**Q-1: Check each and every relation whether the following are symmetric, reflexive and transitive:**

**(i) The relation R of the set S = {2, 3, 4, 5, 6, 7, 8, 9. . . . . . . . 15} is defined as**

**R = {(a, b): 2a – b = 0}**

**(ii) The relation R of the set S having natural numbers is defined as**

**R = {(a, b): b = 2a + 6 and a < 5}**

**(iii) The relation R of the set S = {2, 3, 4, 5, 6, 7} is defined as**

**R = {(a, b): b is divisible by a}**

**(iv) The relation R of the set S having only the integers is defined as**

**R = {(a, b): a – b is an integer}**

**(v) The relation R from the set H having human beings at a particular time in the town is given by:**

**(a) R = {(a, b): a and b is working at the same place}**

**(b) R = {(a, b): a and b are living in the same society}**

**(c) R = {(a, b): a is exactly 6 cm taller than b}**

**(d) R = {(a, b): b is husband of a}**

**(e) R = {(a, b): a is the father of b}**

**Sol: **

**(i)** **S = {2, 3, 4, 5, 6 . . . . . . . 13, 14, 15}**

**R = {(a, b) : 2a – b = 0}**

Therefore, R = {(2, 4), (3, 6), (4, 8), (5, 10), (6, 12), (7, 14)}

Now,

(2, 4) **ɛ** R, but (4, 2) \(\notin\) R, i.e.,

[2 (4) – 2 ≠ 0]

So, **R is not symmetric**.

(2, 2), (3, 3)………..,(15, 15) \(\notin\) R.

So, **R is not reflexive**.

Since, (2, 4), (4, 8) **ɛ** R, but (2, 4) \(\notin\) R

As, [2(2) – 8 ≠ 0]

So, **R is not even transitive.**

**Therefore, R, in this case, is neither reflexive nor transitive and nor symmetric. **

** **

**(ii)** R = {(a, b) : b = 2a + 6 and a < 5}

Therefore, R = {(1, 8), (2, 10), (3, 12), (4, 14)}

Now,

(1, 8) **ɛ** R, but (8, 1) \(\notin\) R, i.e.,

So, **R is not symmetric**.

(1, 1) , (2, 2), . . . . . . . . . . . . . . . \(\notin\) R.

So, **R is not reflexive.**

Since, here there isn’t any pair in the form of (a, b) and (b, c) **ɛ** R, then (a, b) \(\notin\) R

So, **R is not even transitive**.

**Therefore, R in this case is neither reflexive, nor transitive and nor symmetric. **

** **

**(iii)** S = {2, 3, 4, 5, 6, 7}

R = {(a, b): b is divisible by a}

(2, 4) **ɛ** R [as 4 is divided by 2 completely]

(4, 2) \(\notin\) R [as 2 can’t be divided by 4]

So, **R is not symmetric**.

Each and every number is divisible at least by itself.

So, (a, a) **ɛ** R.

So, **R is reflexive**.

Now,

As (a, b) and (b, c) **ɛ** R.

Then clearly, ‘b’ will be divisible by ‘a’ and ‘c’ will be divisible by ‘b’.

Therefore, c is divisible by b.

So, (a, b) **ɛ** R.

Hence, **R is transitive.**

**Therefore, R in this case is reflexive and transitive, but not symmetric. **

**(iv) R = {(a, b) : a – b will be any integer} **

Now,

For each a, b **ɛ** S, if (a, b) ɛ R, then

a – b is an integer.

\(\boldsymbol{\Rightarrow }\) – (a – b) will also be an integer.

\(\boldsymbol{\Rightarrow }\) (b – a) will also be an integer.

Therefore, (a, b) **ɛ** R.

Hence, **R will be symmetric to each other**.

For each a **ɛ** S, (a, a) ɛ R as, a – a = 0 which is also an integer.

Hence, **R is reflexive**.

Let, a pair in the form of (a, b) and (b, c) **ɛ** R, then (a, b, c) **ɛ** Z.

\(\boldsymbol{\Rightarrow }\) (a – b) and (b – c) is an integer.

\(\boldsymbol{\Rightarrow }\) a – c = (x – y) + (y – z) will also be an integer.

Hence, **R is transitive**.

**Therefore, R in this case is reflexive, transitive and symmetric.**

** **

**(v) **

**(a)** R = {(a, b) : a and b is working at the same place}

Now,

If (a, b)** ɛ** R, then a and b works at the same place

\(\boldsymbol{\Rightarrow }\) b and a will also work at the same place.

\(\boldsymbol{\Rightarrow }\) (b, a)** ɛ** R

Hence, **R will be symmetric to each other**.

For each a **ɛ** S, (a, a) **ɛ** R as, ‘a’ and ‘a’ is working at the same place.

Hence, **R is reflexive**.

Let, a pair in the form of (a, b) and (b, c) **ɛ** R.

\(\boldsymbol{\Rightarrow }\) a and b is working at the same place, also b and c is working at the same place.

\(\boldsymbol{\Rightarrow }\) a and c is also working at the same place.

\(\boldsymbol{\Rightarrow }\) (a, c) **ɛ** R

Hence, **R is transitive**.

**Therefore, R in this case is reflexive, transitive and symmetric**.

**(b)** R = {(a, b) : a and b both are living in the same society}

Now,

If (a, b) ɛ R, then ‘a’ and ‘b’ is living in the same society.

\(\boldsymbol{\Rightarrow }\) b and a will also live in the same society.

\(\boldsymbol{\Rightarrow }\) (b, a) ɛ R

Hence, **R will be symmetric to each other**.

For each a ɛ S, (a, a) ɛ R as, ‘a’ and ‘a’ is the same human being.

Hence, **R is reflexive**.

Let, a pair in the form of (a, b) and (b, c) ɛ R.

\(\boldsymbol{\Rightarrow }\) a and b is living in the same society, also b and c is living in the same society.

\(\boldsymbol{\Rightarrow }\) a and c is also living in the same society.

\(\boldsymbol{\Rightarrow }\) (a, c) ɛ R

Hence, **R is transitive**.

**Therefore, R in this case is reflexive, transitive and symmetric.**

**(c)** R = {(a, b) : a is 6 cm taller than b}

Now,

If (a, b) ɛ R, then a is 6 cm taller than b.

\(\boldsymbol{\Rightarrow }\) b cannot ever will be taller than a.

\(\boldsymbol{\Rightarrow }\) (b, a) \(\notin\) R

Hence, **R is not symmetric to each other**.

For each a ɛ S, (a, a) \(\notin\) R.

As, the same human being can’t be taller than himself.

So, ‘a’ can’t be taller than ‘a’.

Hence, **R is not reflexive**.

Let, a pair in the form of (a, b) and (b, c) ɛ R.

\(\boldsymbol{\Rightarrow }\) a is 6 cm taller than b, then b will be 6 cm taller than c.

\(\boldsymbol{\Rightarrow }\) a is atleast 12 cm taller than c.

\(\boldsymbol{\Rightarrow }\) (a, c) \(\notin\) R

Hence, **R is not transitive**.

**Therefore, R in this case neither reflexive, nor transitive and nor symmetric.**

**(d)** R = {(a, b) : b is the husband of a}

Now,

If (a, b) ɛ R,

\(\boldsymbol{\Rightarrow }\) b is the husband of a.

\(\boldsymbol{\Rightarrow }\) a can’t be husband of b.

\(\boldsymbol{\Rightarrow }\) (b, a) \(\notin\) R

Instead of this, if ‘b’ is the husband of a, then ‘a’ is the wife of b.

Hence, **R is not symmetric to each other**.

(a, a) \(\notin\) R

Since, b cannot be the husband of himself.

Hence, **R is not reflexive.**

If (a, b), (b, c) ɛ R

\(\boldsymbol{\Rightarrow }\) b is the husband of a and a is the husband of c.

Which can’t ever be possible? Also, it is not so that b is the husband of c.

\(\boldsymbol{\Rightarrow }\) (a, c) \(\notin\) R

Hence, **R is not transitive**.

**Therefore, R in this case neither reflexive, nor transitive and nor symmetric.**

**(e)** R = {(a, b) : a is the father of b}

Now,

If (a, b) ɛ R,

\(\boldsymbol{\Rightarrow }\) a is the father of b.

\(\boldsymbol{\Rightarrow }\) b can’t be the father of a.

Thus, b is either the son or daughter of a.

\(\boldsymbol{\Rightarrow }\) (b, a) \(\notin \) R

Hence, **R is not symmetric to each other**.

(a, a) \(\notin\) R

Since, b cannot be the father of himself.

Hence, **R is not reflexive**.

Since, (a, b) ɛ R and (b, c) \(\notin \) R

\(\boldsymbol{\Rightarrow }\) a is the father of b, then b is the father of c.

\(\boldsymbol{\Rightarrow }\) a is not the father of c.

Indeed, a will be the grandfather of c.

\(\boldsymbol{\Rightarrow }\) (a, c) \(\notin\) R

Hence, **R is not transitive**.

**Therefore, R in this case neither reflexive, nor transitive and nor symmetric.**

**Q-2: Prove that the relation M in the set M of the real numbers which is defined as**

**M = {(x, y): x ≤ y ^{2}} which is neither reflexive, nor transitive, nor symmetric.**

** **

**Sol: **

M = {(x, y): x ≤ y^{2}}

We can see that (\(\frac{1}{2}, \; \frac{1}{2} \)) \(\notin\) R

As, \(\frac{1}{2} > \left (\frac{1}{2} \right)^{2}\)

Hence, **M is not reflexive**.

(1, 4) ɛ M as 1 < 4^{2}

But, 4 is not lesser than 1

So, (4, 1) \(\notin\) M

Hence, **M is not symmetric**.

(3, 2) and (2, 1.5) ɛ M **[as 3 < 2 ^{2} = 4 and 2 < (1.5)^{2} = 2.25]**

But, 3 > (1.5)^{2} = 2.25

So, (3, 1.5) \(\notin\) M

Hence, **M is not transitive**.

**Therefore, M is neither transitive, nor symmetric, nor reflexive.**

**Q-3: Check whether the relation given below is reflexive, symmetric and transitive:**

**The relation M is defined in the set {2, 3, 4, 5, 6, 7} as M = {(x, y): y = x + 1}.**

**Sol: **

Let us assume that:

S {2, 3, 4, 5, 6, 7}.

The relation M is defined on set S as:

M = {(x, y): y = x + 1}

Therefore, M = {(2, 3), (3, 4), (4, 5), (5, 6), (6, 7)}

We can observe that (2, 3) \(\notin \) M, but (3, 2) ɛ M.

Hence, **M is not symmetric**.

We need to find (x, x) \(\notin \) M, where x ɛ M.

(2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7) ɛ M

Hence, **M is not reflexive**.

(2, 3), (3, 4) ɛ M

But, (2, 4) \(\notin \) M

Hence, **M is not transitive**.

**Therefore, M is neither symmetric, nor reflexive, nor symmetric.**

**Q-4: Prove that the relation M in M which is defined as M = {(x, y): x ≤ y} is transitive and reflexive, but not symmetric.**

**Sol: **

M = {(x, y): x ≤ y}

So, (x, x) ɛ M ** [as x = x]**

Hence, **M is reflexive**.

Then, (4, 6) ɛ M (as 4 < 6)

But, (4, 6) \(\notin \) M since, 6 is greater than 4.

Hence, **M is not symmetric**.

Let, (x, y), (y, z) ɛ M

Then, x ≤ y and y ≤ z

\(\boldsymbol{\Rightarrow }\) x ≤ z

\(\boldsymbol{\Rightarrow }\) (x, z) \(\notin \) M

**Hence, M is transitive.**

**Therefore, M is reflexive and transitive, but not symmetric.**

** **

** **

**Q-5: Check that whether the relation M in M which is defined as M = {(x, y): x ≤ y ^{3}} is transitive, reflexive and symmetric.**

**Sol: **

M = {(x, y): x ≤ y}

We can see that (\(\frac{1}{2}, \; \frac{1}{2} \)) \(\notin\)

As, \(\frac{1}{2} > \left (\frac{1}{2} \right)^{3}\)

Hence, **M is not reflexive.**

(1, 2) ɛ M as 1 < 2^{3 }= 8

But, (2, 1) \(\notin\) M (as 2^{3 }> 1)

Hence, **M is not symmetric**.

We have,

(3, \(\frac{3}{2}\)), (\(\frac{3}{2}, \; \frac{6}{5} \)), as 3 < \(\left(\frac{3}{2} \right)^{3} \)) and (\(\frac{3}{2} < \left(\frac{6}{5} \right)^{3} \)

But, (3, \(\frac{6}{5} \)) \(\notin\) M as 3 > \(\left(\frac{6}{5} \right)^{3}\)

Hence, **M is not transitive.**

**Therefore, M is neither symmetric, nor reflexive, nor transitive. **

**Q-6: Prove that the relation M from the set {2, 3, 4} which is given by M = {(2, 3), (3, 2)} is not reflexive nor transitive, but it is symmetric.**

**Sol: **

As per the data given in the question, we have

M = {(2, 3), (3, 2)}

The set, say S, = {2, 3, 4}

Any relation M on the set S will be defined as M = {(2, 3), (3, 2)}

Thus, (2, 2), (3, 3), (4, 4) \(\notin\) M

Hence, **M is not reflexive**.

We know that,

(2, 3) ɛ M and (3, 2) ɛ M.

Hence**, M is symmetric**.

Now,

Since, (2, 3) as well as (3, 2) ɛ M

But, (2, 2) \(\notin\) M

Hence, **M is not transitive.**

**Therefore, M is symmetric, but it is neither reflexive nor transitive.**

**Q-7: Prove that the relation M in the set S for all the books in a library of the college BET, the relation given for it is M = {(a, b): a and b have the same number of pages in the book} which is the equivalence relation.**

**Sol: **

Let, the set S be the set of all the books in the library of the college BET.

Relation M = {(a, b): a and b have equal number of pages in the book}

**M is reflexive** as (a, a) ɛ M as x and x will have the equal number of pages in the book.

Let, (a, b) ɛ M

\(\boldsymbol{\Rightarrow }\) a and b have the same/ equal number of pages.

\(\boldsymbol{\Rightarrow }\) b and a will also have the equal number of pages in the book.

\(\boldsymbol{\Rightarrow }\) As, (a, b) ɛ M. So, (b, a) ɛ M

Hence, **M is symmetric**.

Let, (a, b) ɛ M and (b, c) ɛ M

\(\boldsymbol{\Rightarrow }\) Since, a and b have equal number of pages in the book so, b and c will also have the equal number of pages in the book.

\(\boldsymbol{\Rightarrow }\) a and b will also have the equal number of pages.

\(\boldsymbol{\Rightarrow }\) (a, b) ɛ M.

Hence, **M is transitive**.

Thus, M is symmetric, transitive and also, reflexive.

**Therefore, M is an equivalence relation.**

**Q-8: Prove that the relation M of the set S = {2, 3, 4, 5, 6} which is given by M = {(x, y): | x – y | is even}, is an equivalence relation. Also, prove that all the elements are related to each other of the set {3, 5} and the elements of (2, 4, 6} are inter- related with each other. But, elements of {3, 5} and {2, 4, 6} are not related to each other nor their any of the elements are interlinked.**

**Sol: **

As per the data given in the question, we have

S = {2, 3, 4, 5, 6, 7} and M = {(x, y): |x – y | is even}

So,

For any of the element x ɛ S, we have | x – x | = 0 and we know that 0 is an even number.

Hence, **R is reflexive**.

Let us assume that, (x, y) ɛ M.

\(\boldsymbol{\Rightarrow }\) | x – y | is even

\(\boldsymbol{\Rightarrow }\) – | -(x – y) | = | y – a | will also be even.

\(\boldsymbol{\Rightarrow }\) (y, x) ɛ M.

Hence, **M is symmetric**.

Let us now assume that, (x, y) ɛ M and (y, z) ɛ M.

\(\boldsymbol{\Rightarrow }\) | x – y | is even and | y – z | is even

\(\boldsymbol{\Rightarrow }\) (x – y) is even and also, (y – z) is even.

\(\boldsymbol{\Rightarrow }\) (x – z) = (x – y) + (y – z) will also be even. [As we know that the sum of the two integers is even]

\(\boldsymbol{\Rightarrow }\) | x – y | is even.

\(\boldsymbol{\Rightarrow }\) (x, z) ɛ M

Hence, **M is transitive**.

Therefore, M is an equivalence relation**. **

**As per the condition given in the question, all the elements are related to each other in the set {3, 5} as each of the elements in the set is odd. Therefore, the mod of the difference of any of the two elements will always be even.**

**Also, all the elements of the set {2, 4, 6} are inter- related as each and every element is an even number in this subset. **

**Also, elements of the subset {3, 5} and {2, 4, 6} are not related to each other in any way as all of the elements of {3, 5} are odd and all the elements of {2, 4, 6} are even. Therefore, the modulus of the difference of the elements (for each of the two subsets) won’t be even always, such that 2 – 3, 3 – 4, 2 – 5, 3 – 6, 4 – 3, 4 – 5, 5 – 2, 5 – 4, 5 – 6, 6 – 3 and 6 – 5 all are an odd numbers**.

** **

** **

**Q-9: Prove that all the relation M of the set S = {a ɛ P : 0 ≤ a ≤ 12}, which is given by**

**(a) M = {(x, y): | x – y | is a multiple of 3}**

**(b) M = {(x, y): x = b} is an equivalence relation. Get all the sets of elements which are related to 1 in every case.**

**Sol: **

S = {a ɛ P: 2 ≤ a ≤ 14} = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}

**(i)**

M = {(x, y): | x – y | will be the multiple of 3}

Let, (x, y) ɛ M

\(\boldsymbol{\Rightarrow }\) | x – y | will be the multiple of 3.

\(\boldsymbol{\Rightarrow }\) |- (x – y)| = | y – x | will also be the multiple of 3.

\(\boldsymbol{\Rightarrow }\) (y, x) ɛ M

Hence, **M is symmetric**.

For any of the element x ɛ S, we have (x, x) ɛ M as | x – x | = 0 which is a multiple of 3.

Hence, **M is reflexive**.

Let, (x, y), (y, z) ɛ M

\(\boldsymbol{\Rightarrow }\) | x – y | will be the multiple of 3 and | y – z | will be the multiple of 3.

\(\boldsymbol{\Rightarrow }\) (x – y) will be the multiple of 3 and (y – z) will be the multiple of 3.

\(\boldsymbol{\Rightarrow }\) (x – z) = (x – y) + (y – z) will be the multiple of 3.

\(\boldsymbol{\Rightarrow }\) | x – z | will be the multiple of 3.

\(\boldsymbol{\Rightarrow }\) (x, z) ɛ M

Hence, **M is transitive.**

Therefore, M is an equivalence relation.

The set of elements related to 1 is {1, 4, 7, 10, 13} as:

**| 1 -1 | = 0 which is a multiple of 0.**

**| 4 – 1 | = 0 which is a multiple of 3.**

**| 7 – 1 | = 6 which is a multiple of 3.**

**| 10 – 1 | = 9 which is a multiple of 3.**

**| 13 – 1 | = 12 which is a multiple of 3.**

**(b) **

M = {(x, y): x = y}

Let, (x, y) ɛ M

\(\boldsymbol{\Rightarrow }\) x = y

\(\boldsymbol{\Rightarrow }\) y = x

\(\boldsymbol{\Rightarrow }\) (y, x) ɛ M

Hence, **M is symmetric**.

For every element x ɛ M, as we have (x, x) ɛ M, as x = x.

Hence, **M is reflexive**.

Let, (x, y) Let, (x, y) ɛ M and (y, z) ɛ M

\(\boldsymbol{\Rightarrow }\) x = y and y = z

\(\boldsymbol{\Rightarrow }\) x = z

\(\boldsymbol{\Rightarrow }\) (x, z) ɛ M

Hence, **M is transitive**.

Therefore, M is an equivalence relation.

**All the elements in M which are related to 1 can be those elements from the set M which is equal to 1.**

**Therefore, the set of elements related to 1 is {1}.**

**Q-10: Give an example for each of the relation, which is**

**(a) Symmetric but, neither transitive nor reflexive.**

**(b) Transitive but, neither reflexive nor symmetric.**

**(c) Symmetric and reflexive but, not transitive.**

**(d) Transitive and reflexive but, not symmetric.**

**(e) Transitive and symmetric but, not reflexive.**

**Sol: **

**(a)** Let, S = {7, 8, 9}

A relation M in S as S = {(7, 8), (8, 7)}

The relation **M is not reflexive** because (7, 7),(8, 8), (9, 9) \(\notin\) M.

(7, 8) ɛ M and also, (8, 7) ɛ M

Hence, **M is symmetric**.

Now,

(7, 8), (8, 7) ɛ M but, (7, 7) \(\notin\) M

Hence, **M is not transitive**.

**Therefore, the relation M is symmetric but, neither reflexive nor transitive.**

** **

**(b)** Consider a relation M which is defined as:

M = {(x, y): x < y}

For a ɛ M, we have (x, x) \(\notin\) M as x won’t be less than itself ever.

Also, x = x

Hence, **M is not reflexive**.

As per the given condition, x < y

Let us take an e.g.(2, 3) ɛ M since, 2 < 3.

But, 3 won’t ever be less than 2.

Therefore, (3, 2) \(\notin\) M

Hence, **M is not symmetric**.

Let, (x, y), (y, x) ɛ M

\(\boldsymbol{\Rightarrow }\) x < y and y < z

\(\boldsymbol{\Rightarrow }\) x < z

\(\boldsymbol{\Rightarrow }\) (x, z) ɛ M

Hence, **M is not transitive.**

**Therefore, the relation R is neither symmetric nor it is reflexive but, it is transitive. **

**(c)** Let, S = {2, 4, 6}

Let us define the relation M on S as

S = {(2, 2), (4, 4), (6, 6), (2, 4), (4, 2), (4, 6), (6, 4)}

The relation S will be reflexive in this case as for each x ɛ S, (x, x) ɛ M

i.e., {(2, 2), (4, 4), (6, 6)} ɛ M

The relation S will be symmetric in this case as (x, y) ɛ M

\(\boldsymbol{\Rightarrow }\) (y, x) ɛ M for each x, y ɛ M

The relation S won’t be transitive in this case as (2, 4), (4, 6) ɛ M, but (2, 6) \(\notin \) M.

**Therefore, the relation M is not transitive but, it is symmetric and reflexive.**

** **

**(d)** Let us define any relation S in S as,

S = {(x, y): x^{3} ≥ y^{3}}

(x, x) ɛ M since, x^{3} = x^{3}

Hence, **M is reflexive**.

(3, 2) ɛ M since, 3^{3} ≥ 2^{3}

But, (2, 3) \(\notin \) M since, 2^{3} ≤ 3^{3}

Hence, **M is not symmetric**.

Let, (x, y), (y, z) ɛ M

\(\boldsymbol{\Rightarrow }\) x^{3} ≥ y^{3} and y^{3} ≥ z^{3}

\(\boldsymbol{\Rightarrow }\) x^{3} ≥ z^{3}

\(\boldsymbol{\Rightarrow }\) (x, z) ɛ M

Hence, **M is transitive**.

**Therefore, the relation M is not symmetric but, it is transitive and reflexive.**

** **

**(e)** Let, S = {-6, -7}

Let us define a relation M in S as,

S = {(-6, -7), (-7, -6), (-6, -6)}

The relation **M is not reflexive** as (-7, -7) \(\notin \) M

The relation **M is symmetric** as (-6, -7) ɛ M and (-7, -6) ɛ M

We can see that,

(-6, -7), (-7, -6) ɛ M and also, (-6, -6) ɛ M

**Hence, the relation M is transitive also.**

**Therefore, the relation M is symmetric as well as transitive but, it is not reflexive.**

**Q-11: Prove that the relation A in the set S for the points in the plane which is given by A = {(M, N): The distance between the point M from (0, 0) will be the same as the distance between the point N from (0, 0)}, is an equivalence relation. Now, also prove that the set of all the points which is related to the point M ≠ (0, 0) is a circle which is passing from the point P with having centre at origin.**

**Sol: **

A = {(M, N): The distance between the point M from (0, 0) will be the same as the distance between the point N from (0, 0)}

(M, M) ɛ A, as the distance between the point M from (0, 0) will be the same as the distance between the point N from (0, 0).

Hence, **R is reflexive**.

Let, (M, N) ɛ A

\(\boldsymbol{\Rightarrow }\) The distance between the point M from (0, 0) will be the same as the distance between the point N from (0, 0).

\(\boldsymbol{\Rightarrow }\) The distance between the point N from (0, 0) will be the same as the distance between the point M from (0, 0).

\(\boldsymbol{\Rightarrow }\) (N, M) ɛ A

Hence, **A is symmetric**.

Let, (M, N), (N, P) ɛ A

\(\boldsymbol{\Rightarrow }\) The distance between the points M and N from the origin is the same, then also, the distance between the points N and P from the origin is the same.

\(\boldsymbol{\Rightarrow }\) The distance of the points M and P is the same from the origin.

\(\boldsymbol{\Rightarrow }\) (M, P) ɛ A

Hence, **the relation A is transitive**.

Thus, the relation A is an equivalence relation.

The set of all the points which are related to M ≠ (0, 0) can be those points which have the same distance from the origin as the distance of the point M from (0, 0).

Simply, If O be the origin and OM = x, then the set of all of the points which are related to the point M is at the distance x from the origin point.

**Therefore, the set of points hence forms a circle which having the centre at the origin and this circle is passing through the point M. **

**Q-12: Prove that the relation A which is defined in the set S for all the triangles as A = {(P _{1}, P_{2}): P_{1} is similar to P_{2}}, is an equivalence relation. Assume three right angled triangles, say, triangle P_{1} having sides 4, 5, 6, triangle P_{2} having sides 6, 14, 15 and triangle P_{3 }having sides 7, 9, 11. Find which triangle among P_{1}, P_{2} and P_{3} will be related?**

**Sol: **

M = {(P_{1}, P_{2}): P_{1} is similar to P_{2}}

**M can be reflexive** in this case as its obvious that, each triangle is similar to itself.

Now,

If (P_{1}, P_{2}) ɛ A, so P_{1} is completely similar to P_{2}.

\(\boldsymbol{\Rightarrow }\) P_{2 } is similar to P_{1}

\(\boldsymbol{\Rightarrow }\) (P_{2}, P_{1)} ɛ A

Hence, **M is symmetric**.

Let, (P_{1}, P_{2)}, (P_{2}, P_{3}) ɛ A

\(\boldsymbol{\Rightarrow }\) P_{1} is similar to P_{2} and also, P_{2} is similar to P_{3 }

\(\boldsymbol{\Rightarrow }\) P_{1} is similar to P_{3 }

\(\boldsymbol{\Rightarrow }\) (P_{1}, P_{3}) ɛ A

Hence, **M is transitive**.

Therefore, R is an equivalence relation.

We can see that,

\(\frac{3}{6} = \frac{4}{8} = \frac{5}{10} \left(\frac{1}{2} \right)\)Hence, the corresponding sides of the triangle P_{1} and P_{3 }is in the same proportion (ratio).

Thus, the triangle P_{1} will be similar to the triangle P_{3}.

**Therefore, the triangle P _{1} is inter- related to the triangle P_{3}**

**Q-13: Prove that the relation M is defined in the set S of every polygon in such a way that M = {(R _{1}, R_{2}): R_{1} and R_{2} must have the equal number of sides}, is an equivalence relations. Find all the sets of all the elements in S which is related to the right angled triangle T having sides 4, 5 and 6.**

** **

**Sol: **

M = {(R_{1}, R_{2}): R_{1} and R_{2} must have the equal number of sides}

**M will be reflexive** in this case, as (R_{1}, R_{2}) ɛ M, since a same polygon has same number of the sides among itself.

Let, (R_{1}, R_{2}) ɛ M

\(\boldsymbol{\Rightarrow }\) R_{1} and R_{2} has same number of its sides.

\(\boldsymbol{\Rightarrow }\) R_{2} and R_{1} has same number of its sides.

\(\boldsymbol{\Rightarrow }\) (R_{2}, R_{1}) ɛ M

Hence, **M is symmetric**.

Let, (R_{1}, R_{2}), (R_{2}, R_{3}) ɛ M

\(\boldsymbol{\Rightarrow }\) R_{1} and R_{2} has same number of its sides.

Also, R_{2} and R_{3} have same number of its sides.

\(\boldsymbol{\Rightarrow }\) R_{1} and R_{3} has same number of its sides.

\(\boldsymbol{\Rightarrow }\) (R_{1}, R_{3}) ɛ M

Hence, **M is transitive**.

Therefore, M will be an equivalence relation.

All the elements in the set S is related to the right angled triangle (T) having sides 4, 5 and 6 are the polygon who have exactly 3 sides.

**Therefore, all the elements in the set S are related to the right triangle T is the set of all of the triangles. **

**Q-14: Assume that, Q is the set of all of the lines in the XY- plane and M is the relation in Q which is defined as M = {(P _{1}, P_{2}): P_{1} is parallel to P_{2}}. Prove that the relation M is an equivalence relation. Hence, find the set of all of the lines which are related to the line b = 2a + 4.**

** **

**Sol: **

M = {(P_{1}, P_{2}): P_{1} is parallel to P_{2}}

**M will be reflexive** as any line can always be at least parallel to itself, i.e., (P_{1}, P_{1}) ɛ M.

Let, (P_{1}, P_{2}) ɛ M

\(\boldsymbol{\Rightarrow }\) P_{1} is parallel to P_{2 }and also, P_{2} will be parallel to P_{1 }

\(\boldsymbol{\Rightarrow }\) (P_{2}, P_{1}) ɛ M

Hence, **M is symmetric**.

Let, (P_{1}, P_{2}), (P_{2}, P_{3}) ɛ M

\(\boldsymbol{\Rightarrow }\) P_{1} is parallel to P_{2 }and also, P_{2} will be parallel to P_{3 }

\(\boldsymbol{\Rightarrow }\) P_{1} is parallel to P_{3}

\(\boldsymbol{\Rightarrow }\) (P_{2}, P_{1}) ɛ M

Hence, **M is transitive**.

Therefore, M is an equivalence relation.

The set of all of the lines which are related to the given line which is b = 2a + 4 will be the set of all of the lines which are parallel to the line b = 2a + 4.

Here, slope of the given line, say m, b = 2a + 4 is 2.

We know that the slope of the parallel lines is the same.

The line which is parallel to the given line is in the form b = 2a + c, where c ɛ M

**Therefore, the set of all of the lines which are related to the given line which is given by b = 2a + c, where c ɛ M.**

**Q-15: Let, M be the given relation in the set S = {2, 3, 4, 5} which is given by-**

**M = {(2, 3), (3, 3), (2, 2), (5, 5), (2, 4), (4, 4), (4, 3)}. Select the correct answer:**

**(a) M is symmetric and reflexive, but it is not transitive.**

**(b) M is transitive and reflexive, but it is not symmetric.**

**(c) M is transitive and symmetric, but it is not reflexive.**

**(d) M is an equivalence relation.**

** **

**Sol: **

Given relation is:

M = {(2, 3), (3, 3), (2, 2), (5, 5), (2, 4), (4, 4), (4, 3)}

We can see that, (x, x) ɛ M, for each x ɛ {2, 3, 4, 5}

Hence, **M is reflexive**.

We can note that, (2, 3) ɛ M, but (3, 2) \(\notin \) M

Hence, **M is not symmetric**.

We can observe that (x, y), (y, z) ɛ M

\(\boldsymbol{\Rightarrow }\) (x, z) ɛ M for each a, b, c ɛ {2, 3, 4, 5}.

Hence, **M is transitive**.

Therefore, M is transitive and reflexive, but it is not symmetric.

**The correct answer is B. **

**Q-16: Let us consider M be a relation in any set M which is given by**

**M = {(x, y): x = y – 2, y > 6}.**

**Select the correct choice from the following:**

**(a) (2, 4) ɛ M (b) (3, 8) ɛ M**

**(c) (6, 8) ɛ M (d) (8, 7) ɛ M**

** **

**Sol: **

M = {(x, y): x = y – 2, y > 6}

As, y > 6

So, (2, 4) \(\notin \) M

3 ≠ 8 – 2 as well,

So, (3, 8) \(\notin \) M

8 ≠ 7 – 2 as well,

So, (8, 7) \(\notin \) M

Let us consider (6, 8)

6 > 8 and also, we have 6 = 8 – 2

Hence,

(6, 8) ɛ M

**Therefore, the correct answer is C. **