Ncert Solutions For Class 12 Maths Ex 1.1

Ncert Solutions For Class 12 Maths Chapter 1 Ex 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 ≤ y2} which is neither reflexive, nor transitive, nor symmetric.

Sol:

M = {(x, y): x ≤ y2}

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 < 42

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 < 22 = 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 ≤ y3} 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 < 23 = 8

But, (2, 1) \(\notin\) M (as 23 > 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): x3 ≥ y3}

(x, x) ɛ M since, x3 = x3

Hence, M is reflexive.

(3, 2) ɛ M since, 33 ≥ 23

But, (2, 3) \(\notin \) M since, 23 ≤ 33

Hence, M is not symmetric.

Let, (x, y), (y, z) ɛ M

\(\boldsymbol{\Rightarrow }\) x3 ≥ y3 and y3 ≥ z3

\(\boldsymbol{\Rightarrow }\) x3 ≥ z3

\(\boldsymbol{\Rightarrow }\) (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 = {(P1, P2): P1 is similar to P2}, is an equivalence relation. Assume three right angled triangles, say, triangle P1 having sides 4, 5, 6, triangle P2 having sides 6, 14, 15 and triangle P3 having sides 7, 9, 11. Find which triangle among P1, P2 and P3 will be related?

 

Sol:

M = {(P1, P2): P1 is similar to P2}

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

Now,

If (P1, P2) ɛ A, so P1 is completely similar to P2.

\(\boldsymbol{\Rightarrow }\) P2 is similar to P1

\(\boldsymbol{\Rightarrow }\) (P2, P1) ɛ A

Hence, M is symmetric.

Let, (P1, P2), (P2, P3) ɛ A

\(\boldsymbol{\Rightarrow }\) P1 is similar to P2 and also, P2 is similar to P3

\(\boldsymbol{\Rightarrow }\) P1 is similar to P3

\(\boldsymbol{\Rightarrow }\) (P1, P3) 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 P1 and P3 is in the same proportion (ratio).

Thus, the triangle P1 will be similar to the triangle P3.

Therefore, the triangle P1 is inter- related to the triangle P3

 

 

Q-13: Prove that the relation M is defined in the set S of every polygon in such a way that M = {(R1, R2): R1 and R2 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 = {(R1, R2): R1 and R2 must have the equal number of sides}

M will be reflexive in this case, as (R1, R2) ɛ M, since a same polygon has same number of the sides among itself.

Let, (R1, R2) ɛ M

\(\boldsymbol{\Rightarrow }\) R1 and R2 has same number of its sides.

\(\boldsymbol{\Rightarrow }\) R2 and R1 has same number of its sides.

\(\boldsymbol{\Rightarrow }\) (R2, R1) ɛ M

Hence, M is symmetric.

Let, (R1, R2), (R2, R3) ɛ M

\(\boldsymbol{\Rightarrow }\) R1 and R2 has same number of its sides.

Also, R2 and R3 have same number of its sides.

\(\boldsymbol{\Rightarrow }\) R1 and R3 has same number of its sides.

\(\boldsymbol{\Rightarrow }\) (R1, R3) 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 = {(P1, P2): P1 is parallel to P2}. 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 = {(P1, P2): P1 is parallel to P2}

M will be reflexive as any line can always be at least parallel to itself, i.e., (P1, P1) ɛ M.

Let, (P1, P2) ɛ M

\(\boldsymbol{\Rightarrow }\) P1 is parallel to P2 and also, P2 will be parallel to P1

\(\boldsymbol{\Rightarrow }\) (P2, P1) ɛ M

Hence, M is symmetric.

Let, (P1, P2), (P2, P3) ɛ M

\(\boldsymbol{\Rightarrow }\) P1 is parallel to P2 and also, P2 will be parallel to P3

\(\boldsymbol{\Rightarrow }\) P1 is parallel to P3

\(\boldsymbol{\Rightarrow }\) (P2, P1) 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.