**EXERCISE – 1.1**

**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. **

**EXERCISE – 1. 2**

**Q-1: Prove that the function f: \(R_{*}\rightarrow R_{*}\) which is defined by f(a) = \(\frac{1}{a}\) which is one- one and onto, where \(R_{*}\) is the set of all of the non- zero real numbers. Check whether the result is true or not, if the domain, say, \(R_{*}\) is replaced by M having co- domain as same as \(R_{*}\)?**

** **

**Sol: **

As per the data given in the question,

\(R_{*}\rightarrow R_{*}\) which is defined by f(a) = \(\frac{1}{a}\)

For one- one condition:

Let, a and b \(\epsilon R_{*} \) such that, f(a) = f(b)

\(\Rightarrow \frac{1}{a} = \frac{1}{b}\)\(\boldsymbol{\Rightarrow }\) a = b

**Hence, the function f is one- one.**

For onto condition:

Clearly, for b\(\epsilon R_{*} \), there must exists a = \(\frac{1}{b} \; \epsilon R_{*} \) [since, b ≠ 0], so that

f(x) = \(\frac{1}{\frac{1}{b}}\) = y

**Hence, the function f is onto. **

Therefore, the given function f(a) is one- one as well as onto.

Let us consider a another function g: M → \( R_{*} \) which is defined by g(a) = \(\frac{1}{a}\).

g(x_{1}) = g(x_{2})

\(\boldsymbol{\Rightarrow }\) x_{1} = x_{2}

Hence, the another function g is also one- one.

It is obvious that g is not onto as for 1.2 \(\epsilon R_{*} \), there doesn’t exist any a in M so that:

g(a) = \(\frac{1}{1.2}\)

**Therefore, the function ‘g’ is one- one but, it is not onto.**

**Q-2: Check the following functions for their injectivity and surjectivity:**

**(i) f: N →N which is given by f(a) = a ^{2}**

**(ii) f: Z →Z which is given by f(a) = a ^{2}**

**(iii) f: R → R which is given by f(a) = a ^{2}**

**(iv) f: N →N which is given by f(a) = a ^{3}**

**(v) f: Z → Z which is given by f(a) = a ^{3}**

** **

**Sol: **

**(i)** f: N → N which is given by f(a) = a^{2 }

We can see that for a, b ɛ N

**f(a) = f(y)**

\(\boldsymbol{\Rightarrow }\) a^{2 }= b^{2 }

\(\boldsymbol{\Rightarrow }\) a = b

Hence, the function f is injective.

2 ɛ N

But, there doesn’t exist any of the a in N, so that f(a) = b^{2 }= 2

Hence, the function f is not surjective.

Therefore, the function f is not surjective but, it is injective.

**(ii)** f: Z → Z which is given by f(a) = a^{2 }

We can see that for a, b ɛ Z

f(-1) = f(1) = 1

But, -1≠ 1.

Hence, the function f is not injective.

-2 ɛ Z

But, there doesn’t exist any of the a ɛ Z, so that f(a) = – 2 or a^{2 }= -2

Hence, the function f is not surjective.

**Therefore, the function f is neither surjective nor it is injective.**

**(iii)** f: R → R which is given by f(a) = a^{2 }

We can see that for a, b ɛ R

f(-1) = f(1) = 1

But, -1≠ 1.

Hence, the function f is not injective.

-2 ɛ R

But, there doesn’t exist any of the a ɛ R, so that f(a) = – 2 or a^{2 }= -2

Hence, the function f is not surjective.

Therefore, the function f is neither surjective nor it is injective.

**(iv) f: N → N which is given by f(a) = a ^{3 }**

We can see that for a, b ɛ N

f(a) = f(y)

\(\boldsymbol{\Rightarrow }\) a^{3 }= b^{3 }

\(\boldsymbol{\Rightarrow }\) a = b

Hence, the function f is injective.

2 ɛ N

But, there doesn’t exist any of the element of a ɛ Z, so that f(a) = 2 or a^{3 }= 2

Hence, the function f is not surjective.

**Therefore, the function f is not surjective but it is injective.**

**(v) f: Z → Z which is given by f(a) = a ^{3 }**

We can see that for a, b ɛ Z

f(a) = f(y)

\(\boldsymbol{\Rightarrow }\) a^{3 }= b^{3 }

\(\boldsymbol{\Rightarrow }\) a = b

Hence, the function f is injective.

2 ɛ Z

But, there doesn’t exist any of the element of a ɛ Z, so that f(a) = 2 or a^{3 }= 2

Hence, the function f is not surjective.

**Therefore, the function f is not surjective but it is injective.**

**Q-3: Show that the GIF (Greatest Integer Function) f: R →R which is given by f(a) = [a], which is neither one- one nor onto, where [a] notifies the greatest integer which must be less than or equal to a.**

** **

**Sol: **

f: R → R which is given by f(a) = [a]

f(1.4) = [1.4] = 1, f(1.7) = [1.7] = 1

\(\boldsymbol{\Rightarrow }\) f(1.4) = f(1.7) = 1, but 1.4 ≠ 1.7

Hence, the function f is not one- one.

Let us consider that, 0.9 ɛ R.

We know that, f(a) = [a] will always be any integer.

So, there doesn’t exist any of the element a ɛ R, such that f(a) = 0.9

Hence, the function f is not onto.

**Therefore, the GIF (Greatest Integer Function) is neither onto nor one- one.**

**Q-4: Prove that the Modulus Function f: R → R which is given by f(a) = [a], which is neither one- one nor it is onto, where |a| is a, if a will be positive or 0 and |a| is –a, if a will be negative.**

** **

**Sol: **

f : R → R which is given by

f(a) = |a| = a, if a ≥ 0 and (-a), if a ≤ 0

It is obvious that f(-2) = |-2| = 2 and f(2) = |2| = 2

So, f(-2) = f(2), but -2 ≠ 2

Hence, f is not one- one

Let us consider -2 ɛ R.

We know that, f (a) = |a| will always be non- negative.

Therefore, neither of any element a exist in the domain R, so that

f(a) = |a| = -2.

Hence, f is not onto.

**Hence, the modulus function is neither onto nor one- one.**

**Q-5: Prove that the Signum function f: R → R which is given by f(a) = **

**1, if a > 0**

**0, if a = 0**

**-1, if x < 0 is neither onto nor one- one.**

** **

**Sol: **

f: R → R which is given by f(a) =

1, if a > 0

0, if a = 0

-1, if x < 0

Since, f(a) can only take 3 values (i.e., 1, 0, -1) for the elements -3 in the co- domain R, there does not exist any of the a in domain R, so that f(a) = -3.

Hence, f is not onto.

Here, we can see that:

f(2) = f(3) = 1, but 2 ≠ 3

Hence, f is not one- one.

**Therefore, the Signum function in this case is neither onto nor one- one.**

**Q-6: Let us take P as {2, 3, 4}, Q as {5, 6, 7, 8} and let f = {(2, 5), (3, 6), (4, 7)} which are the function from P and Q. Prove that the function f is one- one.**

** **

**Sol: **

As per the data given in the question, we have

P = {2, 3, 4} and Q = {5, 6, 7, 8}

f : P → Q is defines as f = {(2, 5), (3, 6), (4, 7)}

Thus, f (2) = 5, f(3) = 6 and f(4) = 7

Now,

Here we can see that the images of the distinct elements of P under f are also distinct.

**Therefore, the function f is not one- one.**

** **

** **

**Q-7. Check whether the function in each of the following conditions is one- one, onto or bijective. Justify your answer in each cases.**

**(i) f : R → R which is defines by f(a) = 4 – 5a**

**(ii) f : R → R which is defined by f(a) = 2 + a ^{2}**

** **

**Sol: **

**(i)** f : R → R which is defined as f(a) = 4 – 5a

Let, a_{1}, a_{2} ɛ so that, f(a_{1}) = f(a_{2})

\(\boldsymbol{\Rightarrow }\) 4 – 5a_{1} = 4 – 5a_{2 }

\(\boldsymbol{\Rightarrow }\) – 5a_{1} = – 5a_{2 }

\(\boldsymbol{\Rightarrow }\) a_{1} = a_{2 }

**Hence, the function f is one- one.**

Now,

For any of the real number (b) in R, there exists \(\frac{4 – b}{5}\) in R, so that

f (\(\frac{4 – b}{5}\)) = \( 4 – 5\left(\frac{4 – b}{5} \right) \)

f (\(\frac{4 – b}{5}\)) = 4 – (4 – b)

f (\(\frac{4 – b}{5}\)) = b

**Hence, the function f is onto**.

**Therefore, the function f is bijective**.

**(ii)** f : R → R which is defined as f(a) = 2 + a^{2}

Let, a_{1}, a_{2} ɛ so that, f(a_{1}) = f(a_{2})

\(\boldsymbol{\Rightarrow }\) 2 + a_{1}^{2} = 2 + a_{2}^{2}

\(\boldsymbol{\Rightarrow }\) a_{1}^{2} = a_{2}^{2}

\(\boldsymbol{\Rightarrow }\) a_{1} = ± a_{2 }

Thus, f(a_{1}) = f(a_{2}) which doesn’t imply that a_{1} = a_{2}

Let us illustrate an example, f(2) = f(-2) = 3

**Hence, the function f is not one- one.**

Now,

Let us take an element -3 in the co- domain R.

We can see that, f(a) = 2 + a^{2} which is positive for all of the x ɛ R.

So, there doesn’t exist any a in the domain R so that f(a) = -3.

**Hence, the function f is not onto**.

**Therefore, the function f is neither one- one nor onto so, it is not bijective.**

**Q-8: Let us consider P and Q be the two sets. Prove that f : P × Q → Q × P so that (x, y) = (y, x) is bijective function.**

Sol:

f : P × Q → Q × P which is defined as f(x, y) = y, x.

Let, (x_{1}, y_{1)}, (x_{2}, y_{2)} ɛ P × Q so that, f(x_{1}, y_{1)}= f(x_{2}, y_{2)}

\(\boldsymbol{\Rightarrow }\) (y_{1}, x_{1)} = (y_{2}, x_{2)}

\(\boldsymbol{\Rightarrow }\) y_{1 }= y_{2 } and x_{1 }= x_{2 }

\(\boldsymbol{\Rightarrow }\) (x_{1}, y_{1)} = (x_{2}, y_{2)}

**Hence, the function f is one- one.**

Let us consider that (y, x) ɛ Q × P be any element.

Thus,

There will exist (x, y) ɛ P × Q so that,

f(x, y) = (y, x).

**Hence, f is onto.**

**Therefore, the function f is bijective.**

**Q-9: Let f : N → N which will be defined by f(p) =**

**\( \frac{p + 1}{2}\), if p is odd**

**\( \frac{p}{2}\), if p is even, for all p ɛ N.**

**Check and justify your answer that whether the function f is bijective or not?**

** **

**Sol: **

f : N → N which is defined as f(p) =

\(\frac{p + 1}{2}\), if p is odd

\(\frac{p}{2}\), if p is even

for all n ɛ N

We can see that:

f(3) = \(\frac{3 + 1}{2}\) = 2 and f(4) = \(\frac{4}{2}\) = 2

\(\boldsymbol{\Rightarrow }\) f(3) = f(4), but 3 ≠ 4

**Hence, f is not one- one.**

Let us consider any natural number (p) in the co- domain N.

**Case-1: [ p is odd ]**

\(\boldsymbol{\Rightarrow }\) p = 2r + 1 for few r ɛ N.

Thus, there exists 4r + 1 ɛ N so that,

f(4r + 1) = \(\frac{4r + 1 + 1}{2}\) = \(\frac{4r + 2}{2}\) = 2r + 1

**Case-2: [p is even]**

\(\boldsymbol{\Rightarrow }\) p = 2r for few r ɛ N.

Thus, there exists 4r ɛ N so that,

f(4r) = \(\frac{4r}{2}\) = 2r

**Hence, the function f is onto.**

**Therefore, the function f is not a bijective function. **

**Q-10: Let P = A – {3} and Q = A – {1}. Let us consider the function f : P → Q which defined by f(a) = \(\left(\frac{a – 3}{a – 4} \right)\). Check and Justify your answer, whether the function f is onto and one- one?**

** **

**Sol: **

P = A – {3}, Q = A – {1} and f : P → Q which is defined by

f(a) = \(\left(\frac{a – 3}{a – 4} \right)\)

Let, a, b ɛ P so that f(a) = f(b)

\(\boldsymbol{\Rightarrow }\) \( \frac{a – 3}{a – 4} = \frac{b – 3}{b – 4} \)

\(\boldsymbol{\Rightarrow }\) (a – 3)(b – 4) = (b – 3)(a – 4)

\(\boldsymbol{\Rightarrow }\) ab – 4a – 3b + 12 = ab – 4b – 3a + 12

\(\boldsymbol{\Rightarrow }\) a = b

Hence, the function f is one- one.

Let, b ɛ Q = A – {1}.

Thus, b ≠ 1

Then, the function f will be onto if a ɛ P so that, f(a) = b.

f(a) = b

\(\boldsymbol{\Rightarrow }\) \(\frac{a – 3}{a – 4}\) = b

\(\boldsymbol{\Rightarrow }\) (a – 3) = b(a – 4)

\(\boldsymbol{\Rightarrow }\) a – 3 = ab – 4b

\(\boldsymbol{\Rightarrow }\) a(1 – b) = 3 – 4b

\(\boldsymbol{\Rightarrow }\) a = \(\frac{3 – 4b}{1 – b}\) ɛ [b ≠ 1]

Then,

For any value of b ɛ Q, there will exist \(\frac{3 – 4b}{1 – b}\) ɛ P, so that

f(\(\frac{3 – 4b}{1 – b}\)) = \(\frac{\left (\frac{3 – 4b}{1 – b} \right) – 3}{\left (\frac{3 – 4b}{1 – b} \right) – 4}\)

= \(\frac{\left(3 – 4b \right) – 3\left(1 – b \right)}{\left(3 – 4b \right) – 4\left(1 – b \right)}\)

= \(\frac{3 – 4b – 3 + 3b}{3 – 4b – 4 + 4b}\)

= \(\frac{-b}{-1}\) = b

Hence, the function f is onto.

**Therefore, the function f is onto as well as it is one- one.**

**Q-11: Let us assume, f : R → R will be defined as g(a) = a ^{5}. Select the correct answer from the following:**

**(a) f is many- one onto (b) f is one- one onto**

**(c) f is neither one- one nor onto (d) f is one- one but, not onto**

** **

**Sol: **

f : R → R which is defined as g(a) = a^{4}.

Let, a, b ɛ R so that f(a) = f(b).

\(\boldsymbol{\Rightarrow }\) a^{4} = b^{4 }

\(\boldsymbol{\Rightarrow }\) a = ± b

Thus, f(a) = f(b) doesn’t imply that a = b.

Example:

f(3) = f(-3) = 3

Hence, the function f is not one- one.

Let us consider an element 3 in the co- domain R. It’s obvious that there doesn’t exist any a in the domain R so that, f(a) = 3.

Hence, the function f is not onto.

Therefore, the function f is neither one- one nor it is onto.

**Thus, the correct answer is option (c).**

**Q-12: Let us assume, f : R → R will be defined as g(a) = 4a . Select the correct answer from the following:**

**(a) f is many- one onto (b) f is one- one onto**

**(c) f is neither one- one nor onto (d) f is one- one but, not onto**

** **

**Sol: **

f : R → R which is defined as g(a) = 4a.

Let, a, b ɛ R so that f(a) = f(b).

\(\boldsymbol{\Rightarrow }\) 4a = 4b

\(\boldsymbol{\Rightarrow }\) a = b

Hence, the function f is one- one.

For any of the real number (b) in the co- domain R, there must exist \(\frac{b}{3}\) in R so that, f(\(\frac{b}{3} = 3\left(\frac{b}{3} \right) = y \))

Hence, the function f is onto.

Therefore, the function f is one- one and also, it is onto.

**Thus, the correct answer is option (b).**

**EXERCISE – 1.3 **

** **

** Q-1: Let us consider, g : {3, 5, 6} → {2, 4, 7} and f : {2, 4, 7} → {3, 5} will begiven by g = {(3, 4), (5, 7), (6, 3)} and f = {(3, 5), (4, 5), (7, 3)}. Find fog. **

** **

**Sol: **

The given functions g : {3, 5, 6} → {2, 4, 7} and f : {2, 4, 7} → {3, 5} will be defined by g = {(3, 4), (5, 7), (6, 3)} and f = {(3, 5), (4, 5), (7, 3)}

fog(3) = f[g(3)] = f(4) = 5 [as, g(3) = 4 and f(4) = 5]

fog(5) = f[g(5)] = f(7) = 3 [as, g(5) = 7 and f(7) = 3]

fog(6) = f[g(6)] = f(3) = 5 [as, g(6) = 3 and f(3) = 5]

Hence,

**fog = {(3, 5), (5, 3), (6, 5)}**

**Q-2: Let us consider **

**f, g and h be the functions from R to R. Prove that:**

**(g + f)oh = goh + foh**

**(g.f)oh = (goh).(foh)**

** **

**Sol: **

We need to prove that,

(g + f)oh = foh + goh

\(\boldsymbol{\Rightarrow }\) {(g + f)oh}(x) = {goh + foh}(x)

Now,

LHS = {(g + f)oh}(x)

= (g + f)[h(x)]

= g[h(x)] + f[h(x)]

= (goh)(x) + (foh)(x)

= {(goh) + (foh)}{x)

**= RHS**

Therefore, {(g + f)oh} = goh + foh

Hence, proved.

We need to prove:

(g.f)oh = (goh).(foh)

LHS = [(g.f)oh](x)

= (g.f)[h(x)]

= g[h(x)]. f[h(x)]

= (goh)(x) . (foh)(x)

= {(goh).(foh)}(x)

= (goh).(foh)

**Hence, LHS = RHS**

**Therefore, (g.f)oh = (goh).(foh)**

**Q-3: Find fog and gof, if**

**(i) g(a) = |a| and f(a) = |5a – 2|**

**(ii) g(a) = 8a ^{3} and f(a) = \(x^{\frac{1}{3}}\) **

**Sol: **

**(i)** g(a) = |a| and f(a) = |5a – 2|

Therefore, the binary operation * is not associative.

fog(a) = f(g(a)) = f(|a|) = \(\left | 5\left | x \right | – 2 \right |\)

Therefore, gof(a) = g(f(a)) = g(|5x – 2|) = \(\left |\left | 5x – 2 \right | \right |\) = \( \left | 5x – 2 \right | \)

**(ii)** g(a) = 8a^{3 }and f(a) = a^{\(\frac{1}{3}\)}

Therefore, fog(a) = f(g(a)) = f(8a^{3}) = (8a^{3})^{\(\frac{1}{3}\) }= (2^{3} a^{3})^{ \(\frac{1}{3}\)} = 2a

Therefore, gof(a) = g(f(a)) = g(\(a^{\frac{1}{3}}\)) = 8(\(a^{\frac{1}{3}}\))^{3} = 8a

**Q-4: If g(a) = \(\frac{\left(4a + 3 \right)}{\left(6a – 4 \right)}\), a ≠ \(\frac{2}{3} \). Prove that gog(a) = a, for every a ≠ \(\frac{2}{3} \). What will be the inverse of the function g?**

**Sol: **

As per the data given in the question, we have

g(a) = \(\frac{4a + 3}{6a – 4}\), a ≠ \(\frac{2}{3} \)

(gog)(x) = g(g(x)) = g(\(\frac{4a + 3}{6a – 4}\)) = \(\frac{4\left (\frac{4a + 3}{6a – 4} \right) + 3}{6 \left (\frac{4a + 3}{6a – 4} \right) – 4}\)

= \(\frac{16a + 12 + 18a – 12}{24a + 18 – 24a + 16} \)

= \(\frac{34a}{34}\)

Therefore, gog(a) = a, for all of the a ≠ \(\frac{2}{3} \)

So, gog = I_{a}

Therefore, the function f given is invertible and the inverse of the function f is f itself.

**Q-5: Explain with suitable reason that, whether the following functions has any inverse**

**(i) g : {2, 3, 4, 5} → {11} with**

**g = {(2, 11), (3, 11), (4, 11), (5, 11)}**

**(ii) f : {6, 7, 8, 9} → {2, 3, 4, 5} with**

**f = {(6, 5), (7, 4), (8, 5), (9, 3)}**

**(iii) h : {3, 4, 5, 6} → {8, 10, 12, 14} with**

**h = {(3, 8), (4, 10), (5, 12), (6, 14)}**

** **

**Sol: **

**(i)** g : {2, 3, 4, 5} → {11} which is defined as g = {(2, 11), (3, 11), (4, 11), (5, 11)}

From the definition of f, given in the question, we note that, the function f is a many one function as,

g(2) = g(3) = g(4) = g(5) = 11

Hence, the function g is not one- one.

**Therefore, in this case, the function g doesn’t have any inverse.**

**(ii)** f : {6, 7, 8, 9} → {2, 3, 4, 5} which is defined as f = {(6, 5), (7, 4), (8, 5), (9, 3)}

From the definition of f, given in the question, we note that, the function f is a many one function as,

f(6) = f(7) = 4

Hence, the function f is not one- one.

**Therefore, in this case, the function f doesn’t have any inverse.**

**(iii)** h : {3, 4, 5, 6} → {8, 10, 12, 14} which is defined as

h = {(3, 8), (4, 10), (5, 12), (6, 14)}

Here, we can see that, all the distinct elements from the set {3, 4, 5, 6} have distinct images under h.

Hence, the function h is one- one.

Also, the function h is onto as each element y from the set {8, 10, 12, 14}, there exists an element, say a, in the set {3, 4, 5, 6} so that h(a) = b.

Hence, h is one- one as well as onto function.

**Therefore, the function h will have an inverse.**

**Q-6: Prove that: **

**g : [-2, 2] → R, **

**which is given by g(a) = \(\frac{a}{\left(a + 2 \right)}\) will be one- one. What will be the inverse of the function g : [-2, 2] → Range g.**

** **

**Sol: **

g : [-2, 2] → R which is given by

g(a) = \(\frac{a}{\left(a + 2 \right)}\)

For one- one:

Let, g(a) = g(b)

\(\boldsymbol{\Rightarrow }\) \(\frac{a}{\left(a + 2 \right)}\) = \(\frac{b}{\left(b + 2 \right)}\)

\(\boldsymbol{\Rightarrow }\) a(b + 2) = b (a + 2)

\(\boldsymbol{\Rightarrow }\) ab + 2a = ab + 2b

\(\boldsymbol{\Rightarrow }\) 2a = 2b

\(\boldsymbol{\Rightarrow }\) a = b

Hence, the function g is one- one function.

We can see that, g : [-2, 2] → Range g is onto.

Hence, g : [-2, 2] → Range g here is one- one and onto, and thus, the inverse of the function f : [-2, 2] → Range g exists.

Let us consider f : Range g → [-2, 2] be the inverse for g.

Let, b be any arbitrary element of the range f.

As, f : [-2, 2] → Range g is onto, here we have

b = g(a) for some values of a ɛ [-2, 2]

\(\boldsymbol{\Rightarrow }\) b = \(\frac{a}{\left(a + 2 \right)}\)

\(\boldsymbol{\Rightarrow }\) b(a + 2) = a

\(\boldsymbol{\Rightarrow }\) ab + 2b = a

\(\boldsymbol{\Rightarrow }\) a(1 – b) = 2b

\(\boldsymbol{\Rightarrow }\) a = \(\frac{2b}{\left(1 – b \right)}\), b ≠ 1

Then,

Let we define f: Range g → [-2, 2] as

f(b) = \(\frac{2b}{\left(1 – b \right)}\), b ≠ 1

Thus,

(fog)(a) = f(g(a)) = f\(\left(\frac{a}{a + 2} \right)\)= \(\frac{2\left (\frac{a}{a + 2} \right)}{1 – \left (\frac{a}{a + 2} \right)} = \frac{2a}{a + 2 – a} = \frac{2a}{2} = a\)

and,

(gof)(a) = g(f(a)) = g \(\left (\frac{2b}{1 – b} \right) = \frac{\left (\frac{2b}{1 – b} \right)}{\left (\frac{2b}{1 – b} \right) + 2} = \frac{2b}{2b + 2 – 2b} = \frac{2b}{2} = b\)

Hence,

fog = a = \(I_{\left \lfloor -1, 1 \; \right \rfloor}\) and gof = b = I_{Range }f

Thus,

g^{-1} = f

**g ^{-1}(b) = \( \frac{2b}{2}\), b ≠ 1.**

**Q-7: Let us consider g : R → R which is given by g(a) = 4a + 3. Prove that the function g is invertible. Also find the inverse of g.**

** **

**Sol: **

g : R → R which is given by,

g(a) = 4a + 3

For one- one function,

Let, g(a) = g(b)

\(\boldsymbol{\Rightarrow }\) 4a + 3 = 4b + 3

\(\boldsymbol{\Rightarrow }\) 4a = 4b

\(\boldsymbol{\Rightarrow }\) a = b

Hence, the function g is a one- one function.

For onto function,

For b ɛ R, let b = 4a + 3.

\(\boldsymbol{\Rightarrow }\) a = \(\frac{b – 3}{4}\) ɛ R.

Hence, for every b ɛ R, there does exist a = \(\frac{b – 3}{4}\) ɛ R, so that

g(a) = g \(\left (\frac{b – 3}{4} \right)\) = 4 \(\left (\frac{b – 3}{4} \right)\) + 3 = b

Hence, the function g is onto.

Therefore, the function g is one- one and onto and hence, g^{-1} will exist.

Let us consider, f: R → R by f(x) = \(\frac{b – 3}{4}\)

Then,

(fog)(a) = f(g(a)) = f(4a + 3) = \(\frac{\left (4a + 3 \right) – 3}{4} = \frac{4x}{4} = x\)

And,

(gof)(b) = g(f(x)) = g\(g\left (\frac{b – 3}{4} \right) = 4\left (\frac{b – 3}{4} \right) + 3 = b – 3 + 3 = b\)

So, fog = gof = I_{R}

**Therefore, the function f will be invertible and have inverse which is given by g ^{-1}(b) = f(b) = \(\frac{b – 3}{4}\)**

**Q-8: Let us consider a function, g : R _{+} → [4, **

**) which is given by g(a) = a**

^{2}+ 4. Prove that the function g is invertible with the inverse g^{-1}of the given function g by g^{-1 }(b) = \(\sqrt{b – 4}\), where R_{+ }will be the set of all the non- negative real number.** **

**Sol: **

g : R_{+ }→ [4, ∞) which is given by g(a) = a^{2} + 4

For one- one function,

Let, g(a) = g(b)

\(\boldsymbol{\Rightarrow }\) a^{2} + 4 = b^{2} + 4

\(\boldsymbol{\Rightarrow }\) a^{2} = b^{2 }

\(\boldsymbol{\Rightarrow }\) a = b

Hence, the function f is one- one function.

For onto function,

For b ɛ [4, ∞), let b = a^{2} + 4

\(\boldsymbol{\Rightarrow }\) a^{2} = b – 4 ≥ 0

\(\boldsymbol{\Rightarrow }\) a = \(\sqrt{b – 4}\) ≥ 0

Hence, for every b ɛ [4, ∞), there exists a = \(\sqrt{b – 4}\) ɛ R_{+}, so that

G(a) = g(\(\sqrt{b – 4}\)) = (\(\sqrt{b – 4}\))^{2 } + 4 = b – 4 + 4 = b

Hence, the function g is onto.

So, the function g is one- one and onto and hence, g^{-1} exists.

Let we define, f : [4, ∞) → R_{+ }by f(b) = \(\sqrt{b – 4}\)

Then,

(fog)(a) = f(g(a)) = f(\(x^{2} + 4\)) = \(\sqrt{\left (a^{2} + 4 \right) – 4} = \sqrt{a^{2}}\) = a

And,

(gof)(b) = g(f(b)) = g\(\left (\sqrt{b – 4} \right)\) = \(\left (\sqrt{b – 4} \right)^{2}\) + 4 = b – 4 + 4 = b

Hence, gof = fog = I_{R}

**Therefore, the function g is invertible and it will be the inverse of g which is given by g ^{-1}(b) = f(b) = \(\sqrt{b – 4}\)**

** **

** **

**Q-9: Let us consider a function, g: R _{+ }→ [-5, ∞) which is given by g(a) = 9a^{2} + 6a – 5. Prove that, the function g is invertible with g^{-1}(b) = \(\left (\frac{\left (\sqrt{b + 6} \right) – 1}{3} \right)\).**

** **

**Sol: **

g: R_{+ }→ [-5, ∞) which is given by g(a) = \(9a^{2} + 6a – 5 \)

Let, b be an arbitrary element of [-5, ∞)

Let,

b = \(9a^{2} + 6a – 5 \)

\(\boldsymbol{\Rightarrow }\) b = (3a + 1)^{2} – 1 – 5 = (3a + 1)^{2} – 6

\(\boldsymbol{\Rightarrow }\) b + 6 = (3a + 1)^{2}

\(\boldsymbol{\Rightarrow }\) 3a + 1 = \(\sqrt{b + 6}\) [as b ≥ -5 \(\boldsymbol{\Rightarrow }\) b + 6 > 0]

\(\boldsymbol{\Rightarrow }\) a = \(\frac{\left (\sqrt{b + 6} \right) – 1}{3}\)

Hence, the function g is onto, which means having range g = [-5, ∞)

Let us consider, f : [-5, ∞) → R_{+} as f(b) = \(\frac{\left (\sqrt{b + 6} \right) – 1}{3}\)

Then,

(fog)(a) = f(g(a)) = f(9a^{2} + 6a – 5) = g ((3a + 1)^{2} – 6) = \(\frac{\sqrt{\left (3a + 1 \right)^{2} – 6 + 6} – 1}{3}\)

=\(\frac{3a + 1 – 1}{3}\) = \(\frac{3a}{3}\) = a

And,

(gof)(b) = g(f(b)) = g\(\left (\frac{\sqrt{b + 6} – 1}{3} \right) = \left [3 \left (\frac{\sqrt{b + 6} – 1}{3} \right) + 1 \right]^{2} – 6\)

= \(\left (\sqrt{b + 6} \right)^{2} – 6\) = b + 6 – 6 = b

Hence, fog = a = I_{R }and gof = b = I_{Range }g

Therefore, g is invertible and the inverse of g will be given by:

g^{-1}(b) = f(b) = \(\left (\frac{\left (\sqrt{y + 6} \right)^{2} – 1}{3} \right)\)

**Q-10: Let us consider a function g : P → Q be an invertible function. Prove that the function g have a unique inverse. **

** **

**Sol: **

Let, the function, g : P → Q be an invertible function.

Let, function g have two inverses (say, g_{1} and g_{2})

Thus, for every b ɛ Q, we have,

gof_{1 }(b) = I_{Y}(b) = gof_{2}(b)

\(\boldsymbol{\Rightarrow }\) g(f_{1}(b)) = g(f_{2}(b))

\(\boldsymbol{\Rightarrow }\) f_{1}(b) = f_{2}(b) [as g is invertible \(\boldsymbol{\Rightarrow }\) g is on- one]

\(\boldsymbol{\Rightarrow }\) f_{1}= f_{2} [as f is one- one]

**Therefore, the function g has a unique inverse.**

** **

** **

**Q-11. Let us consider the function g: {2, 3, 4} → {x, y, z}, generally given by g(2) = x, g(3) = y, g(4) = z. Find g ^{-1} and prove that (g^{-1})^{-1} = g.**

** **

**Sol: **

The given function g : {2, 3, 4} → {x, y, z} which is given by g(2) = , g(3) = y, and g(4) = z

If we will define f: {x, y, z} → {2, 3, 4}, since f(x) = 2, f(y) = 3, f(z) = 4.

(gof)(x) = g(f(x)) = g(2) = x

(gof)(y) = g(f(y)) = g(3) = y

(gof)(z) = g(f(z)) = g(4) = z

And,

(fog)(2) = f(g(2)) = f(x) = 2

(fog)(3) = f(g(3)) = f(y) = 3

(fog)(4) = f(g(4)) = f(x) = 4

Thus,

fog = I_{X } and gof = I_{Y}, such that P = {2, 3, 4} and Q = {x, y, z}

Hence, the inverse of the function g exists and g^{-1} = f

Thus,

g^{-1} :{x, y, z} → {2, 3, 4} which is given by g^{-1}(x) = 2, g^{-1}(y) = 3 and g^{-1}(z) = 4

Let us now obtain the inverse for g^{-1}, i.e., finding the inverse of the function f.

Let us define h : {2, 3, 4} → {x, y, z} such that h(2) = x, h(3) = y and h(4) = z

Now,

(foh)(2) = f(h(2)) = f(x) = 2

(foh)(3) = f(h(3)) = f(x) = 3

(foh)(4) = f(h(4)) = f(x) = 4

And,

(hof)(x) = h(f(x)) = h(2) = x

(hof)(y) = h(f(y)) = h(3) = y

(hof)(z) = h(f(z)) = h(4) = z

Hence, foh = I_{X }and hof = I_{Y}, P = {2, 3, 4} and Q = {x, y, z}.

So, the inverse of the function of f exists and f^{-1} = h \(\boldsymbol{\Rightarrow }\) (g^{-1})^{-1} = h

We can see that, h = g

**Hence, (f ^{-1})^{-1} = f.**

**Q-12: Let us consider g : P → Q be an invertible function. Prove that the inverse of g ^{-1} is g, i.e., (g^{-1})^{-1} = g.**

** **

**Sol: **

As per the given condition,

g : P → Q is an invertible function.

Thus, there exists a function say, f : Q → P so that, fog = I_{P} and gof = I_{Q}

So, g^{-1} = f

Then, fog = I_{P} and gof = I_{Q }

\(\boldsymbol{\Rightarrow }\) g^{-1}og= I_{P} and fof^{-1} = I_{Q}

**Therefore, g ^{-1} : Q → P will be invertible and the function g is the inverse of g^{-1}, i.e., (g^{-1})^{-1} = g**

**Q-13: If g : R → R is given by g(a) = \(\left (3 – a^{3} \right)^{\frac{1}{3}}\), then gog(a) is**

**(a) \(\frac{1}{a^{3}}\) **

**(b) a ^{3}**

**(c) a **

**(d) (3 – a ^{3)}**

** **

**Sol: **

g : R → R which is given by g(a) = \(\left (3 – a^{3} \right)^{\frac{1}{3}}\)

Then,

gog(a) = g(g(a)) = g\(\left (\left (3 – a^{3} \right)^{\frac{1}{3}} \right)\) = \(\left [3 – \left (\left (3 – a^{3} \right)^{\frac{1}{3}} \right)^{3} \right]^{\frac{1}{3}}\)

= \(\left [3 – \left (3 – a^{3} \right) \right]^{\frac{1}{3}}\) = \(\left (a^{3} \right)^{\frac{1}{3}} = a \)

Hence, gog(a) = a

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

**Q-14: Let us consider a function g: R – \(\left \{-\frac{4}{3} \right \}\) → R as g(a) = \(\frac{4a}{3a + 4}\). The inverse of the function g is map f: Range g → R – \(\left \{-\frac{4}{3} \right \}\) is given by**

**(a) f(b) = \(\frac{3b}{3 – 4b}\) (b) f(b) = \(\frac{4b}{4 – 3b}\)**

**(c) f(b) = \(\frac{4b}{3 – 4b}\) (d) f(b) = \(\frac{3b}{4 – 3b}\)**

** **

**Sol: **

As per the data given in the question, we have

g : R – \(\left \{-\frac{4}{3} \right \}\) → R which is a function defined as g(a) = \(\frac{4a}{3a + 4}\).

Let, b be the arbitrary element having range g.

Thus, there exists a ɛ R – \(\left \{-\frac{4}{3} \right \}\) so that, b = f(a).

\(\boldsymbol{\Rightarrow }\) b = \(\frac{4a}{3a + 4}\)

\(\boldsymbol{\Rightarrow }\) b(3a + 4) = 4a

\(\boldsymbol{\Rightarrow }\) 3ab + 4b = 4a

\(\boldsymbol{\Rightarrow }\) a = \(\frac{4b}{4 – 3b}\)

Let, f : Range g → R – \(\left \{-\frac{4}{3} \right \}\) so that, f(b) = \(\frac{4b}{4 – 3b}\)

Thus,

fog(a) = f(g(a) = f\(\left (\frac{4a}{3a + 4} \right) = \frac{4\left (\frac{4a}{3a + 4} \right)}{4 – 3\left (\frac{4a}{3a + 4} \right)}\)

= \(\frac{16a}{12a + 16 – 12a} = \frac{16a}{16}\) = a

And,

gof(b) = g(f(b)) = g\(\left (\frac{4b}{4 – 3b} \right) = \frac{4\left (\frac{4b}{4 – 3b} \right)}{3\left (\frac{4b}{4 – 3b} \right) + 4}\)

= \(\frac{16b}{12b + 16 – 12b} = \frac{16b}{16} = b\)

Thus, fog = I_{R – \(\left \{- \frac{4}{3} \right \}\)} and gof = I_{Range f}

Therefore, the inverse of the function g is the map f: Range g → R – \(\left \{- \frac{4}{3} \right \}\), which will be given by

g(b) = \(\frac{4b}{4 – 3b}\)

**Hence, the correct answer is b.**

**EXERCISE – 1.4**

** **

**Q-1: Check whether each of the following definitions gives a binary operation or not.**

**In each of the event, that * will not be a binary operation, give explanation for this condition.**

**(i) On Z ^{+}, define * by x * y = x – y**

**(ii) On Z ^{+}, define * by x * y = xy**

**(iii) On R, define * by x * y = xy ^{2}**

**(iv) On Z ^{+}, define * by x * y = |x – y|**

**(v) On Z ^{+}, define * by x * y = x**

**Sol: **

**(i)** On Z^{+}, * will be defined by x * y = x – y

It’s not a binary operation,

Such that the image of (2, 3) under * is 2 * 3 = 2 – 3 = -1 \(\notin \; Z^{+}\)

**(ii)** On Z^{+}, * will be defined by x * y = xy.

We can see that for every x, y ɛ \(\; Z^{+}\), there will be a unique element xy in Z^{+ }.

It means that * carries every pair (x, y) from the unique element x * y = xy in Z^{+}.

**Hence, * is the binary operation.**

**(iii)** On R, * will be defined as x * y = xy^{2}.

We can see that for every x, y ɛ R, there will be a unique element xy^{2} in R.

It means that * will carry each pair (x, y) to a unique element x * y = xy^{2} in R.

**Hence, * is a binary operation.**

**(iv)** On Z^{+}, * will be defined as x * y =|x – y|.

We can see that for every x, y ɛ Z^{+}, there will be a unique element |x – y| in Z^{+}.

It means that * will carry each pair (x, y) to a unique element x * y = |x – y| in Z^{+}.

**Hence, * is a binary operation.**

**(v)** On Z^{+}, * will be defined as x * y = x.

We can see that for every x, y ɛ Z^{+}, there will be a unique element x in Z^{+}.

It means that * will carry each pair (x, y) to a unique element x * y = x in Z^{+}.

**Hence, * is a binary operation.**

**Q-2: For every binary operations * which is defined below, check whether * is associative or commutative in the following cases.**

**(i) On Z, explain x * y = x – y**

**(ii) On Q, explain x * y = xy + 1**

**(iii) On Q, explain x * y = \(\frac{xy}{2}\)**

**(iv) On Z ^{+}, explain x * y = 2^{xy}**

**(v) On Z ^{+}, explain x * y = x^{y}**

**(vi) On R – {-1}, explain x * y = \(\frac{x}{y + 1}\) **

** **

**Sol: **

**(i)** On Z, * will be defined as x * y = x – y

(2 * 3) * 4 = (2 – 3)* 4 = -1 * 4 = -1 – 4 = -5

2*(3 * 4) = 2 *(3 – 4) = 2 * -1 = 2 – (-1) = 3

Thus,

(2 * 3) * 4 ≠ 2*(3 * 4), where 2, 3, 4 ɛ Z.

Therefore, the given operation * is not associative.

Now, also

We can observe that 2 * 3 = 2 – 3 = -1 and 3 * 2 = 3 – 2 = 1

Since, 2 * 3 ≠ 3 * 2, where 2, 3 ɛ Z

**Therefore, the given operation * is not associative.**

** **

**(ii) On Q, * will be defined as x * y = xy + 1**

We can observe here that,

(2 * 3)* 4 = (2 × 3 + 1) * 3 = 7 * 3 = 7 × 3 + 1 = 22

2 *(3 * 4) = 2 * (3 × 4 + 1) = 2 * 13 = 2 × 13 + 1 = 27

Thus,

(2 * 3)* 4 ≠ 2 *(3 * 4), where 2, 3, 4 ɛ Q.

Therefore, the given operation * is not associative.

Now, also

We know that: xy = ya for every x, y ɛ Q.

\(\boldsymbol{\Rightarrow }\) xy + 1 = yx + 1 for every x, y ɛ Q

\(\boldsymbol{\Rightarrow }\) x * y = x * y for every x, y ɛ Q

**Therefore, the given operation * is commutative. **

(iii) **On Q, * will be defined by x * y = \(\frac{xy}{2}\)**

For every x, y, z ɛ Q, we have

(x * y)* z = \(\left (\frac{xy}{2} \right) * z = \frac{\left (\frac{xy}{2} \right)z}{2} = \frac{xyz}{4}\)

And,

x *(y * z) = \( x * \left (\frac{yz}{2} \right) = \frac{x \left (\frac{yz}{2} \right)}{2} = \frac{xyz}{4}\)

Thus,

(x * y)* z = x *(y * z), where x, y, z ɛ Q

Hence, the given operation * is associative.

We know that, xy = yx for every x, y ɛ Q

\(\boldsymbol{\Rightarrow }\) \(\frac{xy}{2} = \frac{yx}{2}\) for every x, y ɛ Q.

\(\boldsymbol{\Rightarrow }\) x * y = y * x for every x, y ɛ Q.

**Hence, the given operation * is commutative.**

**(iv)** **On Z ^{+}, * will be defined by x * y = 2^{xy}**

For every x, y, z ɛ Q, we have

(1 * 3)* 4 = 2^{1 × 3 } * 4 = 8 * 4 = 2^{8 × 4 }

And,

1 *(3 * 4) = 1 * 2^{3 × 4 }= 1 * 4096 = 2^{1 × 4096 }

Thus,

(1 * 3)* 4 ≠ 1 *(3 * 4), where x, y, z ɛ Q

Hence, the given operation * is not associative.

We know that, xy = yx for every x, y ɛ Q

\(\boldsymbol{\Rightarrow }\) 2^{xy} = 2^{yx} for every x, y ɛ Q.

\(\boldsymbol{\Rightarrow }\) x * y = y * x for every x, y ɛ Q.

**Hence, the given operation * is commutative.**

**(v)** **On Z ^{+}, * will be defined by x * y = x^{y}**

We can see that,

For every x, y, z ɛ Z^{+}, we have

(1 * 3)* 4 = 1^{3 }* 4 = 1 * 4 = 1^{ 4 }= 1

And,

1 *(3 * 4) = 1 * 3^{ 4 }= 1 * 81 = 1^{81} = 1

^{ }Thus,

(1 * 3)* 4 = 1 *(3 * 4), where x, y, z ɛ Z^{+}

Hence, the given operation * is associative.

We can also see that,

2 * 3 = 2^{3} = 8 and 3 * 2 = 3^{2} = 9

Hence, 2 * 3 ≠ 3 * 2, where x, y, z ɛ Z^{+}

**Therefore, the given operation * is not commutative.**

**(vi)** **On R, * – {-1} which is defined as x * y = \(\frac{x}{y + 1}\)**

We can see that

(2 * 3)* 4 = \(\left (\frac{2}{3 + 1} \right) * 4 = \left (\frac{2}{4} \right) * 4 = \frac{\frac{1}{2}}{4 + 1} = \frac{1}{2 \times 4} = \frac{1}{8}\)

And,

2 *(3 * 4) = \(2 * \left (\frac{3}{4 + 1} \right) = 2 * \left (\frac{3}{5} \right) = \frac{2}{\frac{3}{5} + 1} = \frac{1}{\frac{3 + 5}{5}} = \frac{5}{8}\)

Thus,

(2 * 3)* 4 ≠ 2 *(3 * 4), where 2, 3, 4 ɛ R – {-1}

**Hence, the given operation * is not associative.**

**Q-3: Let us consider a binary operation ^ for the set {2, 3, 4, 5, 6} which will be defined as x ^ y = min {x, y}.**

**Draw the operational table for the operation ^.**

** **

**Sol: **

As per the data given in the question, we have

A binary operation ^ for the set {2, 3, 4, 5, 6} which will be defined as x ^ y = min {x, y}.

Then,

x, y ɛ {2, 3, 4, 5, 6}

**Hence, the operational table for the operation ^ given in the question will be given as below:**

^ |
2 |
3 |
4 |
5 |
6 |

2 |
2 | 2 | 2 | 2 | 2 |

3 |
2 | 3 | 3 | 3 | 3 |

4 |
2 | 3 | 4 | 4 | 4 |

5 |
2 | 3 | 4 | 5 | 5 |

6 |
2 | 3 | 4 | 5 | 6 |

** **

** **

**Q-4: Let us consider a binary operation * for the set {2, 3, 4, 5, 6} which is given by the multiplication table given below:**

**(i) Find the value of (3 * 4) * 5 and 3 *(4 * 5).**

**(ii) Check whether * is commutative.**

**(iii) Find the value of (3 * 4)*(5 * 6)**

* |
2 |
3 |
4 |
5 |
6 |

2 |
2 | 2 | 2 | 2 | 2 |

3 |
2 | 3 | 2 | 2 | 3 |

4 |
2 | 2 | 4 | 2 | 2 |

5 |
2 | 2 | 2 | 5 | 5 |

6 |
2 | 3 | 2 | 5 | 6 |

** **

**Sol: **

**(i)** (3 * 4)* 5 = 2 * 5 **= 2**

3 *(4 * 5) = 3 * 2 = 2

**(ii)** For each x, y ɛ {2, 3, 4, 5, 6}, we have x * y = y * x

**Hence, the given operation is commutative.**

**(iii)** (3 * 4)*(5 * 6) = 2 * 5 **= 2**

** **

** **

**Q-5: Let us consider *’ be a binary operation for the set {2, 3, 4, 5, 6} which will be defined by x *’ y = HCF of x * y. Check whether the operation *’ is the same as the operation * which is defined and used previously? Also, do justify your answer.**

** **

**Sol: **

A binary operation for the set {2, 3, 4, 5, 6} which will be defined by x * y = HCF of x and y.

The operational table for the operation *’ is given as the table below:

*’ |
2 |
3 |
4 |
5 |
6 |

2 |
2 | 2 | 2 | 2 | 2 |

3 |
2 | 3 | 2 | 2 | 3 |

4 |
2 | 2 | 4 | 2 | 2 |

5 |
2 | 2 | 2 | 5 | 5 |

6 |
2 | 3 | 2 | 5 | 6 |

Here, we will observe that the operational table for the operation * and *’ is the same.

**Hence, the operation *’ is the same as * operation.**

**Q-6: Let us consider a binary operation * on N which is given by x * y = LCM of x and y. Find the following:**

**(i) 6 * 7, 21 * 18 (ii) Is the operation * commutative?**

**(iii) Is the operation * associative? (iv) Obtain the identity of * in N.**

**(v) Which of the element of N are invertible for the operation *?**

** **

**Sol: **

As per the data given in the question, we have

A binary operation * on N which is given by x * y = LCM of x and y.

**(i)** 6 * 7 = LCM of 6 and 7 = 42

**21 * 18 = LCM of 21 and 18 = 126 **

**(ii)** We know that,

LCM of x and y = LCM of y and x for every x, y ɛ N.

Hence, x * y = y * x

**Therefore, the given operation * is commutative.**

** **

**(iii)** For x, y, z ɛ N, we have

(x * y) * z = (LCM of x and y) * z = LCM of x, y and z

x * (y * z) = x * (LCM of y and z) = LCM of x, y and z

Hence,

(x * y) * z = x * (y * z)

**Therefore, the given operation * is associative.**

**(iv)** We know that:

LCM of x and 1 = LCM of 1 and x = x for all x ɛ N

\(\boldsymbol{\Rightarrow }\) x * 1 = x = 1 * x for all x ɛ N

**Therefore, 1 is an identity of * in N.**

**(v)** The element x in N will be invertible with respect to the operation *, if there will exist an element y in N, so that x * y = e = y * x

Here, e = 1

This means,

LCM of x and y = 1 = LCM of y and x

This case will only be possible when x and y is equal to 1.

**Therefore, 1 is the only element which is invertible to N with respect to the operation ***

**Q-7: Check whether the operation * is defined for the set {1, 2, 3, 4, 5} by x * y = LCM of x and y, is a binary operation? Also, do justify your answer.**

** **

**Sol: **

The operation * for the set R = {1, 2, 3, 4, 5} which is defines as x * y = LCM of x and y.

Thus,

The operational table for the given operation * will be given by:

* |
1 |
2 |
3 |
4 |
5 |

1 |
1 | 2 | 3 | 4 | 5 |

2 |
2 | 4 | 6 | 8 | 10 |

3 |
3 | 6 | 9 | 12 | 15 |

4 |
4 | 8 | 12 | 16 | 20 |

5 |
5 | 10 | 15 | 20 | 25 |

We can observe from the table defined above that,

2 * 3 = 3 * 2 = 6 \(\notin \) R,

2 * 5 = 5 * 2 = 10 \(\notin \) R,

3 * 4 = 4 * 3 = 12 \(\notin \) R,

3 * 5 = 5 * 3 = 15 \(\notin \) R,

4 * 5 = 5 * 4 = 20 \(\notin \) R

**Therefore, the operation * is not a binary operation.**

** **

** **

**Q- 8: Let us consider a binary operation * on N which is defined by x * y = HCF of x and y. Check whether * is commutative? Check whether * is associative? Check whether there exists any identity for the binary operation * on N?**

** **

**Sol: **

A binary operation * on N which is defined by x * y = HCF of x and y

We know that,

HCF of x and y = HCF of y and x for every x, y ɛ N

Thus,

x * y = y * x

Hence, the binary operation * is commutative.

For x, y, z \(\epsilon \), we have

(x * y)* z = (HCF of x and y)* z = HCF of x, y and z

x *(y * z) = x *(HCF of y and z) = HCF of x, y and z.

Hence, (x * y)* z = x *(y * z)

Thus, the binary operation * is associative.

Then,

An element e ɛ N is the identity for the binary operation * if, x * e = x = e * x for every x ɛ N.

But, this is not true for any x ɛ N.

**Therefore, the given binary operation * don’t have any identity in N.**

**Q-9: Let us consider a binary operation * for the set Q of the rational numbers for the following:**

**(i) x * y = x – y (ii) x * y = x ^{2 }+ y^{2}**

**(iii) x * y = x + xy (iv) x * y = (x – y) ^{2}**

**(v) x * y = \(\frac{xy}{4}\) (vi) x * y = xy ^{2}**

**Determine which of the binary operations are commutative and which are associative.**

** **

**Sol: **

**(i)** On Q, the binary operation * is defined by x * y = x – y. It will be observed that:

And,

\(\frac{1}{4} * \frac{1}{3} = \frac{1}{4} – \frac{1}{3} = \frac{3 – 4}{12} = -\frac{1}{12}\)Therefore, \(\frac{1}{3} * \frac{1}{4} \neq \frac{1}{4} * \frac{1}{3}\), where, \(\frac{1}{3}, \; \frac{1}{4}\) ɛ Q.

Hence, the binary operation * is not commutative.

It will be observed that,

\(\left (\frac{1}{3} * \frac{1}{4} \right) * \frac{1}{5} = \left (\frac{1}{3} – \frac{1}{4} \right) * \frac{1}{5} = \left (\frac{4 – 3}{12} \right) * \frac{1}{5} = \frac{1}{12} * \frac{1}{5} = \frac{1}{12} – \frac{1}{5} = \frac{5 – 12}{60} = \frac{-7}{60}\)And,

\(\frac{1}{3} * \left (\frac{1}{4} * \frac{1}{5} \right) = \frac{1}{3} * \left (\frac{1}{4} – \frac{1}{5} \right) = \frac{1}{3} * \left (\frac{5 – 4}{20} \right) = \frac{1}{3} * \frac{1}{20} = \frac{20 – 3}{60} = \frac{17}{20}\)Thus,

\(\left (\frac{1}{3} * \frac{1}{4} \right) * \frac{1}{5} = \frac{1}{3} * \left (\frac{1}{4} * \frac{1}{5} \right)\), where \(\frac{1}{3}, \frac{1}{4} \; and \; \frac{1}{4} \; \epsilon \; Q\)

**Therefore, the binary operation * is not associative. **

**(ii)** On Q, the binary operation * is defined as x * y = x^{2} + y^{2}.

For x, y ɛ Q, we have

x * y = x^{2} + y^{2} = y^{2} + x^{2} = y * x

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

Hence, the binary operation * is commutative.

We can observe that,

(2 * 3)* 4 = (2^{2} + 3^{2}) * 4 = (4 + 9)* 4 = 13 * 4 = 13^{2} + 4^{2} = 185,

And,

2 * (3 * 4) = 2 * (3^{2} + 4^{2}) = 2 * (3 + 4) = 2 * 7 = 2^{2} + 7^{2 }= 53

**Thus, (2 * 3)* 4 ≠ 2 * (3 * 4), where 2, 3, 4 ɛ Q.**

**(iii)** On Q, the binary operation * which is defined by x * y = x + xy

It will be observed that,

2 * 3 = 2 + 2 × 3 = 2 + 6 = 8

3 * 2 = 3 + 3 × 2 = 3 + 6 = 9

Thus, the binary operation * is not commutative.

It will be observed that,

(2 * 3)* 4 = (2 + 2 × 3)* 4 = (2 + 6)* 4 = 8 * 4 = 8 + 8 × 4 = 40

And,

2 * (3 * 4) = 2 * (3 + 3 × 4) = 2 * (3 + 12) = 2 * 15 = 2 + 2 × 15 = 32

Thus, (2 * 3)* 4 ≠ 2 * (3 * 4), where 2, 3, 4 ɛ Q.

**Therefore, the binary operation * is not associative.**

**(iv)** On Q, the binary operation * will be defined by x * y = (x – y)^{2}

For x, y ɛ Q, we have

x * y = (x – y)^{2}

y * x = (y – a)^{2} = [-(x – y)]^{2} = (x – y)^{2}

Hence, the binary operation * is commutative.

We can also observe that,

(2 * 3)* 4 = (2 – 3)^{2 }* 4 = 1 * 4 = (1 – 4)^{2} = (-3)^{2} = 9

And,

2 * (3 * 4) = 2 * (3 – 4)^{2 }= 2 * 1 = (2 – 1)^{2} = (-1)^{2} = 1

Thus, (2 * 3)* 4 ≠ 2 * (3 * 4), where 2, 3, 4 ɛ Q.

**Therefore, the binary operation * is not associative.**

**(v)** On Q, the binary operation * will be defined as x * y = \(\frac{xy}{4}\)

For x, y ɛ Q, we will get

x * y = \(\frac{xy}{4}\) = \(\frac{yx}{4}\) = y * x

Hence, x * y = y * x

Therefore, the binary operation * is commutative.

For x, y, z \(\epsilon \)Q, we will get

(x * y) * z = \( \left (\frac{xy}{4} \right) * z = \frac{\left (\frac{xy}{4} \right). z}{4} = \frac{xyz}{16}\)

And,

x * (y * z) = \( x * \left (\frac{yz}{4} \right) = \frac{x * \left (\frac{yz}{4} \right)}{4} = \frac{xyz}{16}\)

Hence, (x * y) * z = x * (y * z), where x, y, z \(\epsilon \)

**Therefore, the binary operation * is associative.**

** **

**(vi)** On Q, the binary operation * will be defined as x * y = xy^{2}.

We can observe that,

\(\frac{1}{3} * \frac{1}{4} = \frac{1}{3}.\left (\frac{1}{4} \right)^{2} = \frac{1}{3} . \frac{1}{16} = \frac{1}{48}\)And,

\(\frac{1}{4} * \frac{1}{3} = \frac{1}{4}.\left (\frac{1}{3} \right)^{2} = \frac{1}{4} . \frac{1}{9} = \frac{1}{36}\)Hence, \(\frac{1}{3} * \frac{1}{4}\) ≠ \(\frac{1}{4} * \frac{1}{3}\), where \(\frac{1}{3} \; and \; \frac{1}{4} \) ɛ Q.

Therefore, the binary operation * is not commutative.

We can also observe that:

\(\left (\frac{1}{3} * \frac{1}{4} \right) * \frac{1}{5} = \left [\frac{1}{3}\left (\frac{1}{4} \right)^{2} \right] * \frac{1}{5} = \frac{1}{48} * \frac{1}{5} = \frac{1}{48} * \left (\frac{1}{5} \right)^{2} = \frac{1}{48 \times 25} = \frac{1}{1200}\)And,

\(\frac{1}{3} * \left (\frac{1}{4} * \frac{1}{5} \right) = \frac{1}{3}\left [\frac{1}{4}\left (\frac{1}{5} \right)^{2} \right] = \frac{1}{3} * \frac{1}{100} = \frac{1}{3} * \left (\frac{1}{100} \right)^{2} = \frac{1}{3 \times 10000} = \frac{1}{30000}\)Hence, \(\left (\frac{1}{3} * \frac{1}{4} \right) * \frac{1}{5} ≠ \frac{1}{3} * \left (\frac{1}{4} * \frac{1}{5} \right)\), where \(\frac{1}{3}, \frac{1}{4}, \frac{1}{5} \; \epsilon \; Q \)

**Therefore, the binary operation * is not associative.**

**Thus, the operations which are defined in (ii), (iv), (v) are commutative and the operation defined in (v) is only associative.**

** **

** **

**Q-10: Check the identity of the operations which are equated above.**

**Sol: **

Let us consider an element e ɛ Q which can be the identity element for the operation * if,

x * e = x = e * x, for every x ɛ Q.

Here, among each of the six operations, there is no such element e ɛ Q, which satisfies the above condition.

**Hence, none of the six operations evaluated above have identity.**

**Q-11: Let us consider M = N × N and * be the binary operation on M which can be defined by**

**(p, q)* (r, s) = (p + r, q + s).**

**Prove that the operation * is commutative and associative. If there is any possibility of having an identity element, then what will be the identity element for the operation * on M?**

**Sol: **

M = N × N and * be the binary operation on M which can be defined by

(p, q)* (r, s) = (p + r, q + s)

Let, (p, q), (r, s) ɛ M

Then,

p, q, r, s ɛ N

Here, we have

(p, q)*(r, s) = (p + r, q + s)

(r, s)*(p, q) = (r + p, s + q) = (p + r, q + s)

**[Addition will be commutative in the set of the natural numbers]**

Thus, (p, q)*(r, s) = (r, s)*(p, q)

Hence, the operation * is commutative here.

Let, (p, q), (r, s), (t, u) ɛ M

Then, p, q, r, s, t, u ɛ N

Now,

[(p, q)*(r, s)] * (t, u) = (p + r, q + s)*(t, u) = (p + r + t, q + s + u)

And,

(p, q) * [(r, s) * (t, u)] = (p, q)*(r + t, s + u) = (p + r + t, q + s + u)

Thus, [(p, q)*(r, s)] * (t, u) = (p, q) * [(r, s) * (t, u)]

Hence, the operation * is associative.

Let us consider an element e = (e_{1}, e_{2}) ɛ M which can be an identity element for * operation, if

x * e = x = e * x, for every x = (x_{1}, x_{2}) ɛ M

Then,

(x_{1 }+ e_{1}, x_{2 }+ e_{2}) = (x_{1}, x_{2}) = (e_{1 }+ x_{1}, e_{2 }+ x_{2}),

This cannot be true for every element in M.

**Hence, the operation * will not have any identity element.**

**Q-12: Check whether the statements given below are true or false. Also, justify your answer.**

**(i) For any arbitrary binary operation * for the set N, x * x = x for every x ɛ N.**

**(ii) If the operation * is a commutative binary operation for N, then x *(y * z) = (z * y)* x.**

** **

**Sol: **

**(i)** Explain a binary operation * on N as x * y = x + y for every x ɛ N.

Thus, for y = x = 4, we have

4 * 4 = 4 + 4 = 8 ≠ 3.

Hence, the statement (i) is false.

**(ii)** RHS = (z * y)* x

= (y * z)* x **[* is commutative]**

= x * (y * z) ** [Again, as * is commutative]**

**= LHS**

Therefore, x *(y * z) = (z * y)* x

**Hence, statement (ii) is true.**

**Q-13: Let us consider a binary operation * for N defined as x * y = x ^{3} + y^{3}. Select the correct answer.**

**(a) Is * commutative but, not associative?**

**(b) Is * both commutative and associative?**

**(c) Is * neither commutative nor associative?**

**(d) Is * associative but, not commutative?**

** **

**Sol: **

On N, the binary operation * will be defined as x * y = x^{3} + y^{3}.

For x, y ɛ N, we have

x * y = x^{3} + y^{3} = y^{3} + x^{3 }= y * x **[Addition is commutative in N]**

Hence, the binary operation * is commutative.

We can observe that,

(2 * 3)* 4 = (2^{3} + 3^{3})* 4 = (8 + 27)* 4 = 35 * 4 = 35^{3} + 4^{4} = 42875 + 16 = 42891

And,

2 *(3 * 4) = 2 *(3^{3} + 4^{3}) = 2 *(27 + 64) = 2 * 91 = 2^{3} + 91^{3} = 8 + 753571 = 753579

(2 * 3)* 4 ≠ 2 *(3 * 4), where 2, 3, 4 ɛ N

Thus, the operation * is not associative.

Therefore, the operation * is commutative but, not associative.

**Hence, the correct statement is a.**