Ncert Solutions Maths class 12th

The given relation to set A of all triangles is defined as

R= $\{\left(T_1,T_2\right):T_1$ is similar to $T_2\}$

For $T_1\in A$ ,

$T_1$ is always similar to $T_1$

So, $\left(T_1,T_1\right)\in R$ . Hence R is reflexive.

For $T_{1,}{T}_2\in A$ and $\left(T_1,T_2\right)\in R$ we have

$T_1\sim T_2\left(similar\right)$

$T_2\sim T_1$ i.e., $\left(T_2,T_1\right)\in R$

so, R is symmetric.

for, $T_1,T_2,T_3\in A$ and $\left(T_1,T_2\right)\in R$ and $\left(T_2,T_3\right)\in R$

$T_1\sim T_2$ and $T_2\sim T_3$

i.e., $T_1\sim T_3$ $\to \left(T_1,T_3\right)\in R$

so, R is transitive

$\therefore$ R is an equivalence relation.

Given, sides of $T_1$ are 3,4,5

Sides of $T_2$ are 5,12,13

Sides of $T_3$ are 6,8,10

As $\frac{3}{5}\ne \frac{4}{12}\ne \frac{5}{13}$ we conclude that $T_1$ is not similar to $T_2$

As $\frac{5}{6}\ne \frac{12}{8}\ne \frac{13}{10}$ we conclude that $T_2$ is not similar to $T_3$

But as $\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{1}{2}$ we conclude that $T_1\sim T_3$

Hence, $T_1$ is related to $T_3$

We have,

A= $\{x\in 2,0\le x\le 12\}$

The relation in set A is defined by

R= { $\left(a,b\right):\left|a-b\right|$ is a multiple of 4}

For all $a\in A$ ,

$\left|a-a\right|=0$ is a multiple of 4

So, $\left(a,a\right)\in R$ i.e., R is reflexive

For $a,b\in A\&\left(a,b\right)\in R$ we have,

$\left|a-b\right|$ is multiple of 4

$\left|-\left(b-a\right)\right|$ is multiple of 4

$\left|b-a\right|$ is multiple of 4

So, $\left(b,a\right)\in R$

i.e., R is symmetric

for $a,b,c\in A$ $\&\left(a,b\right)\in R\&\left(b,c\right)\in R$

$\left|a-b\right|$ & $\left|b-c\right|$ is a multiple of 4

So $\left|a-b\right|+\left|b-c\right|$ is also a multiple of 4

$\left|a-b+b-c\right|$ is a multiple of 4

$\left|a-c\right|$ is a multiple of 4

So, $\left(a,c\right)\in R$

i.e., R is transitive

Hence, R is an equivalence relation.

Finding all set of elements related to 1

For $a\in A$

Then, $\left(a,1\right)\in R$ i.e., $\left|a-1\right|$ is a multiple of 4

So, a can be 0 ≤ a ≤ 12

Only,

$\left|1-1\right|=0$

$\left|5-1\right|=4$ is a multiple of 4

$\left|9-1\right|=8$

Hence, sets of elements related to A= $\{1,5,9\}$

The relation in set A is defined as R= $\{\left(a,b\right):a=b\}$

For all $a\in A$ ,

a=a is true

so, $\left(a,a\right)\in R$

i.e., R is reflexive

for $a,b\in A\&\left(a,b\right)\in R$ we have,

a=b

b=a i.e., $\left(b,a\right)\in R$

so, R is symmetric

for $a,b,c\in A$ , $\left(a,b\right)\in R$ and $\left(b,c\right)\in R$ . We have,

a=b and b=c

a=c i.e., $\left(a,c\right)\in R$

$\therefore$ R is transitive

Hence, R is an equivalence relation

Find all sets of elements related to 1

For $a\in A$ , we need $\left(a,1\right)\in R$

i.e., a=1

so, set of elements related to 1 in A is {1}.

We have,

R= $\{\left(a,b\right)\therefore \left|a-b\right|$ is even $\}$ is a relation in set A= $\{1,2,3,4,5\}$

For all $a\in A$ , $\left|a-a\right|=0$ is even.

So, $\left(a,a\right)\in R$ . Hence R is reflexive

For $a,b\in A$ and $\left(a,b\right)\in R$

$\left|a-b\right|$ is even

$\left|-b+a\right|$ is even $|$

$\left|-\left(b-a\right)\right|$ is even

$\left|b-a\right|$ is even

i.e., $\left(b,a\right)\in R$

Hence, R is symmetric.

For $a,b,c\in A$ and $\left(a,b\right)\in R$ and $\left(b,c\right)\in R$

We have $\left|a-b\right|$ is even

and $\left|b-c\right|$ is even

then, $\left|a-b\right|+\left|b-c\right|$ is even as even + even=even

$\left|a-b+b-c\right|$ is even

$\left|a-c\right|$ is even

$\therefore$ $\left(a,c\right)\in R$

So, R is transitive.

$\therefore$ R is an equivalence relation

All elements of [1,3,5] are odd positive numbers and its subset are odd and their difference given an even number. Hence, they are related to each other.

Similarly, all elements of [2,4] are even positive numbers and its subset are even and their difference gives an even number. Hence, they are related to each other.

However, elements of [1,3,5] are related to elements of [2,4] since the difference of the two subsets is never even.