃

∴ R₃is not transitive.

4. Question

Let A = {1, 2, 3}, and let R₁= {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3), (2, 2), (2, 1), (3, 3)}, R₂={(2,2), (3,1), (1, 3)}, R₃= {(1, 3),(3, 3)}. Find whether or not each of the relations R₁, R₂, R₃on A is (i) reflexive (ii) symmetric (iii) transitive.

Answer

We have been given,

A = {1, 2, 3}

Here, R₁, R_2,and R₃are the binary relations on A.

So, recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

So, using these results let us start determining given relations.

Let us take R₁.

R₁= {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}

(i). Reflexive:

∀ 1, 2, 3 ∈ A [∵ A = {1, 2, 3}]

(1, 1) ∈ R₁

(2, 2) ∈ R₂

(3, 3) ∈ R₃

So, for a ∈ A, (a, a) ∈ R₁

∴ R₁is reflexive.

(ii). Symmetric:

∀ 1, 2, 3 ∈ A

If (1, 3) ∈ R₁, then (3, 1) ∈ R₁

[∵ R₁= {(1, 1), (1, 3), (3, 1) , (2, 2), (2, 1), (3, 3)}] But if (2, 1) ∈ R₁, then (1, 2) ∉ R₁

[∵ R₁= {(1, 1), (1, 3), (3, 1), (2, 2),(2, 1), (3, 3)}] So, if (a, b) ∈ R₁, then (b, a) ∉ R₁

∀ a, b ∈ A

∴ R₁is not symmetric.

(iii). Transitivity:

∀ 1, 2, 3 ∈ A

If (1, 3) ∈ R₁and (3, 3) ∈ R₁

Then, (1, 3) ∈ R₁

[∵ R₁= {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1),(3, 3)}] But, if (2, 1) ∈ R₁and (1, 3) ∈ R₁

Then, (2, 3) ∉ R₁

So, if (a, b) ∈ R₁and (b, c) ∈ R₁, then (a, c) ∉ R₁ ∀ a, b, c ∈ A

∴ R₁is not transitive.

Now, take R₂.

R₂= {(2, 2), (3, 1), (1, 3)}

(i). Reflexive:

∀ 1, 2, 3 ∈ A [∵ A = {1, 2, 3}]

(1, 1) ∉ R₂

(2, 2) ∈ R₂

(3, 3) ∉ R₂

So, for a ∈ A, (a, a) ∉ R₂

∴ R₂is not reflexive.

(ii). Symmetric:

∀ 1, 2, 3 ∈ A

If (1, 3) ∈ R₂, then (3, 1) ∈ R₂

[∵ R₂= {(2, 2), (3, 1), (1, 3)}]

If (2, 2) ∈ R₂, then (2, 2) ∈ R₂

[∵ R₂= {(2, 2), (3, 1), (1, 3)}]

So, if (a, b) ∈ R₂, then (b, a) ∈ R₂

∀ a, b ∈ A

∴ R₂is symmetric.

(iii). Transitivity:

∀ 1, 2, 3 ∈ A

If (1, 3) ∈ R₂and (3, 1) ∈ R₂

Then, (1, 1) ∉ R₂

[∵ R₂= {(2, 2), (3, 1), (1, 3)}]

So, if (a, b) ∈ R₂and (b, c) ∈ R₂, then (a, c) ∉ R₂

∀ a, b, c ∈ A

∴ R₂is not transitive.

Now take R₃.

R₃= {(1, 3), (3, 3)}

(i). Reflexive:

∀ 1, 3 ∈ A [∵ A = {1, 2, 3}]

(1, 1) ∉ R₃

(3, 3) ∈ R₃

So, for a ∈ A, (a, a) ∉ R₃

∴ R₃is not reflexive.

(ii). Symmetric:

∀ 1, 3 ∈ A

If (1, 3) ∈ R₃, then (3, 1) ∉ R₃

[∵ R₃= {(1, 3), (3, 3)}]

So, if (a, b) ∈ R₃, then (b, a) ∉ R₃

∀ a, b ∈ A

∴ R₃is not symmetric.

(iii). Transitivity:

∀ 1, 3 ∈ A

If (1, 3) ∈ R₃and (3, 3) ∈ R₃

Then, (1, 3) ∈ R₃

[∵ R₃= {(1, 3), (3, 3)}]

So, if (a, b) ∈ R₃and (b, c) ∈ R₃, then (a, c) ∈ R₃

∀ a, b, c ∈ A

∴ R₃is transitive.

5 A. Question

The following relations are defined on the set of real numbers : aRb if a – b > 0

Find whether these relations are reflexive, symmetric or transitive. Answer

Let set of real numbers be ℝ.

So, recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz. We have

aRb if a – b > 0

Check for Reflexivity:

For a ∈ ℝ

If aRa,

⇒ a – a > 0

⇒ 0 > 0

But 0 > 0 is not possible.

Hence, aRa is not true.

So, ∀ a ∈ ℝ, then aRa is not true.

⇒ R is not reflexive.

∴ R is not reflexive.

Check for Symmetry:

∀ a, b ∈ ℝ

If aRb,

⇒ a – b > 0

Replace a by b and b by a, we get

⇒ b – a > 0

[Take a = 12 and b = 6.

a – b > 0

⇒ 12 – 6 > 0

⇒ 6 > 0, which is a true statement.

Now, b – a > 0

⇒ 6 – 12 > 0

⇒ –6 > 0, which is not a true statement as –6 is not greater than 0.] ⇒ bRa is not true.

So, if aRb is true, then bRa is not true.

∀ a, b ∈ ℝ

⇒ R is not symmetric.

∴ R is not symmetric.

Check for Transitivity:

∀ a, b, c ∈ ℝ

If aRb and bRc.

⇒ a – b > 0 and b – c > 0

⇒ a – c > 0 or not.

Let us check.

a – b > 0 means a > b.

b – c > 0 means b > c.

a – c > 0 means a > c.

If a > b and b > c,

⇒ a > b, b > c

⇒ a > b > c

⇒ a > c

Hence, aRc is true.

So, if aRb is true and bRc is true, then aRc is true.

∀ a, b, c ∈ ℝ

⇒ R is transitive.

∴ R is transitive.

5 B. Question

The following relations are defined on the set of real numbers :

aRb if 1 + ab > 0

Find whether these relations are reflexive, symmetric or transitive.

Answer

Let set of real numbers be ℝ.

So, recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

We have

aRb if 1 + ab > 0

Check for Reflexivity:

For a ∈ ℝ

If aRa,

⇒ 1 + aa > 0

⇒ 1 + a²> 0

If a is a real number.

[Positive or negative, large or small, whole numbers or decimal numbers are all Real Numbers. Real numbers are so called because they are ‘Real’ not ‘imaginary’.]

This means, even if ‘a’ was negative.

a²= positive.

a²+ 1 = positive

And any positive number is greater than 0. Hence, 1 + a²> 0

⇒ aRa is true.

So, ∀ a ∈ ℝ, then aRa is true.

⇒ R is reflexive.

∴ R is reflexive.

Check for Symmetry:

∀ a, b ∈ ℝ

If aRb,

⇒ 1 + ab > 0

Replace a by b and b by a, we get ⇒ 1 + ba > 0

Whether we write ab or ba, it is equal. ab = ba

So, 1 + ba > 0

⇒ bRa is true.

So, if aRb is true, then bRa is true. ∀ a, b ∈ ℝ

⇒ R is symmetric.

∴ R is symmetric.

Check for Transitivity:

∀ a, b, c ∈ ℝ

If aRb and bRc.

⇒ 1 + ab > 0 and 1 + bc > 0

⇒ 1 + ac > 0 or not.

Let us check.

1 + ab > 0 means ab > –1.

1 + bc > 0 means bc > –1.

1 + ac > 0 means ac > –1.

If ab > –1 and bc > –1.

⇒ ac > –1 should be true.

Take a = –1, b = 0.9 and c = 1

ab > –1

⇒ (–1)(0.9) > –1

⇒ –0.9 > –1, is true on the number line. bc > –1

⇒ (0.9)(1) > –1

⇒ 0.9 > –1, is true on the number line.

ac > –1

⇒ (–1)(1) > –1

⇒ –1 > –1, is not true as –1 cannot be greater than itself.

⇒ ac > –1 is not true.

⇒ 1 + ac > 0 is not true.

⇒ aRc is not true.

So, if aRb is true and bRc is true, then aRc is not true.

∀ a, b, c ∈ ℝ

⇒ R is not transitive.

∴ R is not transitive.

5 C. Question

The following relations are defined on the set of real numbers :

aRb if | a | ≤ b.

Find whether these relations are reflexive, symmetric or transitive.

Answer

Let set of real numbers be ℝ.

So, recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

We have

aRb if |a| ≤ b

Check for Reflexivity:

For a ∈ ℝ

If aRa,

⇒ |a| ≤ a, is true

If a is a real number.

[Positive or negative, large or small, whole numbers or decimal numbers are all Real Numbers. Real numbers are so called because they are ‘Real’ not ‘imaginary’.]

This means, even if ‘a’ was negative.

|a| = positive.

& |a| ≤ a

Hence, |a| ≤ a.

⇒ aRa is true.

So, ∀ a ∈ ℝ, then aRa is true.

⇒ R is reflexive.

∴ R is reflexive.

Check for Symmetry:

∀ a, b ∈ ℝ

If aRb,

⇒ |a| ≤ b

Replace a by b and b by a, we get

⇒ |b| ≤ a, which might be true or not.

Let a = 2 and b = 3.

|a| ≤ b

⇒ |2| ≤ 3, is true

|b| ≤ a

⇒ |3| ≤ 2, is not true

⇒ bRa is not true.

So, if aRb is true, then bRa is not true.

∀ a, b ∈ ℝ

⇒ R is not symmetric.

∴ R is not symmetric.

Check for Transitivity:

∀ a, b, c ∈ ℝ

If aRb and bRc.

⇒ |a| ≤ b and |b| ≤ c

⇒ |a| ≤ c or not.

Let us check.

If |a| ≤ b and |b| ≤ c

b ≠ |b|

Say, if b = –2

⇒ –2 ≠ |–2|

⇒ –2 ≠ 2

But, from |a| ≤ b

b ≥ 0 in every case otherwise the statement would not hold true. ⇒ b can only accept positive values including 0.

⇒ b is a whole number.

∴ if |a| ≤ b and |b| ≤ c

⇒ |a| ≤ b, b ≤ c

⇒ |a| ≤ b ≤ c

⇒ |a| ≤ c

⇒ aRc is true.

So, if aRb is true and bRc is true, then aRc is true.

∀ a, b, c ∈ ℝ

⇒ R is transitive.

∴ R is transitive.

6. Question

Check whether the relation R defined on the set A={1,2,3,4,5,6} as R= {(a, b) : b = a + 1} is reflexive, symmetric or transitive.

Answer

We have the set A = {1, 2, 3, 4, 5, 6}

So, recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

We have

R = {(a, b): b = a + 1}

∵ Every a, b ∈ A.

And A = {1, 2, 3, 4, 5, 6}

The relation R on set A can be defined as:

Put a = 1

⇒ b = a + 1

⇒ b = 1 + 1

⇒ b = 2

⇒ (a, b) ≡ (1, 2)

Put a = 2

⇒ b = 2 + 1

⇒ b = 3

⇒ (a, b) ≡ (2, 3)

Put a = 3

⇒ b = 3 + 1

⇒ b = 4

⇒ (a, b) ≡ (3, 4)

Put a = 4

⇒ b = 4 + 1

⇒ b = 5

⇒ (a, b) ≡ (4, 5)

Put a = 5

⇒ b = 5 + 1

⇒ b = 6

⇒ (a, b) ≡ (5, 6)

Put a = 6

⇒ b = 6 + 1

⇒ b = 7

⇒ (a, b) ≠ (6, 7) [∵ 7 ∉ A]

Hence, R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

Check for Reflexivity:

For 1, 2, …, 6 ∈ A [∵ A = {1, 2, 3, 4, 5, 6}]

(1, 1) ∉ R

(2, 2) ∉ R

…

(6, 6) ∉ R

So, ∀ a ∈ A, then (a, a) ∉ R.

⇒ R is not reflexive.

∴ R is not reflexive.

Check for Symmetry:

∀ 1, 2 ∈ A [∵ A = {1, 2, 3, 4, 5, 6}]

If (1, 2) ∈ R

Then, (2, 1) ∉ R

[∵ R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}]

So, if (a, b) ∈ R, then (b, a) ∉ R

∀ a, b ∈ A

⇒ R is not symmetric.

∴ R is not symmetric.

Check for Transitivity:

∀ 1, 2, 3 ∈ A

If (1, 2) ∈ R and (2, 3) ∈ R

Then, (1, 3) ∉ R

[∵ R = {(1, 2), (2, 3) , (3, 4), (4, 5), (5, 6)}]

So, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∉ R.

∀ a, b, c ∈ A

⇒ R is not transitive.

∴ R is not transitive.

7. Question

Check whether the relation R on R defined by R = {(a, b) : a ≤ b³} is reflexive, symmetric or transitive. Answer

We have the set of real numbers, R.

So, recall that for any binary relation R on set A. We have, R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx. R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz. We have

R = {(a, b): a ≤ b³}

Check for Reflexivity:

For a ∈ R

If (a, a) ∈ R,

⇒ a ≤ a³, which is not true.

Say, if a = – 2.

a ≤ a³

⇒ – 2 ≤ – 8

⇒ –2 ≤ –8, which is not true as – 2 > – 8.

Hence, (a, a) ∉ R

So, ∀ a ∈ R, then (a, a) ∉ R.

⇒ R is not reflexive.

∴ R is not reflexive.

Check for Symmetry:

∀ a, b ∈ R

If (a, b) ∈ R

⇒ a ≤ b³

Replace a by b and b by a, we get

⇒ b ≤ a³

[Take a = –2 and b = 3.

a ≤ b³

⇒ –2 ≤ 3³

⇒ –2 ≤ 27, which is a true statement.

Now, b ≤ a³

⇒ 3 ≤ (–2)³

⇒ 3 ≤ –8, which is not a true statement as 3 > –8] ⇒ (b, a) ∉ R

So, if (a, b) ∈ R, then (b, a) ∉ R

∀ a, b ∈ R

⇒ R is not symmetric.

∴ R is not symmetric.

Check for Transitivity:

∀ a, b, c ∈ R

If (a, b) ∈ R and (b, c) ∈ R

⇒ a ≤ b³and b ≤ c³

⇒ a ≤ c³or not.

Let us check.

Take a = 3, and .

a ≤ b³

⇒ 3 ≤ 3.37, which is true.

b ≤ c³

⇒ 1.5 ≤ 1.728

a ≤ c³

⇒ 3 ≤ 1.728, which is not true as 3 > 1.728.

Hence, (a, c) ∉ R.

So, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∉ R.

∀ a, b, c ∈ ℝ

⇒ R is not transitive.

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

8. Question

Prove that every identity relation on a set is reflexive, but the converse is not necessarily true. Answer

To Prove: Every identity relation on a set is reflexive, but every reflexive relation is not identity relation. Proof:

Let us first understand what ‘Reflexive Relation’ is and what ‘Identity Relation’ is.

Reflexive Relation: A binary relation R over a set A is reflexive if every element of X is related to itself. Formally, this may be written as ∀ x ∈ A: xRx.

Identity Relation: Let A be any set.

Then the relation R= {(x, x): x ∈ A} on A is called the identity relation on A. Thus, in an identity relation, every element is related to itself only.

Let A = {a, b, c} be a set.

Let R be a binary relation defined on A.

Let R_A= {(a, a): a ∈ A} is the identity relation on A.

Hence, every identity relation on set A is reflexive by definition.

Converse: Let A = {a, b, c} is the set.

Let R = {(a, a), (b, b), (c, c), (a, b), (c, a)} be a relation defined on A.

R is reflexive as per definition.

[∵ (a, a) ∈ R, (b, b) ∈ R & (c, c) ∈ R]

But, (a, b) ∈ R

(c, a) ∈ R

⇒ R is not identity relation by definition.

Hence, proved that every identity relation on a set is reflexive, but the converse is not necessarily true.

9 A. Question

If A = {1, 2, 3, 4}, define relations on A which have properties of being

reflexive, transitive but not symmetric.

Answer

Recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

Using these properties, we can define R on A.

A = {1, 2, 3, 4}

We need to define a relation (say, R) which is reflexive, transitive but not symmetric. Let us try to form a small relation step by step.

The relation must be defined on A.

Reflexive relation:

R = {(1, 1), (2, 2), (3, 3), (4, 4)} …(1)

Transitive relation:

R = {(1, 2), (2, 1), (1, 1)}, is transitive but also symmetric.

So, let us define another relation.

R = {(1, 3), (3, 2), (1, 2)}, is transitive and not symmetric. …(2)

Let us combine (i) and (ii) relation.

R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 3), (3, 2), (1, 2)} …(A)

(A) can be shortened by eliminating (3, 2) and (1, 2) from R.

R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 3)} …(B)

Further (B) can be shortened by eliminating (2, 2) and (4, 4).

R = {(1, 1), (3, 3), (1, 3)} …(C)

All the results (A), (B) and (C) is correct.

Thus, we have got the relation which is reflexive, transitive but not symmetric.

9 B. Question

If A = {1, 2, 3, 4}, define relations on A which have properties of being

symmetric but neither reflexive nor transitive.

Answer

Recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

Using these properties, we can define R on A.

A = {1, 2, 3, 4}

We need to define a relation (say, R) which is symmetric but neither reflexive nor transitive. The relation R must be defined on A.

Symmetric relation:

R = {(1, 2), (2, 1)}

Note that, the relation R here is neither reflexive nor transitive, and it is the shortest relation that can be form.

Similarly, we can also write:

R = {(1, 3), (3, 1)}

Or R = {(3, 4), (4, 3)}

Or R = {(2, 3), (3, 2), (1, 4), (4, 1)}

And so on…

All of these are right answers.

Thus, we have got the relation which is symmetric but neither reflexive nor transitive. 9 C. Question

If A = {1, 2, 3, 4}, define relations on A which have properties of being

reflexive, symmetric and transitive.

Answer

Recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

Using these properties, we can define R on A.

A = {1, 2, 3, 4}

We need to define a relation (say, R) which is reflexive, symmetric and transitive.

The relation must be defined on A.

Reflexive Relation:

R = {(1, 1), (2, 2), (3, 3), (4, 4)}

Or simply shorten it and write,

R = {(1, 1), (2, 2)} …(1)

Symmetric Relation:

R = {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3)}

Or simply shorten it and write,

R = {(1, 2), (2, 1)} …(2)

Combine results (1) and (2), we get

R = {(1, 1), (2, 2), (1, 2), (2, 1)}

It is reflexive, symmetric as well as transitive as per definition.

Similarly, we can find other combinations too.

Thus, we have got the relation which is reflexive, symmetric as well as transitive.

10. Question

Let R be a relation defined on the set of natural numbers N as

R = {(x, y) : x, y ∈ N, 2x + y = 41}

Find the domain and range of R. Also, verify whether R is

(i) reflexive,

(ii) symmetric

(iii) transitive.

Answer

First let us define what range and domain are.

Range: The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually) after we have substituted the domain. In plain English, the definition means: The range is the resulting y-values we get after substituting all the possible x-values.

Domain: The domain of definition of a function is the set of “input” or argument values for which the function is defined. That is, the function provides an “output” or value for each member of the domain.

We have been given that, R is a relation defined on N.

N = set of natural numbers

R = {(x, y): x, y ∈ N, 2x + y = 41}

We have the function as

2x + y = 41

⇒ y = 41 – 2x

As y ∈ N (Natural number)

⇒ 41 – 2x ≥ 1

⇒ –2x ≥ 1 – 41

⇒ –2x ≥ –40

⇒ 2x ≤ 40

⇒ x ≤ 20

So, the domain is first 20 natural numbers. As, 2x + y = 41

⇒ 2x = 41 – y

As x ∈ N (Natural number)

⇒ 41 – y ≥ 2

⇒ –y ≥ 2 – 41

⇒ –y ≥ –39

⇒ y ≤ 39

So, the range is first 39 natural numbers. We have relation R defined on set N. R = {(x, y): x, y ∈ N, 2x + y = 41}

Check for Reflexivity:

∀ x ∈ N

If (x, x) ∈ R

⇒ 2x + x = 41

⇒ 3x = 41

⇒ x = 13.67

But, x ≠ 13.67 as x ∈ N.

⇒ (x, x) ∉ R

So, ∀ x ∈ N, then (x, x) ∉ R

⇒ R is not reflexive.

∴ R is not reflexive.

Check for Symmetry:

∀ x, y ∈ N

If (x, y) ∈ R

⇒ 2x + y = 41 …(i)

Now, replace x by y and y by x, we get ⇒ 2y + x = 41 …(ii)

Take x = 20 and y = 1.

Equation (i) ⇒ 2(20) + 1 = 41

⇒ 40 + 1 = 41

⇒ 41 = 41, holds true.

Equation (ii) ⇒ 2(1) + 20 = 41

⇒ 2 + 20 = 41

⇒ 22 = 41, which is not true as 22 ≠ 41.

⇒ (y, x) ∉ R

So, if (x, y) ∈ R, then (y, x) ∉ R.

∀ x, y ∈ N

⇒ R is not symmetric.

∴ R is not symmetric.

Check for Transitivity:

∀ x, y, z ∈ N

If (x, y) ∈ R and (y, z) ∈ R

⇒ 2x – y = 41 and 2y – z = 41

⇒ 2x – z = 41, may be true or not.

Let us sole these to find out.

We have

2x – y = 41 …(iii)

2y – z = 41 …(iv)

Multiply 2 by equation (i), we get

4x – 2y = 82 …(v)

Adding equation (v) and (iv), we get

(4x – 2y) + (2y – z) = 82 + 41

⇒ 4x – z = 123

⇒ 2x + 2x – z = 123

⇒ 2x – z = 123 – 2x

Take x = 40 (as x ∈ N)

⇒ 2x – z = 123 – 2(40)

⇒ 2x – z = 123 – 80

⇒ 2x – z = 43 ≠ 41

⇒ (x, z) ∉ R

So, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∉ R.

∀ x, y, z ∈ N

⇒ R is not transitive.

∴ R is not transitive.

11. Question

Is it true that every relation which is symmetric and transitive is also reflexive? Give reasons. Answer

It is not true that every relation which is symmetric and transitive is also reflexive.

Take for example:

Take a set A = {1, 2, 3, 4}

And define a relation R on A.

Symmetric relation:

R = {(1, 2), (2, 1)}, is symmetric on set A.

Transitive relation:

R = {(1, 2), (2, 1), (1, 1)}, is the simplest transitive relation on set A.

⇒ R = {(1, 2), (2, 1), (1, 1)} is symmetric as well as transitive relation.

But R is not reflexive here.

If only (2, 2) ∈ R, had it been reflexive.

Thus, it is not true that every relation which is symmetric and transitive is also reflexive. 12. Question

An integer m is said to be related to another integer n if m is a multiple of n. Check if the relation is symmetric, reflexive and transitive.

Answer

According to the question,

m is related to n if m is a multiple of n.

∀ m, n ∈ I (I being set of integers)

The relation comes out to be:

R = {(m, n): m = kn, k ∈ ℤ}

Recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

Check for Reflexivity:

∀ m ∈ I

If (m, m) ∈ R

⇒ m = k m, holds.

As an integer is always a multiple of itself, So, ∀ m ∈ I, then (m, m) ∈ R.

⇒ R is reflexive.

∴ R is reflexive.

Check for Symmetry:

∀ m, n ∈ I

If (m, n) ∈ R

⇒ m = k n, holds.

Now, replace m by n and n by m, we get

n = k m, which may or not be true.

Let us check:

If 12 is a multiple of 3, but 3 is not a multiple of 12.

⇒ n = km does not hold.

So, if (m, n) ∈ R, then (n, m) ∉ R.

∀ m, n ∈ I

⇒ R is not symmetric.

∴ R is not symmetric.

Check for Transitivity:

∀ m, n, o ∈ I

If (m, n) ∈ R and (n, o) ∈ R

⇒ m = kn and n = ko

Where k ∈ ℤ

Substitute n = ko in m = kn, we get

m = k(ko)

⇒ m = k²o

If k ∈ ℤ, then k²∈ ℤ.

Let k²= r

⇒ m = ro, holds true.

⇒ (m, o) ∈ R

So, if (m, n) ∈ R and (n, o) ∈ R, then (m, o) ∈ R.

∀ m, n ∈ I

⇒ R is transitive.

∴ R is transitive.

13. Question

Show that the relation “≥” on the set R of all real numbers is reflexive and transitive but not symmetric. Answer

We have

The relation “≥” on the set R of all real numbers.

Recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

So, let the relation having “≥” be P.

We can write

P = {(a, b): a ≥ b, a, b ∈ R}

Check for Reflexivity:

∀ a ∈ R

If (a, a) ∈ P

⇒ a ≥ a, which is true.

Since, every real number is equal to itself.

So, ∀ a ∈ R, then (a, a) ∈ P.

⇒ P is reflexive.

∴ P is reflexive.

Check for Symmetry:

∀ a, b ∈ R

If (a, b) ∈ P

⇒ a ≥ b

Now, replace a by b and b by a. We get

b ≥ a, might or might not be true.

Let us check:

Take a = 7 and b = 5.

a ≥ b

⇒ 7 ≥ 5, holds.

b ≥ a

⇒ 5 ≥ 7, is not true as 5 < 7.

⇒ b ≥ a, is not true.

⇒ (b, a) ∉ P

So, if (a, b) ∈ P, then (b, a) ∉ P

∀ a, b ∈ R

⇒ P is not symmetric.

∴ P is not symmetric.

Check for Transitivity:

∀ a, b, c ∈ R

If (a, b) ∈ P and (b, c) ∈ P

⇒ a ≥ b and b ≥ c

⇒ a ≥ b ≥ c

⇒ a ≥ c

⇒ (a, c) ∈ P

So, if (a, b) ∈ P and (b, c) ∈ P, then (a, c) ∈ P

∀ a, b, c ∈ R

⇒ P is transitive.

∴ P is transitive.

Thus, shown that the relation “≥” on the set R of all the real numbers are reflexive and transitive but not symmetric.

14 A. Question

Give an example of a relation which is

reflexive and symmetric but not transitive.

Answer

Recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

Let there be a set A.

A = {1, 2, 3, 4}

We need to define a relation on A which is reflexive and symmetric but not transitive. Let there be a set A.

A = {1, 2, 3, 4}

Reflexive relation:

R = {(1, 1), (2, 2), (3, 3), (4, 4)} …(1)

Symmetric relation:

R = {(3, 4), (4, 3)} …(2)

Combine results (1) and (2), we get

R = {(1, 1), (2, 2), (3, 3), (4, 4), (3, 4), (4, 3)}

Check for Transitivity:

If (3, 4) ∈ R and (4, 3) ∈ R

Then, (3, 3) ∈ R

∀ 3, 4 ∈ A [∵ A = {1, 2, 3, 4}]

So eliminate (3, 3) from R, we get

R = {(1, 1), (2, 2), (4, 4), (3, 4), (4, 3)}

Check for Transitivity:

If (4, 3) ∈ R and (3, 4) ∈ R

Then, (4, 4) ∈ R

∀ 3, 4 ∈ A

So, eliminate (4, 4) from R, we get

R = {(1, 1), (2, 2), (3, 4), (4, 3)}

Thus, the relation which is reflexive and symmetric but not transitive is: R = {(1, 1), (2, 2), (3, 4), (4, 3)}

14 B. Question

Give an example of a relation which is

reflexive and transitive but not symmetric.

Answer

Recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

Let there be a set A.

A = {1, 2, 3, 4}

We need to define a relation on A which is reflexive and transitive but not symmetric. Let there be a set A.

A = {1, 2, 3, 4}

Reflexive relation:

R = {(1, 1), (2, 2), (3, 3), (4, 4)} …(1)

Transitive relation:

R = {(3, 4), (4, 1), (3, 1)} …(2)

Combine results (1) and (2), we get

R = {(1, 1), (2, 2), (3, 3), (4, 4), (3, 4), (4, 1), (3, 1)}

Check for Symmetry:

If (3, 4) ∈ R

Then, (4, 3) ∉ R

∀ 3, 4 ∈ A [∵ A = {1, 2, 3, 4}]

One example is enough to prove that R is not symmetric.

Thus, the relation which is reflexive and transitive but not symmetric is:

R = {(1, 1), (2, 2), (3, 3), (4, 4), (3, 4), (4, 1), (3, 1)}

14 C. Question

Give an example of a relation which is

symmetric and transitive but not reflexive.

Answer

Recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

Let there be a set A.

A = {1, 2, 3, 4}

We need to define a relation on A which is symmetric and transitive but not reflexive.

It is not possible to define such relation which is symmetric and transitive but not reflexive. As every relation which is symmetric and transitive will use identity ordered pair of the form (x, x) to balance the relation (to make the relation symmetric and transitive). Without such identity pair both, symmetry and transitivity will not be possible.

14 D. Question

Give an example of a relation which is

symmetric but neither reflexive nor transitive.

Answer

Recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

Let there be a set A.

A = {1, 2, 3, 4}

We need to define a relation which is symmetric but neither reflexive nor transitive. Let there be a set A.

A = {1, 2, 3, 4}

Symmetric Relation:

{(1, 3), (3, 1)}

This is neither reflexive nor transitive.

∵ (1, 1) ∉ R

(3, 3) ∉ R

Hence, R is not reflexive.

∵ (1, 3) ∈ R and (3, 1) ∈ R

Then, (1, 1) ∉ R

Hence, R is not transitive.

Thus, the relation which is symmetric but neither nor transitive is: R = {(1, 3), (3, 1)}

14 E. Question

Give an example of a relation which is

transitive but neither reflexive nor symmetric.

Answer

Recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

Let there be a set A.

A = {1, 2, 3, 4}

We need to define a relation which is transitive but neither reflexive nor symmetric. Let there be a set A.

A = {1, 2, 3}

Transitive Relation:

R = {(2, 4), (4, 1), (2, 1)}

This is neither reflexive nor symmetric.

∵ (1, 1) ∉ R

(2, 2) ∉ R

(4, 4) ∉ R

Hence, R is not reflexive.

∵ if (2, 4) ∈ R

Then, (4, 2) ∉ R

Hence, R is not symmetric.

Thus, the relation which is transitive but neither reflexive nor symmetric is:

R = {(2, 4), (4, 1), (2, 1)}

15. Question

Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, add a minimum number ordered pairs so that the enlarged relation is symmetric, transitive and reflexive.

Answer

Given is:

R = {(1, 2), (2, 3)} on the set A.

A = {1, 2, 3}

Right now, we have

R = {(1, 2), (2, 3)}

Symmetric Relation:

We know (1, 2) ∈ R

Then, (2, 1) ∈ R

Also, (2, 3) ∈ R

Then, (3, 2) ∈ R

So, add (2, 1) and (3, 2) in R, so that we get

R’ = {(1, 2), (2, 1), (2, 3), (3, 2)}

Transitive Relation:

We need to make the relation R’ transitive.

So, we know (1, 2) ∈ R and (2, 1) ∈ R

Then, (1, 1) ∈ R

Also, (2, 3) ∈ R and (3, 2)

Then, (2, 2) ∈ R

Also, (2, 1) ∈ R and (1, 2) ∈ R

Then, (2, 2) ∈ R

Also, (3, 2) ∈ R and (2, 3) ∈ R

Then, (3, 3) ∈ R

Add (1, 1), (2, 2) and (3, 3) in R’, we get

R’’ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}

Thus, we have got a relation which is reflexive, symmetric and transitive.

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}

The ordered pair added are (1, 1), (2, 2), (3, 3), (3, 2).

16. Question

Let A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)} be a relation on A. What minimum number of ordered pairs may be added to R so that it may become a transitive relation on A.

Answer

We have the relation R such that

R = {(1, 2), (1, 1), (2, 3)}

R is defined on set A.

A = {1, 2, 3}

Recall that,

A relation R defined on a set A is called transitive if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R, ∀ a, b, c ∈ A. For transitive relation:

Note in R,

(1, 2) ∈ R and (2, 3) ∈ R

Then, (1, 3) ∈ R

So, add (1, 3) in R.

R = {(1, 2), (1, 1), (2, 3), (1, 3)}

Now, we can see that R is transitive.

Hence, the ordered pair to be added is (1, 3).

17. Question

Let A = {a, b, c} and the relation R be defined on A as follows R={(a,a), (b, c), (a, b)}. Then, write a minimum number of ordered pairs to be added in R to make it reflexive and transitive.

Answer

Recall that,

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

We have relation R = {(a, a), (b, c), (a, b)} on A.

A = {a, b, c}

For Transitive:

If (a, b) ∈ R and (b, c) ∈ R

Then, (a, c) ∈ R

∀ a, b, c ∈ A

For Reflexive:

∀ a, b, c ∈ R

Then, (a, a) ∈ R

(b, b) ∈ R

(c, c) ∈ R

We need to add (b, b), (c, c) and (a, c) in R.

We get

R = {(a, a), (b, b), (c, c), (a, b), (b, c), (a, c)}

18 A. Question

Each of the following defines a relation on N :

x > y, x, y ∈ N

Determine which of the above relations are reflexive, symmetric and transitive. Answer

Recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

We have

x > y, x, y ∈ N

This relation is defined on N (set of Natural Numbers)

The relation can also be defined as

R = {(x, y): x > y} on N

Check for Reflexivity:

∀ x ∈ N

We should have, (x, x) ∈ R

⇒ x > x, which is not true.

1 can’t be greater than 1.

2 can’t be greater than 2.

16 can’t be greater than 16.

Similarly, x can’t be greater than x.

So, ∀ x ∈ N, then (x, x) ∉ R

⇒ R is not reflexive.

Check for Symmetry:

∀ x, y ∈ N

If (x, y) ∈ R

⇒ x > y

Now, replace x by y and y by x. We get

y > x, which may or not be true.

Let us take x = 5 and y = 2.

x > y

⇒ 5 > 2, which is true.

y > x

⇒ 2 > 5, which is not true.

⇒ y > x, is not true as x > y

⇒ (y, x) ∉ R

So, if (x, y) ∈ R, but (y, x) ∉ R ∀ x, y ∈ N

⇒ R is not symmetric.

Check for Transitivity:

∀ x, y, z ∈ N

If (x, y) ∈ R and (y, z) ∈ R

⇒ x > y and y > z

⇒ x > y > z

⇒ x > z

⇒ (x, z) ∈ R

So, if (x, y) ∈ R and (y, z) ∈ R, and then (x, z) ∈ R

∀ x, y, z ∈ N

⇒ R is transitive.

Hence, the relation is transitive but neither reflexive nor symmetric. 18 B. Question

Each of the following defines a relation on N :

x + y = 10, x, y ∈ N

Determine which of the above relations are reflexive, symmetric and transitive. Answer

Recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

We have

x + y = 10, x, y ∈ N

This relation is defined on N (set of Natural Numbers)

The relation can also be defined as

R = {(x, y): x + y = 10} on N

Check for Reflexivity:

∀ x ∈ N

We should have, (x, x) ∈ R

⇒ x + x = 10, which is not true everytime.

Take x = 4.

x + x = 10

⇒ 4 + 4 = 10

⇒ 8 = 10, which is not true.

That is 8 ≠ 10.

So, ∀ x ∈ N, then (x, x) ∉ R

⇒ R is not reflexive.

Check for Symmetry:

∀ x, y ∈ N

If (x, y) ∈ R

⇒ x + y = 10

Now, replace x by y and y by x. We get

y + x = 10, which is as same as x + y = 10.

⇒ y + x = 10

⇒ (y, x) ∈ R

So, if (x, y) ∈ R, and then (y, x) ∈ R ∀ x, y ∈ N

⇒ R is symmetric.

Check for Transitivity:

∀ x, y, z ∈ N

If (x, y) ∈ R and (y, z) ∈ R

⇒ x + y = 10 and y + z = 10

⇒ x + z = 10, may or may not be true.

Let us take x = 6, y = 4 and z = 6

x + y = 10

⇒ 6 + 4 = 10

⇒ 10 = 10, which is true.

y + z = 10

⇒ 4 + 6 = 10

⇒ 10 = 10, which is true.

x + z = 10

⇒ 6 + 6 = 10

⇒ 12 = 10, which is not true

That is, 12 ≠ 10

⇒ x + z ≠ 10

⇒ (x, z) ∉ R

So, if (x, y) ∈ R and (y, z) ∈ R, and then (x, z) ∉ R

∀ x, y, z ∈ N

⇒ R is not transitive.

Hence, the relation is symmetric but neither reflexive nor transitive. 18 C. Question

Each of the following defines a relation on N :

xy is square of an integer, x, y ∈ N

Determine which of the above relations are reflexive, symmetric and transitive. Answer

Recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

We have

xy is the square of an integer. x, y ∈ N.

This relation is defined on N (set of Natural Numbers)

The relation can also be defined as

R = {(x, y): xy = a², a = √(xy), a ∈ N} on N

Check for Reflexivity:

∀ x ∈ N

We should have, (x, x) ∈ R

⇒ xx = a², where a = √(xx)

⇒ x²= a², where a = √(x²)

which is true every time.

Take x = 1 and y = 4

xy = a²

⇒ 1 × 4 = (√(1 × 4))²[∵ a = √(xy)]

⇒ 4 = (√4)²

⇒ 4 = (2)²

⇒ 4 = 4

So, ∀ x ∈ N, then (x, x) ∈ R

⇒ R is reflexive.

Check for Symmetry:

∀ x, y ∈ N

If (x, y) ∈ R

⇒ xy = a², where a = √(xy)

Now, replace x by y and y by x. We get

yx = a², which is as same as xy = a²

where a = √(yx)

⇒ yx = a²

⇒ (y, x) ∈ R

So, if (x, y) ∈ R, and then (y, x) ∈ R ∀ x, y ∈ N

⇒ R is symmetric.

Check for Transitivity:

∀ x, y, z ∈ N

If (x, y) ∈ R and (y, z) ∈ R

⇒ xy = a²and yz = a²

⇒ xz = a², may or may not be true.

Let us take x = 8, y = 2 and z = 50

xy = a², where a = √(xy)

⇒ (8)(2) = (√(8 × 2))²

⇒ 16 = (4)²

⇒ 16 = 16, which is true.

yz = a²

⇒ (2)(50) = (√(2 × 50))²

⇒ 100 = (10)²

⇒ 100 = 100, which is true

xz = a²

⇒ (8)(50) = (√(8 × 50))²

⇒ 400 = (20)²

⇒ 400 = 400

We won’t be able to find a case to show a contradiction.

⇒ xz = a²

⇒ (x, z) ∈ R

So, if (x, y) ∈ R and (y, z) ∈ R, and then (x, z) ∈ R

∀ x, y, z ∈ N

⇒ R is transitive.

Hence, the relation is symmetric and transitivity, but not reflexive. 18 D. Question

Each of the following defines a relation on N :

x + 4y = 10, x, y ∈ N

Determine which of the above relations are reflexive, symmetric and transitive. Answer

Recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

We have

x + 4y = 10, x, y ∈ N

This relation is defined on N (set of Natural Numbers) The relation can also be defined as

R = {(x, y): 4x + y = 10} on N

Check for Reflexivity:

∀ x ∈ N

We should have, (x, x) ∈ R

⇒ 4x + x = 10, which is obviously not true everytime. Take x = 4.

4x + x = 10

⇒ 16 + 4 = 10

⇒ 20 = 10, which is not true.

That is 20 ≠ 10.

So, ∀ x ∈ N, then (x, x) ∉ R

⇒ R is not reflexive.

Check for Symmetry:

∀ x, y ∈ N

If (x, y) ∈ R

⇒ 4x + y = 10

Now, replace x by y and y by x. We get

4y + x = 10, which may or may not be true. Take x = 1 and y = 6

4x + y = 10

⇒ 4(1) + 6 = 10

⇒ 4 + 6 = 10

⇒ 10 = 10

4y + x = 10

⇒ 4(6) + 1 = 10

⇒ 24 + 1 = 10

⇒ 25 = 10, which is not true.

⇒ 4y + x ≠ 10

⇒ (y, x) ∉ R

So, if (x, y) ∈ R, and then (y, x) ∉ R ∀ x, y ∈ N ⇒ R is not symmetric.

Check for Transitivity:

∀ x, y, z ∈ N

If (x, y) ∈ R and (y, z) ∈ R

Then, (x, z) ∈ R

We have

4x + y = 10

⇒ y = 10 – 4x

Where x, y ∈ N

So, put x = 1

⇒ y = 10 – 4(1)

⇒ y = 10 – 4

⇒ y = 6

Put x = 2

⇒ y = 10 – 4(2)

⇒ y = 10 – 8

⇒ y = 2

We can’t take y >2, because if we put y = 3

⇒ y = 10 – 4(3)

⇒ y = 10 – 12

⇒ y = –2

But, y ≠ –2 as y ∈ N

So, only ordered pairs possible are

R = {(1, 6), (2, 2)}

This relation R can never be transitive.

Because if (a, b) ∈ R, then (b, c) ∉ R

⇒ R is not reflexive.

Hence, the relation is neither reflexive nor symmetric nor transitive.

Exercise 1.2

1. Question

Show that the relation R defined by R = {(a, b): a – b is divisible by 3; a, b ∈ Z} is an equivalence relation. Answer

We have,

R = {(a,b) : a–b is divisible by 3; a, b ∈ Z}

To prove : R is an equivalence relation

Proof :

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-

(a, a) ∈ R

Let a ∈ Z

⇒ a – a = 0

⇒ a – a is divisible by 3

(∵ 0 is divisible by 3).

⇒ (a, a) ∈ R

⇒ R is reflexive

Symmetric : For Symmetric, we need to prove that

If (a, b) ∈ R, then (b, a) ∈ R

Let a, b ∈ Z and (a, b) ∈ R

⇒ a – b is divisible by 3

⇒ a – b = 3p(say) For some p ∈ Z

⇒ –( a – b) = –3p

⇒ b – a = 3 × (–p)

⇒ b – a ∈ R

⇒ R is symmetric

Transitive : : For Transitivity, we need to prove that

If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R

Let a, b, c ∈ Z and such that (a, b) ∈ R and (b, c) ∈ R

⇒ a – b = 3p(say) and b – c = 3q(say) For some p, q ∈ Z

⇒ a – c = 3 (p + q)

⇒ (a, c) ∈ R

⇒ R is transitive

Since, R is reflexive, symmetric and transitive

⇒ R is an equivalence relation.

2. Question

Show that the relation R on the set Z of integers, given by

R = {(a, b) : 2 divides a – b}, is an equivalence relation.

Answer

We have,

R = {(a, b) : a – b is divisible by 2; a, b ∈ Z}

To prove : R is an equivalence relation

Proof :

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-

(a, a) ∈ R

Let a ∈ Z

⇒ a – a = 0

⇒ a – a is divisible by 2

⇒ (a, a) ∈ R

⇒ R is reflexive

Symmetric : For Symmetric, we need to prove that

If (a, b) ∈ R, then (b, a) ∈ R

Let a, b ∈ Z and (a, b) ∈ R

⇒ a – b is divisible by 2

⇒ a – b = 2p For some p ∈ Z

⇒ b – a = 2 × (–p)

⇒ b – a ∈ R

⇒ R is symmetric

Transitive : : For Transitivity, we need to prove that

If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R

Let a, b, c ∈ Z and such that (a, b) ∈ R and (b, c) ∈ R

⇒ a – b = 2p(say) and b – c = 2q(say) , For some p, q ∈ Z

⇒ a – c = 2 (p + q)

⇒ a – c is divisible by 2

⇒ (a, c) ∈ R

⇒ R is transitive

Now, since R is symmetric, reflexive as well as transitive-

⇒ R is an equivalence relation.

3. Question

Prove that the relation R on Z defined by

(a, b) ∈ R ⇔ a – b is divisible by 5

is an equivalence relation on Z.

Answer

We have,

R = {(a, b) : (a – b) is divisible by 5} on Z.

We want to prove that R is an equivalence relation on Z.

Proof :

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-

(a, a) ∈ R

Let a ∈ Z

⇒ a – a = 0

⇒ a – a is divisible by 5.

∴ (a, a) ∈ R so R is reflexive

Symmetric : For Symmetric, we need to prove that-

If (a, b) ∈ R, then (b, a) ∈ R

Let (a, b) ∈ R

⇒ a – b = 5p(say) For some p ∈ Z

⇒ b – a = 5 × (–p)

⇒ b – a is divisible by 5

⇒ (b, a) ∈ R, so R is symmetric

Transitive : : For Transitivity, we need to prove that

If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R

Let (a, b) ∈ R and (b, c) ∈ R

⇒ a – b = 5p(say) and b – c = 5q(say), For some p, q ∈ Z

⇒ a – c = 5 (p + q)

⇒ a – c is divisible by 5.

⇒ R is transitive

∴ R being reflexive, symmetric and transitive on Z.

⇒ R is equivalence relation on Z.

4. Question

Let n be a fixed positive integer. Define a relation R on Z as follows :

(a, b) ∈ R ⇔ a – b is divisible by n.

Show that R is an equivalence relation on Z.

Answer

R = {(a, b) : a – b is divisible by n} on Z.

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-

(a, a) ∈ R

Let a ∈ Z

⇒ a – a = 0 × n

⇒ a – a is divisible by n

⇒ (a, a) ∈ R

⇒ R is reflexive

Symmetric : For Symmetric, we need to prove that

If (a, b) ∈ R, then (b, a) ∈ R

Let (a, b) ∈ R

⇒ a – b = np For some p ∈ Z

⇒ b – a = n(–p)

⇒ b – a is divisible by n

⇒ (b, a) ∈ R

⇒ R is symmetric

Transitive : : For Transitivity, we need to prove that

If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R

Let (a, b) ∈ R and (b, c) ∈ R

⇒ a – b = np and b – c = nq For some p, q ∈ Z

⇒ a – c = n (p + q)

⇒ a – c = is divisible by n

⇒ (a, c) ∈ R

⇒ R is transitive

∴ R being reflexive, symmetric and transitive on Z.

⇒ R is an equivalence relation on Z

5. Question

Let Z be the set of integers. Show that the relation R = {(a, b) : a, b∈ Z and a + b is even} is an equivalence relation on Z.

Answer

We have,

Z = set of integers and

R = {(a, b) : a, b ∈ Z and a + b is even} be a relation on Z.

To prove: R is an equivalence relation on Z.

Proof :

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-

(a, a) ∈ R

Let a ∈ Z

⇒ a + a is even

⇒ (a, a) ∈ R

⇒ R is reflexive

Symmetric: For Symmetric, we need to prove that

If (a, b) ∈ R, then (b, a) ∈ R

Let a, b ∈ Z and (a, b) ∈ R

⇒ a + b is even

⇒ b + a is even

⇒ (b, a) ∈ R

⇒ R is symmetric

Transitive : : For Transitivity, we need to prove that

If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R

Let (a, b) ∈ R and (b, c) ∈ R For some a, b, c ∈ Z

⇒ a + b is even and b + c is even

[if b is odd, then a and c must be odd ⇒ a + c is even,

If b is even, then a and c must be even ⇒ a + c is even]

⇒ a + c is even

⇒ (a, c) ∈ R

⇒ R is transitive

Hence, R is an equivalence relation on Z

6. Question

m is said to be related to n if m and n are integers and m – n is divisible by 13. Does this define an equivalence relation?

Answer

To check that relation is equivalence, we need to check that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-

(a, a) ∈ R

Let m ∈ Z

⇒ m – m = 0

⇒ m – m is divisible by 13

⇒ (m, m) ∈ R

⇒ R is reflexive

Symmetric : For Symmetric, we need to prove that

If (a, b) ∈ R, then (b, a) ∈ R

Let m, n ∈ Z and (m, n) ∈ R

⇒ m – n = 13p For some p ∈ Z

⇒ n – m = 13 × (–p)

⇒ n – m is divisible by 13

⇒ (n – m) ∈ R,

⇒ R is symmetric

Transitive:: For Transitivity, we need to prove that

If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R

Let (m, n) ∈ R and (n, q) ∈ R For some m, n, q ∈ Z

⇒ m – n = 13p and n – q = 13s For some p, s ∈ Z

⇒ m – q = 13 (p + s)

⇒ m – q is divisible by 13

⇒ (m, q) ∈ R

⇒ R is transitive

Hence, R is an equivalence relation on Z.

7. Question

Let R be a relation on the set A of ordered pairs of non-zero integers defined by (x, y) R (u, v) iff xv = yu.

Show that R is an equivalence relation.

Answer

(x, y) R (u, v) ⇔ xv = yu

Proof :

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-

(a, a) ∈ R

∵ xy = yu

∴ (x, y) R (x, y)

Symmetric : For Symmetric, we need to prove that

If (a, b) ∈ R, then (b, a) ∈ R

Let (x, y) R (u, v)

TPT (u, v) R (x, y)

Given xv = yu

⇒ yu = xv

⇒ uy = vx

∴ (u, v) R (x, y)

Transitive : : For Transitivity, we need to prove that

If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R

Let (x, y) R (u, v) and (u, v) R (p, q) …(i)

TPT (x, y) R (p, q)

TPT (xq = yp

From (1) xv = yu & uq = vp

xvuq = yuvp

xq = yp

∴ R is transitive

Since R is reflexive, symmetric & transitive

⇒ R is an equivalence relation.

8. Question

Show that the relation R on the set A = {x ∈ Z ; 0 ≤ x ≤ 12}, given by R = {(a, b) : a = b}, is an equivalence relation. Find the set of all elements related to 1.

Answer

We have,

A = {x ∈ Z : 0 ≤ x ≤ 12} be a set and

R = {(a, b) : a = b} be a relation on A

Now,

Proof :

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive.

Reflexivity : For Reflexivity, we need to prove that-

(a, a) ∈ R

Let a ∈ A

⇒ a = a

⇒ (a, a) ∈ R

⇒ R is reflexive

Symmetric : For Symmetric, we need to prove that

If (a, b) ∈ R, then (b, a) ∈ R

Let a, b ∈ A and (a, b) ∈ R

⇒ a = b

⇒ b = a

⇒ (b, a) ∈ R

⇒ R is symmetric

Transitive : : For Transitivity, we need to prove that

If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R

Let a, b & c ∈ A

and Let (a, b) ∈ R and (b, c) ∈ R

⇒ a = b and b = c

⇒ a = c

⇒ (a, c) ∈ R

⇒ R is transitive

Since, R is being reflexive, symmetric and transitive, so R is an equivalence relation. Also, we need to find the set of all elements related to 1.

Since the relation is given by, R = {(a, b) : a = b}, and 1 is an element of A,

R = {(1, 1) : 1 = 1}

The number of all one-one functions from set A = {1, 2, 3} to itself is

(a) 2 (b) 6 (c) 3 (d) 1

Ans b

A relation R from A to B is an arbitrary subset of

(a) A × B (b) B × A (c) A × A (d) B × B

Ans A

A relation R in a set A is said to be an equivalence relation, if R is

(a) symmetric only (b) reflexive only

Ans d

A₁ coincides with the set of all integers in Z which are related to …A…, A₂ coincides with the set of all integers which are related to …B… and A₃ coincides with the set of all integers in Z which are related to …C… . Here, A, B and C are respectively

(a) one, zero, two (b) zero, one, two

Ans b

If (a, b) ÎR, where R is a relation, then

(a) (a, b) is not an ordered pair

(b) a = b only

(d) a is related to b under the relation R

Ans d

The relation R defined in the set A = {1, 2, 3, 4, 5, 6, 7} by R = {(a, b) : both a and b are either odd or even}. Then, R is

(a) symmetric (b) transitive

Ans c

The relation R in the set {1, 2, 3, …, 13, 14} defined as R = {(x, y) : 3x – y = 0} is

(a) reflexive (b) transitive

Ans d

Let R be the relation in the set Z of all integers defined by R = {(x, y) : x – y is an integer}. Then R is

(a) reflexive (b) symmetric

Ans d

If R is the relation defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1}, then R is

(a) reflexive (b) symmetric

Ans d

The relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is

(a) reflexive (b) symmetric

Ans b

If A = {x∊ Z : 0 ≤ x ≤ 12} and R is the relation in A given by R = {(a, b): |a – b| is a multiple of 4}. Then, the set of all elements related to 1 is

(a) {1, 4, 6} (b) {1, 5, 9}

Ans b

The function f : R → R given by f (x) = 2x is

(a) injective (b) surjective (c) many-one (d) bijective

And d

The function f : N → N given by

$\displaystyle f\left( x \right)=\left\{ {\begin{array}{*{20}{c}} {x+1,\text{ if x is odd}} \\ {x-1,\text{ if x is even}} \end{array}} \right.$ is

(a) injective (b) surjective

Ans c

Let f : R → R be defined as f (x) = x⁴. Choose the correct option.

(a) f is one-one onto

(b) f is many-one onto

(d) f is neither one-one nor onto

Ans d

Let A = {1, 2, 3} and B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Then, f is

(a) one-one (b) onto (c) many-one (d) bijective

Ans a

The greatest integer function f : R → R, given by f (x) = [x] is

(a) one-one (b) onto

Ans d

The relation R in the set of natural numbers N defined as R = {(x, y) : y = x + 5and x < 4} is

(a) not reflexive (b) symmetric

Ans a

Let A = {1, 2, 3}. Then, number of relations containing (1, 2) and (1, 3), which are reflexive and symmetric but not transitive, is

(a) 1 (b) 2 (c) 3 (d) 4

Ans A

Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer.

(a) R is reflexive and symmetric but not transitive

(b) R is reflexive and transitive but not symmetric

(d) R is an equivalence relation

Ans b

Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}. Choose the correct answer.

(a) (2, 4) ∊ R (b) (3, 8) ∊ R

Ans C

Let R be the relation on the set R of real numbers defined by R = {(a, b) / 1 + ab > 0}. Then, R is

(a) reflexive, symmetric but not transitive

(b) reflexive, transitive but not symmetric

(d) an equivalence relation

Ans a

If A = {x∊ Z : 0 ≤ x ≤ 12} and R is the relation in A given by R = {(a, b) : a = b}. Then, the set of all elements related to 1 is

(a) {1, 2} (b) {2, 3}

Ans c

Let A = R – {3} and B = R – {1}. Consider the function

f : A → B defined by $\displaystyle f\left( x \right)=\left( {\frac{{x-2}}{{x-3}}} \right)$ . Then, f is

(a) injective (b) bijective

And b

The function f : R → R defined by f (x) = x² + x is

(a) one-one (b) onto

Ans c

The domain of the function f : R → R defined by $\displaystyle f\left( x \right)=\sqrt{{{{x}^{2}}-3x+2}}$ is

(a) (- ∞ ,1] (b) [2, ∞)

Ans c

Let A = {1, 2, 3, 4} and B = {1, 2, 3, 4, 5, 6} are two sets and function f : A → B is defined by f (x) = x + 2″ x ÎA. Then, the function f is

(a) bijective (b) onto

Ans c

Let n(A) = 4 and n(B) = 6. Then, the number of one-one functions from A to B is

(a) 24 (b) 60 (c) 120 (d) 360

And d

If f ,g : R → R defined as $\displaystyle f\left( x \right)=\left\{ {\begin{array}{*{20}{c}} {x,\text{ if x is rational}} \\ {0,\text{ if x is is irrational}} \end{array}} \right.$ and $\displaystyle g\left( x \right)=\left\{ {\begin{array}{*{20}{c}} {0,\text{ if x is rational}} \\ {x,\text{ if x is is irrational}} \end{array}} \right.$ Then, f – g is

(a) one-one (b) neither one-one nor onto

Ans d

The function f : X → Y defined by f (x) = sin x is one-one but not onto, if X and Y respectively equal to

(a) R and R (b) [0, π] and [0, 1]

(c) $\displaystyle \left[ {0,\frac{\pi }{2}} \right]$ and [-1 ,1] (d) $\displaystyle \left[ {-\frac{\pi }{2},\frac{\pi }{2}} \right]$ , and [-1 , 1]

Ans c

Let A = {1, 2, 3,L, n} and B = {a, b}. Then, the number of surjections from A into B isA into B is

(a) ⁿP₂ (b) 2ⁿ – 2

Ans b

Which of the following functions from Z into Z are bijective?

(a) f (x) = x³ (b) f (x) = x + 2

Ans b

A function f : X → Y is said to be invertible, if there exists a function g : Y → X such that

(a) gof = I_X and fog = I_X (b) gof = I_Y and fog = I_Y

Ans d

If Y = {n₂ : n ∊ N} and f : N → Y is a function defined by f (n) = n², then the inverse of f is

(a) $\displaystyle {{f}^{{-1}}}\left( y \right)=\sqrt{y}$ (b) $\displaystyle {{f}^{{-1}}}\left( y \right)=-\sqrt{y}$

Ans a

Let $\displaystyle f:R-\left\{ {-\frac{4}{3}} \right\}$ → R be a function defined as $\displaystyle f\left( x \right)=\frac{{4x}}{{3x+4}}$ The inverse of f is the map g : range $\displaystyle f\to R-\left\{ {-\frac{4}{3}} \right\}$ given by

(a) $\displaystyle g\left( y \right)=\frac{{3y}}{{3-4y}}$

(b) $\displaystyle g\left( y \right)=\frac{{4y}}{{4-3y}}$

(d) $\displaystyle g\left( y \right)=\frac{{3y}}{{4-3y}}$

Ans b

If f (x) = 8x³ and g(x) = x^1/3, then

(a) fog(x) = 2x (b) fog = 8x

Ans b

The function f : R → R defined by f (x) = 2x + sin x, x ∊ R is

(a) one-one (b) onto

Ans b

If f : A → B and g : B → C be two functions, then the composition of functions f and g is given by

(a) gof (x) = g(f (x))

(b) gof (x) = g(x)f (x)

(d) None of these

Ans a

If f : B → A is defined by $\displaystyle f\left( x \right)=\frac{{3x+4}}{{5x-7}}$ and g : A → B is defined by $\displaystyle g\left( x \right)=\frac{{7x+4}}{{5x-3}}$ , where $\displaystyle A=R-\left\{ {\frac{3}{5}} \right\}$ and $\displaystyle B=R-\left\{ {\frac{7}{5}} \right\}$ and I_A is an identity function on A and I_B is identity function on B, then

(a) fog = I_A and gof = I_A

(b) fog = I_A and gof = I_B

(d) fog = I_B and gof = I_A

Ans b

If the function gof is defined and is one-one, then

(a) neither f nor g is one-one

(b) f and g both are necessarily one-one

(d) None of the above

Ans d

Let X = {1, 2, 3} and Y = {a, b, c } and f : X ® Y be one-one and onto function given by f (1) = a, f (2) = b, f (3) = c. Then, the function g : Y → X such that fog = I_Y and gof = I_X is given by

(a) g(a) = 2, g(b) = 1, g(c) = 3

(b) g(a) = 1, g(b) = 2, g(c) = 3

(d) g(a) = 3, g(b) = 2, g(c) = 1

Ans b

Thus, the set of all element related to 1 is 1.

9. Question

Let L be the set of all lines in XY-plane and R be the relation in L defined as R = {(L₁, L₂): L₁is parallel to L₂}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.

Answer

We have, L is the set of lines.

R = {(L₁, L₂) : L₁is parallel to L₂} be a relation on L

Now,

Proof :

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-

(a, a) ∈ R

Since a line is always parallel to itself.

∴ (L₁, L₂) ∈ R

⇒ R is reflexive

Symmetric : For Symmetric, we need to prove that

If (a, b) ∈ R, then (b, a) ∈ R

Let L₁, L₂∈ L and (L₁, L₂) ∈ R

⇒ L₁is parallel to L₂

⇒ L₂is parallel to L₁

⇒ (L₁, L₂) ∈ R

⇒ R is symmetric

Transitive: For Transitivity, we need to prove that

If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R

Let L₁, L₂and L₃∈ L such that (L₁, L₂) ∈ R and (L₂, L₃) ∈ R

⇒ L₁is parallel to L₂and L₂is parallel to L₃

⇒ L₁is parallel to L₃

⇒ (L₁, L₃) ∈ R

⇒ R is transitive

Since, R is reflexive, symmetric and transitive, so R is an equivalence relation. And, the set of lines parallel to the line y = 2x + 4 is y = 2x + c For all c∈ R

where R is the set of real numbers.

10. Question

Show that the relation R, defined on the set A of all polygons as

R = {(P₁, P₂) : P₁and P₂have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?

Answer

R = {(P₁, P₂): P₁and P₂have same the number of sides}

Proof :

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity: For Reflexivity, we need to prove that-

(a, a) ∈ R

R is reflexive since (P₁, P₁) ∈ R as the same polygon has the same number of sides with itself. Symmetric: For Symmetric, we need to prove that

If (a, b) ∈ R, then (b, a) ∈ R

Let (P₁, P₂) ∈ R.

⇒ P₁and P₂have the same number of sides.

⇒ P₂and P₁have the same number of sides.

⇒ (P₂, P₁) ∈ R

∴ R is symmetric.

Transitive: For Transitivity, we need to prove that

If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R

Now, (P1, P2), (P2, P3) ∈ R

⇒ P₁and P₂have the same number of sides. Also, P₂and P₃have the same number of sides. ⇒P₁and P₃have the same number of sides.

⇒ (P₁, P₃) ∈ R

∴ R is transitive.

Hence, R is an equivalence relation.

And, now the elements in A related to the right-angled triangle (T) with sides 3, 4 and 5 are those polygons which have three sides (since T is a polygon with three sides).

Hence, the set of all elements in A related to triangle T is the set of all triangles.

11. Question

Let O be the origin. We define a relation between two points P and Q in a plane if OP = OQ. Show that the relation, so defined is an equivalence relation.

Answer

Let A be set of points on the plane.

Let R = {(P, Q) : OP = OQ} be a relation on A where O is the origin.

To prove R is an equivalence relation, we need to show that R is reflexive, symmetric and transitive on A. Proof :

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-

(a, a) ∈ R

Let p ∈ A

Since OP = OP ⇒ (P, P) ∈ R

⇒ R is reflexive

Symmetric : For Symmetric, we need to prove that

If (a, b) ∈ R, then (b, a) ∈ R

Let (P, Q) ∈ R for P, Q ∈ R

Then OP = OQ

⇒ Op = OP

⇒ (Q, P) ∈ R

⇒ R is symmetric

Transitive: For Transitivity, we need to prove that

If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R

Let (P, Q) ∈ R and (Q, S) ∈ R

⇒ OP = OQ and OQ = OS

⇒ OP = OS

⇒ (P, S) ∈ R

⇒ R is transitive

Thus, R is an equivalence relation on A

12. Question

Let R be the relation defined on the set A = {1, 2, 3, 4, 5, 6, 7} by R = {(a, b) : both a and b are either odd or even}. Show that R is an equivalence relation. Further, show that all the elements of the subset {1, 3, 5, 7} are related to each other, and all the elements of the subset {2, 4, 6} are related to each other, but no element of the subset {1, 3, 5, 7} is related to any element of the subset {2, 4, 6}.

Answer

Given A = {1, 2, 3, 4, 5, 6, 7} and R = {(a, b) : both a and b are either odd or even number} Therefore,

R = {(1, 1), (1, 3), (1, 5), (1, 7), (3, 3), (3, 5), (3, 7), (5, 5), (5, 7), (7, 7), (7, 5), (7, 3), (5, 3), (6, 1), (5, 1), (3, 1), (2, 2), (2, 4), (2, 6), (4, 4), (4, 6), (6, 6), (6, 4), (6, 2), (4, 2)}

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-

(a, a) ∈ R

Here (1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (7,7) all ∈ R

From the relation R it is seen that R is reflexive.

Symmetric: For Symmetric, we need to prove that

If (a, b) ∈ R, then (b, a) ∈ R

From the relation R, it is seen that R is symmetric.

Transitive: For Transitivity, we need to prove that

If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R

[I (a, b) are odd and (b, c) are odd then (a, c) are also odd numbers]

From the relation R, it is seen that R is transitive too.

Also, from the relation R, it is seen that {1, 3, 5, 7} are related with each other only and {2, 4, 6} are related with each other .

13. Question

Let S be a relation on the set R of all real numbers defined by S = {(a, b)∈ R × R : a²+ b²= 1}. Prove that S is not an equivalence relation on R.

Answer

S = {(a, b) : a²+ b²=1}

Proof :

To prove that relation is not equivalence, we need to prove that it is either not reflexive, or not symmetric or not transitive.

Reflexivity : For Reflexivity, we need to prove that-

(a, a) ∈ R

Let a = 1/2, a ∈ R

Then,

⇒ (a, a) ∉ S

⇒ S is not reflexive

Hence, S is not an equivalence relation on R.

14. Question

Let Z be the set of all integers and Z₀be the set of all non-zero integers. Let a relation R on Z × Z₀be defined as follows :

(a, b) R (c, d) ⇔ ad = bc for all (a, b), (c, d) ∈ Z × Z₀

Prove that R is an equivalence relation on Z × Z₀

Answer

We have, Z be set of integers and Z₀be the set of non-zero integers.

R = {(a, b) (c, d) : ad = bc} be a relation on Z and Z₀.

Proof :

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-

(a, a) ∈ R

(a, b) ∈ Z × Z₀

⇒ ab = ba

⇒ ((a, b), (a, b)) ∈ R

⇒ R is reflexive

Symmetric : For Symmetric, we need to prove that

If (a, b) ∈ R, then (b, a) ∈ R

Let ((a, b), (c, d) ∈ R

⇒ ad = bc

⇒ cd = da

⇒ ((c, d), (a, b)) ∈ R

⇒ R is symmetric

Transitive : : For Transitivity, we need to prove that-

If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R

Let (a, b), (c, d) ∈ R and (c, d), (e, f) ∈ R

⇒ ad = bc and cf = de

⇒

⇒ af = be

⇒ (a, c) (e, f) ∈ R

⇒ R is transitive

Hence, R is an equivalence relation on Z × Z₀ 15. Question

If R and S are relations on a set A, then prove the following : (i) R and S are symmetric ⇔ R ⋂ S, and R ⋃ S is symmetric (ii) R is reflexive, and S is any relation ⇔ R ⋃ S is reflexive. Answer

R and S are two symmetric relations on set A (i) To prove: R ⋂ S is symmetric

Symmetric: For Symmetric, we need to prove that If (a, b) ∈ R, then (b, a) ∈ R

Let (a, b) ∈ R ⋂ S

⇒ (a, b) ∈ R and (a, b) ∈ S

⇒ (b, a) ∈ R and (b, a) ∈ S

[∴ R and S are symmetric]

⇒ (b, a) ∈ R ⋂ S

⇒ R ⋂ S is symmetric

To prove: R ⋃ S is symmetric

Symmetric: For Symmetric, we need to prove that If (a, b) ∈ R, then (b, a) ∈ R

Let (a, b) ∈ R ⋃ S

⇒ (a, b) ∈ R or (a, b) ∈ S

⇒ (b, a) ∈ R or (b, a) ∈ S

[∴ R and S are symmetric]

⇒ (b, a) ∈ R ⋃ S

⇒ R ⋃ S is symmetric

(ii) R and S are two relations on a such that R is reflexive. To prove : R ⋃ S is reflexive

Reflexivity : For Reflexivity, we need to prove that- (a, a) ∈ R

Suppose R ⋃ S is not reflexive.

This means that there is a ∈ R ⋃ S such that (a, a) ∉ R ⋃ S

Since a ∈ R ⋃ S,

∴ a ∈ R or a ∈ S

If a ∈ R, then (a, a) ∈ R

[∵ R is reflexive]

⇒ (a, a) ∈ R ⋃ S

Hence, R ⋃ S is reflexive

16. Question

If R and S are transitive relations on a set A, then prove that R ⋃ S may not be a transitive relation on A. Answer

We will prove this using an example.

Let A = {a, b, c} be a set and

R = {(a, a) (b, b) (c, c) (a, b) (b, a)} and

S = {(a, a) (b, b) (c, c) (b, c) (c, d)} are two relations on A

Clearly R and S are transitive relation on A

Now,

R ⋃ S = {(a, a) (b, b) (c, c) (a, b) (b, a) (b, c) (c, b)}

Here, (a, b) ∈ R ⋃ S and (b, c) ∈ R ⋃ S

but (a, c) ∉ R ⋃ S

∴ R ⋃ S is not transitive

17. Question

Let C be the set of all complex numbers and C₀be the set of all non-zero complex numbers. Let a relation R on C₀be defined as

z₁R z₂is real for all z₁, z₂∈ C₀.

Show that R is an equivalence relation.

Answer

We have,

We want to prove that R is an equivalence relation on Z.

Now,

Proof :

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that-

(a, a) ∈ R

Let a ∈ C₀

And, 0 is real

∴ (a, a) ∈ R, so R is reflexive

Symmetric: For Symmetric, we need to prove that If (a, b) ∈ R, then (b, a) ∈ R

Let (a, b) ∈ R

⇒ p is real.

And ∵ p is real

⇒ -p is also a real no.

⇒ (b, a) ∈ R, so R is symmetric

Transitive : : For Transitivity, we need to prove that If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R Let (a, b) ∈ R and (b, c) ∈ R

⇒ p is real no.

….(1)

⇒ q is real.

…..(2)

Dividing (1) by (2), we get

Where, Q is a rational number.

⇒ Q is real number

Now, by componendo dividendo-

⇒ (a, c) ∈ R.

⇒ R is transitive

Thus, R is reflexive, symmetric and, transitive on C₀.

Hence, R is an equivalence relation on C₀.

Very short answer

1. Question

Write the domain of the relation R defined on the set Z of integers as follows: (a, b) ϵR ⬄ a²+ b²= 25

Answer

Given a and b are integers, i.e. a,b ∈ Z.

∴ Domain of R = Set of all first elements in the relation.

= Values of ‘a’ which are in the relation.

=Z (Integers)

Range of R=Set of all second elements in the relation.

=Values of ‘b’ which are in the relation.

=Z(Integers)

Since, a²+ b²= 25 and a, b are integers;

⇒ R= {(5,0), (0,5), (-5,0), (0, -5), (3,4), (4,3), (-3, -4), (-4, -3), (-3,4), (4, -3), (-4,3), (3, -4)}

⇒ Domain of R= {-5, -4, -3,0,3,4,5}

2. Question

If R = {(x,y): x²+ y²≤ 4; x,y ϵ Z} is a relation of Z, write the domain of R. Answer

Given x and y are integers, i.e x,y ∈ Z.

∴ Domain of R = Set of all first elements in the relation.

= Values of ‘x’ which are in the relation.

=Z (Integers)

Range of R=Set of all second elements in the relation.

=Values of ‘y’ which are in the relation.

=Z(Integers)

Since, x²+ y²≤ 4 and x,y are integers;

⇒ R= {(0,0), (1,0), (0,1), (-1,0), (0,-1), (1,1), (-1,-1), (-1,1),

(1,-1), (0,2), (0,-2), (2,0), (-2,0)}

⇒ Domain of R= {-2,-1,0,1,2}

3. Question

Write the identity relation of set A = {a, b, c}.

Answer

⇒ Identity relation of a set refers to the relation in which every element on the set is related to itself. Thus the Identity relation of set A is as under:

⇒ R={(a,a),(b,b),(c,c)}

4. Question

Write the smallest reflexive relation of set A = {1, 2, 3, 4}.

Answer

The smallest reflexive relation of set A = {1, 2, 3, 4} is as under:

As Relation R on a set A is said to be a reflexive relation on A if:

⇒ (a,a) ∈ R ∀ a ∈ A

⇒ R={(1,1),(2,2),(3,3),(4,4)}

5. Question

If R = {(x,y) : x + 2y = 8} is a relation on N, then write the range of R.

Answer

Given x and y are natural numbers, i.e. x,y ∈ N.

∴ Range of R=Set of all second elements in the relation.

=Values of ‘y’ which are in relation.

=N (Natural Numbers)

Since, x+ 2y = 8 and x,y are Natural numbers;

⇒ R= {(2,3), (4,2), (6,1)}

⇒ Range of R= {1,2,3}

NOTE: 0 is a whole number that’s why it is not considered in this set.

6. Question

If R is a symmetric relation on a set A, then write a relation between R and R^–1. Answer

The relation between R and R^–1on a Set A is as under:

⇒ R^-1= {(b,a):(a,b)∈ R}

⇒ Clearly (a,b) ∈ R

⇒ (b,a) ∈ R^-1

i.e Domain(R)=Range(R^-1)

and Domain(R^-1)=Range(R)

7. Question

Let R = {(x,y): |x²– y²| < 1} be a relation on set A = {1, 2, 3, 4, 5}. Write R as a set of ordered pairs. Answer

Given a relation R = {(x,y): |x²– y²| < 1} on set A = {1, 2, 3, 4, 5}.

Now according to the condition |x²– y²| < 1 .

⇒ x²– y²= 0

⇒ Describing R as set of ordered pairs:

⇒ R={(1,1),(2,2),(3,3),(4,4),(5,5)}

8. Question

If A = {2, 3, 4}, B = {1, 3, 7} and R = {(x,y) : xϵ A, y ϵB and x < y} is a relation from A to B, then write R^–1. Answer

Given A = {2, 3, 4}, B = {1, 3, 7} and R = {(x,y) : xϵ A, y ϵB and x < y}

According to the condition x < y

⇒ R= A × B

⇒ R={(2,3), (2,7), (3,7), (4,7)}

⇒ R^-1= B × A

⇒ R^-1={ (3,2), (7,2), (7,3), (7,4)}

9. Question

Let A = {3, 5, 7}, B = {2, 6, 10} and R be a relation from A to B defined by R = {(x, y) : x and y are relatively prime}. Then, write R and R^–1.

Answer

Given A = {3, 5, 7}, B = {2, 6, 10} and R = {(x, y) : x and y are relatively prime } According to the condition, x and y are prime numbers.

⇒ R= A × B

⇒ R= {(3,2), (5,2), (7,2)}

⇒ R^-1= B × A

⇒ R^-1= { (2,3), (2,5), (2,7)}

10. Question

Define a reflexive relation.

Answer

Relation R on a set A is said to be a reflexive relation on A if:

⇒ (a,a) ∈ R ∀ a ∈ A

Eg A={1,2,3}

⇒ R={(1,1),(2,2),(3,3)}

11. Question

Define a symmetric relation.

Answer

Relation R on a set A is said to be a symmetric relation on A if:

(a,b) ∈ R

⇒ (b,a) ∈ R ∀ a,b∈ A

eg A={1,2,3}

⇒ R={(1,2),(2,1),(1,3),(3,1),(2,3),(3,2)}

12. Question

Define a transitive relation.

Answer

Relation R on a set A is said to be a transitive relation on A iff:

(a,b) ∈ R and (b,c) ∈ R

⇒ (a,c) ∈ R ∀

a,b,c ∈ A

A={1,2,3}

⇒ R={(1,2),(2,3),(1,3)}

13. Question

Define an equivalence relation.

Answer

A relation R on a set A is said to be an equivalence relation on A iff:

1. Its reflexive i.e (a,a) ∈ R ∀ a ∈ A

2. Its symmetric i.e (a,b) ∈ R → (b,a) ∈ R ∀ a,b ∈ A

3. Its transitive i.e (a,b) ∈ R and (b,c) ∈ R → (a,c) ∈ R ∀

a,b,c ∈ A

14. Question

If A = {3, 5, 7} and B = {2, 4, 9} and R is a relation given by “is less than”, write R as a set ordered pairs. Answer

Given A = {3, 5, 7} and B = {2, 4, 9}

Now according to the condition:

R is a relation given by “is less than.”

⇒ R= A × B

⇒ R={(3,4),(3,9),(5,9),(7,9)}

15. Question

A = {1, 2, 3, 4, 5, 6, 7, 8} and if R = {(x,y) : y is one half of x; x, yϵA} is a relation on A, then write R as a set of ordered pairs.

Answer

Given A = {1, 2, 3, 4, 5, 6, 7, 8} and a relation R = {(x,y) : y is one half of x; x, yϵA} Now according to the condition

⇒ R={(2,1),(4,2),(6,3),(8,4)}

16. Question

Let A = {2, 3, 4, 5} and B = {1, 3, 4}. If R is the relation from A to B given by a R b iff “a is a divisor of b.” Write R as a set of ordered pairs.

Answer

Given A = {2, 3, 4, 5} and B = {1, 3, 4} and R is a relation from A to B which is true only iff “a is a divisor of b” i.e b is divisible by a.

⇒ R =A × B

⇒ R={(2,4),(4,4),(3,3)}

17. Question

State the reason for the relation R on the set {1, 2, 3} given by R = {(1, 2), (2, 1)} not to be transitive. Answer

Given set {1,2,3} and relation R = {(1, 2), (2, 1)}.

For relation R to be transitive:

R={(1,2),(2,3),(1,3)}

But in the given relation:

⇒ (2,3) and (1,3) ∉ R

Hence, Given Relation R is not transitive.

18. Question

Let R = {(a, a³) : a is a prime number less than 5} be a relation. Find the range of R. Answer

Given a relation R = {(a, a³) : a is a prime number less than 5}

Now according to the question ‘a’ is a prime number < 5:

⇒ a = {2,3}

⇒ R={(2,8),(3,27)}

⇒ Range of R={8,27)

19. Question

Show that R is an equivalence relation on the set Z of integers given by R = {(a, b) : 2 divides a – b}.Write the equivalence class [0].

Answer

Given R = {(a, b) : 2 divides a – b}

For equivance relation we have to check three parameters:

(i) Reflexive:

If (a-b) is divisible by 2 then,

⇒ (a-a)=0 is also divisible by 2

⇒ (a,a) ∈ R

Hence R is Reflexive ∀ (a,b) ∈ Z

(ii)Symmetric:

If (a-b) is divisible by 2 then,

⇒ (b-a)=-(a-b) is also divisible by 2

⇒ (a,b) ∈ R and (b,a) ∈ R

Hence R is Symmetric ∀ (a,b) ∈ Z

(iii)Transitive:

If (a-b) and (b-c) are divisible by 2 then,

⇒ a-c=(a-b)+(b-c) is also divisible by 2

⇒ (a,b) ∈ R, (b,c) ∈ R and (a,c) ∈ R

Hence R is Transitive ∀ (a,b) ∈ Z

⇒ As Relation R is satisfying all the three parameters, hence R is an equivalence relation. Now equivalence class [0] is the set of all those elements in A which are related to 0 under relation. Now,

(a,0) ∈ R

⇒ a – 0 is divisible by 2 and a ∈ A.

⇒ a ∈ A such that 2 divides a.

⇒ a = 0, 2,4

Thus [0] = {0,2,4}

20. Question

For the set A = {1, 2, 3}, define a relation R on the set A as follows:

R = {(1, 1), (2, 2), (3, 3), (1, 3)}

Write the ordered pairs to be added to R to make the smallest equivalence relation. Answer

Given set A = {1, 2, 3} and relation R = {(1, 1), (2, 2), (3, 3), (1, 3)}

A relation is an equivalence relation if and only if it is reflexive, symmetric and transitive. ⇒ R = {(1, 1), (2, 2), (3, 3), (1, 3)} is reflexive and transitive but not symmetric. ⇒ If (3,1) is added to the ordered pairs, R will become symmetric.

Thus the new R:

⇒ R = {(1, 1), (2, 2), (3, 3), (1, 3),(3,1)} is the obtained smallest equivalence relation. 21. Question

Let A = {0, 1, 2, 3} and R be a relation on A defined as

F = {(0, 0) (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}

Is R reflexive? Symmetric” transitive?

Answer

Given A = {0, 1, 2, 3} and a relation R on A defined as

F = {(0, 0) (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}

A relation is an equivalence relation if and only if it is reflexive, symmetric and transitive: (i) The set contains (0,0),(1,1),(2,2),(3,3):

Hence (a,a) ∈ R → R is reflexive.

(ii)The set also contains (0,1),(1,0) and (0,3),(3,0)

Hence (a,b) ∈ R and (b,a) ∈ R → R is symmetric.

(ii) The set also contains (0,0),(0,1) and (0,0),(0,3)

Hence (a,b) ∈ R, (b,c) ∈ R and (a,c) ∈ R → R is transitive.

Hence R is reflexive,symmetric and transitive.

22. Question

Let the relation R be defined on the set A = {1, 2, 3, 4, 5} by R = {(a, b) : |a²– b²| < 8}. Write R as a set of ordered pairs.

Answer

Given a set A = {1, 2, 3, 4, 5} and relation R = {(a, b) : |a²– b²| < 8}

Now according to the question |a²– b²| < 8

⇒ R = {(1,1),(2,2), (3,3), (4,4), (5,5), (1,2), (2,1), (2,3), (3,2), (3,4), (4,3),}

23. Question

Let the relation R be defined on N by a Rb iff 2a + 3b = 30. Then write R as a set of ordered pairs. Answer

Given R = {(a, b) : 2a + 3b = 30}∀ (a,b) ∈ N

Now according to the question 2a + 3b = 30:

⇒ R={(3,8),(6,6),(9,4),(12,2)}

NOTE: 0 is a whole number that’s why its not considered in this set, Although if we consider 0 as a natural number then the answer would be:

⇒ R={(0,10),(3,8),(6,6),(9,4),(12,2),(15,0)}

24. Question

Write the smallest equivalence relation on the set A = {1, 2, 3}.

Answer

A relation is an equivalence relation if and only if it is reflexive, symmetric and transitive: The smallest equivalence relation on the set A={1,2,3} is :

R={(1,1),(1,3),(3,1)}

∵ (1,1) ∈ R → Reflexive

(1,3) ∈ R and (3,1) ∈ R → Symmetric

(1,3) ∈ R and (3,1) ∈ R and (1,1) ∈ R → Transitive

MCQ

1. Question

Let R be a relation on the set N given by R = {a, b) : a = b – 2, b > 6}. Then

A. (2, 4) ϵR

B. (3, 8) ϵR

C. (6, 8) ϵR

D. (8, 7) ϵR

Answer

Given R = {(a, b) : a = b – 2, b > 6}

Now according to the question a = b – 2 and b > 6

⇒ R={(5,7),(6,8),(7,9)…….∞ }

2. Question

Which of the following is not an equivalence relation on Z?

A. a R b ⬄ a + b is an even integer

B. a R b ⬄ a – b is a even integer

C. a R b ⬄ a < b

D. a R b ⬄ a = b

Answer

a R b ⬄ a – b is a even integer

Given R = {(a, b) : a – b is an even integer(i.e divisible by 2)}

For equivance relation we have to check three parameters:

(iii) Reflexive:

If (a-b) is divisible by 2 then,

⇒ (a-a)=0 is also divisible by 2

⇒ (a,a) ∈ R

Hence R is Reflexive ∀ (a,b) ∈ Z

(ii)Symmetric:

If (a-b) is divisible by 2 then,

⇒ (b-a)=-(a-b) is also divisible by 2

⇒ (a,b) ∈ R and (b,a) ∈ R

Hence R is Symmetric ∀ (a,b) ∈ Z

(iii)Transitive:

If (a-b) and (b-c) are divisible by 2 then,

⇒ a-c=(a-b)+(b-c) is also divisible by 2

⇒ (a,b) ∈ R, (b,c) ∈ R and (a,c) ∈ R

Hence R is Transitive ∀ (a,b) ∈ Z

→ As Relation R is satisfying all the three parameters, hence R is an equivalence relation. 3. Question

R is a relation on the set Z of integers and it is given by (x, y) ϵ R ⬄ |x – y| ≤ 1. Then, R is A. reflective and transitive

B. reflexive and symmetric

C. symmetric and transitive

D. an equivalence relation

Answer

∵ According to the condition |x – y| ≤ 1

⇒ R={(1,1),(2,1), (1,2), (2,2), … (n,n), (n+1,n), (n,n+1) …∞ }

⇒ (a,a) ∈ R → Reflexive

⇒ (a,b) ∈ R and (b,a) ∈ R →Symmetric

4. Question

The relation R defined on the set A = {1, 2, 3, 4, 5} by R = is given by A. {(1, 1), (2, 1), (3, 1), (4, 1), (2, 3)}

B. {(2, 2), (3, 2), (4, 2), (2, 4)}

C. {(3, 3), (4, 3), (5, 4), (3, 4)

D. none of these

Answer

Given set A = {1, 2, 3, 4, 5} and relation R ={(a,b):|a²-b²|<16}

According to the condition |a²-b²|<16:

⇒ R={(1, 1), (2, 1), (3, 1), (4, 1), (2, 3),(2, 2), (3, 2), (4, 2), (2, 4),(3, 3), (4, 3), (5, 4), (3, 4)} 5. Question

Let R be the relation over the set of all straight lines in a plane such that ℓ₁R ℓ₂⬄ ℓ₁⊥ ℓ₂. Then, R is A. symmetric

B. reflexive

C. transitive

D. an equivalence relation

Answer

Think of line as a vector quantity:

As ℓ₁⊥ ℓ₂;

And ℓ₂⊥ ℓ₁

Hence R is symmetric.

Also Given a relation R over straight lines such that ℓ₁⊥ ℓ₂

As ℓ₁⊥ ℓ_2:

⇒ ℓ₁. ℓ₂=0(DOT PRODUCT)

∵ cos θ =0 as θ =90°;

⇒ This thing is possible if ℓ₁and ℓ₂are symmetric.

E.g. A= 2i-4j and B=-4i-2j

⇒ A. B=-8+8=0

6. Question

If A = {a, b, c}, then the relation R = {(b, c)} on A is

A. reflexive only

B. symmetric only

C. transitive only

D. reflexive and transitive only

Answer

According to the question:

R={(a,a), (b,b), (c,c), (a,b), (b,c), (a,c)}

Thus R = {(b, c)} can only be transitive.

i.e (a,b)∈ R and (b,c)∈ R → (a,c)∈ R

7. Question

Let A = {2, 3, 4, 5, …, 17, 18}. Let ‘≃’ be the equivalence relation on A × A, cartesian product of A with itself, defined by (a, b) ≃ (c, d) iff ad = bc. Then, the number of ordered pairs of the equivalence class of (3, 2) is

A. 4

B. 5

C. 6

D. 7

Answer

Let (3,2) ≃ (x,y)

⇒ 3y = 2x

This is possible in the cases:

x = 3, y = 2

x = 6, y = 4

x= 9, y =6

x=12, y = 3

x=15, y = 10

x=18,y=12

Hence total pairs are 6.

8. Question

Let A = {1, 2, 3}. Then, the number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is

A. 1

B. 2

C. 3

D. 4

Answer

R={(1,2),(2,1),(1,3),(3,1),(1,1)} is the only relation.

The addition of (2,2) or (3,3) will make R transitive.

9. Question

The relation ‘R’ in N × N such that (a, b) R (c, d)⬄ a + d = b + c is

A. reflexive but of symmetric

B. reflexive and transitive but not symmetric

C. an equivalence relation

D. none of these

Answer

Check:

(a,b)R (a,b) ∈ R

a + b = b + a

hence R is reflexive.

Now,

(a, b) R (c, d) ∈ R

a + d = b + c

⇒ c + b = d + a

⇒ (c,d) R (a,b) ∈ R

R is symmetric

Now,

(a, b) R (c, d) ∈ R

a + d = b + c

(c,d) R (e,f) ∈ R

⇒ c + f = d + e

Now,

a + d + c + f = b + c + d + e

⇒ a + f = b + e

So (a,b) R (e,f)

R is transitive.

Hence R is an equivalence relation.

10. Question

If A = {1, 2, 3}, B = {1, 4, 5, 9} and R is a relation from A to B defined by ‘x is greater than y’. Then, Range of R is

A. {1, 2, 6, 9}

B. {4, 6, 9}

C. {1}

D. none of these

Answer

Here

⇒ R={(2,1), (3,1)}

Hence Range(R)={1}

11. Question

A relation R is defined form {2, 3, 4, 5} to {3, 6, 7, 10} by xRy ⬄ x is relatively prime to y. Then, domain of R is

A. {2, 3 5}

B. {3, 5}

C. {2, 3, 4} D. {2, 3, 4, 5}

Answer

R: x R y⟺ x is relatively prime to y.

Two numbers are relatively prime if their Highest Common Factor is 1.

Then, R= {(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)}

Therefore, the domain of R is {2, 3, 4, 5}

12. Question

A relation ϕ from C to R is defined by x ϕ ⬄ |x| = y. Which one is correct? A. (2 + 3i) ϕ 13

B. 3 ϕ (–3)

C. (1 + i) ϕ 2

D. i ϕ 1

Answer

In complex numbers the y-axis is taken as the imaginary axis,

Thus i ϕ 1

13. Question

Let R be a relation on N defined by x + 2y = 8. The domain of R is

A. {2, 4, 8}

B. {2, 4, 6, 8}

C. {2, 4, 6}

D. {1, 2, 3, 4}

Answer

{2, 4, 6}

∵ R={(2,3),(4,2),(6,1)}

Hence domain(R)={2,4,6}

14. Question

R is a relation from {11, 12, 13} to {8, 10, 12} defined by y = x – 3. Then, R^–1is A. {(8, 11), (10, 13)}

B. {(11, 8), (13, 10)}

C. {(10, 13), (8, 11), (8, 11)}

D. none of these

Answer

{(8, 11), (10, 13)}

Acoording to the question R is a relation from {11, 12, 13} to {8, 10, 12} defined by y = x – 3: ⇒R={(11,8),(13,10)}

⇒R^-1={(8, 11), (10, 13)}

15. Question

Let R = {(a, a), (b, b), (c, c), (a, b)} be a relation on set A = {a, b, c}. Then, R is A. identity relation

B. reflexive

C. symmetric

D. equivalence

Answer

reflexive

∵ R = {(a, a), (b, b), (c, c), (a, b)} is satisfuing only the property of reflexive relation. i.e Its reflexive i.e (a,a) ∈ R ∀ a ∈ A

16. Question

Let A = {1, 2, 3} and R = {(1, 2), (2, 3), (1, 3)} be a relation on A. Then, R is A. neither reflexive nor transitive

B. neither symmetric nor transitive

C. transitive

D. none of these

Answer

∵ R = {(1, 2), (2, 3), (1, 3)} satisfying only the property of transitive relation. i.e Its transitive i.e (a,b) ∈ R and (b,c) ∈ R → (a,c) ∈ R ∀

a,b,c ∈ A

17. Question

If R is the largest equivalence relation on a set A and S is any relation on A, then A. R ⊂ S

B. S ⊂ R

C. R = S

D. none of these

Answer

∵ Understood property

18. Question

If R is a relation on the set A = {1, 2, 3, 4, 5, 6, 7, 8, 9} given by x R y⬄ y = 3x, then R = A. {(3, 1), (6, 2), (8, 2), (9, 3)}

B. {(3, 1), (6, 2), (9, 3)}

C. {(3, 1), (2, 6), (3, 9)}

D. none of these

Answer

∵ for A = {1, 2, 3, 4, 5, 6, 7, 8, 9} the satisfying complete relation is: R={(1, 3), (2, 6), (3, 9)}

19. Question

If R is a relation on the set A = {1, 2, 3} given by R = (1, 1), (2, 2), (3, 3), then R is A. reflexive

B. symmetric

C. transitive

D. all the three options

Answer

∵ An important property of equivalence is that it divides the set into pairwise disjoint subsets called equivalent classes whose collection is called a partition of the set. Note that the union of all equivalence classes gives the complete set.

20. Question

If A = {a, b, c, d}, then a relation R = {(a, b), (b, a), (a, a)} on A is

A. symmetric and transitive only

B. reflexive and transitive only

C. symmetric only

D. none of these

Answer

∵ R is symmetric and reflexive

21. Question

If A = {1, 2, 3}, then a relation R = {(2, 3)} on A is

A. symmetric and transitive only

B. symmetric only

C. transitive only

D. none of these

Answer

∵ R={(1,1),(2,2),(3,3),(1,2),(1,3),(2,3),(3,2),(2,1),(3,1)}

Symmetric- {(2,3),(3,2)}

Transitive- {(1,2),(2,3),(1,3)}

22. Question

Let R be the relation on the set A = {1, 2, 3, 4} given by

R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Then,

A. R is reflexive and symmetric but not transitive

B. R is reflexive and transitive but not symmetric

C. R is symmetric and transitive but not reflexive

D. R is an equivalence relation

Answer

23. Question

Let A = {1, 2, 3}. Then, the number of equivalence relations containing (1, 2) is A. 1

B. 2

C. 3

D. 4

Answer

24. Question

The relation R = {(1, 1), (2, 2), (3, 3)} on the set {1, 2, 3} is

A. symmetric only

B. reflexive only

C. an equivalence relation

D. transitive only

Answer

25. Question

S is a relation over the set R of all real numbers and its is given by (a, b) ϵ S ⬄ ab ≥ 0. Then, S is A. symmetric and transitive only

B. reflexive and symmetric only

C. antisymmetric relation

D. an equivalence relation

Answer

26. Question

If the set Z of all integers, which of the following relation R is not an equivalence relation? A. x R y : if x ≤ y

B. x R y : if x = y

C. x R y : if x – y is an even integer

D. x R y : if x ≡ y (mod 3)

Answer

27. Question

Let A = {1, 2, 3} and consider the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}. Then, R is A. reflexive but not symmetric

B. reflexive but not transitive

C. symmetric and transitive

D. neither symmetric nor transitive.

Answer

28. Question

The relation S defined on the set R of all real number by the rule a Sb iff a ≥ b is A. an equivalence relation

B. reflexive, transitive but not symmetric

C. symmetric, transitive but not reflexive

D. neither transitive nor reflexive but symmetric

Answer

S: a S b ⟺ a ≥ b

Since a=a ∀a ∈ R, therefore a ≥ a always. Hence (a, a) always belongs to S ∀a ∈ R. Therefore, S is reflexive.

If a ≥ b then b ≤ a ⇏ b ≥ a. Hence if (a, b) belongs to S, then (b, a) does not always belongs to S. Hence S is not symmetric.

If a ≥ b and b ≥ c, therefore a ≥ c. Hence if (a, b) and (b, c) belongs to S, then (a, c) will belong to S∀a, b, c∈R. Hence, S is transitive.

29. Question

The maximum number of equivalence relations on the set A = {1, 2, 3} is

A. 1

B. 2

C. 3

D. 5

Answer

A= {1, 2, 3}

Then the equivalence relations would be,

P = {(1, 1), (2, 2), (3, 3)}

Q = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}

R = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)}

S = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 1)}

T = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 1)}

Hence, total 5 equivalence relations.

30. Question

Let R be a relation on the set N of natural numbers defined by n R m iff n divides m. Then, R is A. Reflexive and symmetric

B. Transitive and symmetric

C. Equivalence

D. Reflexive, transitive but Not symmetric

Answer

This question is quite self explanatory:

Reflexive: eg. m=5 and n=5 → 5 is divisible by 5

Transitive: 2 divides 4, 4 divides 40 and 2 also divedes 40.

31. Question

Let L denote the set of all straight lines in a plane. Let a relation R be defined by ℓ R m iff ℓ is perpendicular to m for all ℓ, m ϵ L. Then, R is

A. reflexive

B. symmetric

C. transitive D. none of these

Answer

Consider 6^thpart of MCQs

32. Question

Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as a R b if a is congruent to b for all a, b ϵ T. Then, R is

A. reflexive but not symmetric

B. transitive but not symmetric

C. equivalence

D. none of these

Answer

R: a R b ⟺ a ≅ b

Since, every triangle a∈T is congruent to itself, therefore (a, a)∈R ∀a∈T. Hence, R is reflexive. If a ≅ b, then b ≅ a. Hence if (a, b)∈R, then (b, a)∈R ∀a, b∈T. Hence, R is symmetric.

If a ≅ b and b ≅ c, then a ≅ c. Hence if (a, b) and (b, c) belongs to R, then (a, c) will belong to R∀a, b, c∈T. Hence, R is transitive.

Since R is reflexive, symmetric and transitive, therefore R is equivalence relation.

33. Question

Consider a non-empty set consisting of children in a family and a relation R defined as a R B if a is brother of b. Then, R is

A. symmetric but not transitive

B. transitive but not symmetric

C. Neither symmetric nor transitive

D. both symmetric and transitive

Answer

R: a R b ⟺ a is a brother of b.

If a is a brother of b, then that does not necessarily mean b is a brother of a, since b can be a sister of a too. Hence, if (a, b) ∈ R, then (b, a)∉R always. Hence R is not symmetric.

If a is a brother of b and b is a brother of c, then a has to be a brother of c. Hence if (a, b) and (b, c) belongs to R, then (a, c) will belong to R. Hence, R is transitive.

34. Question

For real numbers x and y, define x R y iff x – y+ √2 is an irrational number. Then the relation R is A. reflexive

B. symmetric

C. transitive

D. none of these

Answer

R: x R y ⟺ is irrational.

Since, x-x=0 always, therefore is irrational. Hence, (x, x)∈R ∀ x ∈ R. Hence, R is reflexive. R is not symmetric. Proof by counter-example:

Let . Then is irrational, Therefore (x, y)∈R. But

is rational, therefore (y, x)∉R. Hence, R is not symmetric. R is not transitive. Proof by counter-example:

Let and , then

is irrational, therefore (x, y)∈R

is irrational, therefore (y, z)∈R

is rational, therefore (x, z)∉R. Hence, R is not transitive.

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