Relations and Functions

Let X={0, 1, 2, 3, 4, 5}.

The operation* on X is defined as:

$a*b=\{\begin{array}{cc}\hfill a+b\hfill & \hfill if,a+b<6\hfill \\ \hfill a+b-6\hfill & \hfill if,a+b\ge 6\hfill \end{array}$

An element $e\in X$ is the identity element for the operation*, if $a*e=a=e*a\forall a\in X$

For $a\in X$ we observed that

$\begin{array}{l}a*0=a+0=a\left[a\in X\Rightarrow a+0<6\right]\\ 0*a=0+a=a\left[a\in X\Rightarrow 0+a<6\right]\\ \therefore a*0=0*a\forall a\in X\end{array}$

Thus, 0 is the identity element for the given operation*.

An element $a\in X$ is invertible if there exists $b\in X$ such that $a*0=0*a.$

$ie\{\begin{array}{cc}\hfill a+b=0=b+a\hfill & \hfill if,a+b<6\hfill \\ \hfill a+6-6=0=b+a-6\hfill & \hfill if,a+b\ge 6\hfill \end{array}$

i.e.,

$a=-b,or,b=6-a$

But, X={0, 1, 2, 3, 4, 5} and $a,b\in X$ . Then, $a\ne -b$ .

$\therefore b=6-a$ is the inverse of $a\&mnForE;a\in X.$

Hence, the inverse of an element $a\in X,a\ne 0$ is 6-a i.e., $a^{-1}=6-a.$

It is given that ∗: P (X) * P (X) → P (X) be defined as

 A * B = (A – B) ∪ (B – A), A, B ∈ P (X).

Now, let A? P (X). Then, we get,

A *? = (A –? ) ∪ (? –A) = A∪? = A

? * A = (? - A) ∪ (A -? ) =? ∪A = A

A *? = A =? * A,     A? P (X)

Therefore? is the identity element for the given operation *.

Now, an element A? P (X) will be invertible if there exists B? P (X) such that

A * B =? = B * A. (as? is an identity element.)

Now, we can see that A * A = (A –A) ∪ (A – A) =? ∪? =? A? P (X).

Therefore, all the element A of P (X) are invertible with A-1 = A. 

It is given that*: $R*R\to$ and $o:R*R\to R$ is defined as

$a*b=\left|a-b\right|and,aob=a,\&mnForE;a,b\in R.$

For $a,b\in R$ , we have:

$\begin{array}{l}a*b=\left|a-b\right|\\ b*a=\left|b-a\right|=\left|-\left(a-b\right)\right|=\left|a-b\right|\\ \therefore a*b=b*a\end{array}$

$\therefore$ The operation* is commutative.

It can be observed that,

$\begin{array}{l}\left(1.2\right).3=\left(\left|1-2\right|\right).3=1.3=\left|1-3\right|=2\\ 1*\left(2*3\right)=1*\left(\left|2-3\right|\right)=1*1=\left|1-1\right|=0\\ \therefore \left(1*2\right)*3=1*\left(2*3\right)\left(where,1,2,3\in R\right)\end{array}$

$\therefore$ The operation* is not associative.

Now, consider the operation o:

It can be observed that $1o2=1,and,2o1=2.$

$\therefore 1o2\ne 2o1\left(where,1,2\in R\right)$

$\therefore$ The operation o is not commutative.

Let, $a,b,c\in R$ . Then we have:

$\begin{array}{l}\left(aob\right)oc=aoc=a\\ ao\left(boc\right)=aob=a\\ \Rightarrow \left(aob\right)oc=ao\left(boc\right)\end{array}$

$\therefore$ The operation o is associative.

Now, $a,b,c\in R$ . Then we have:

$\begin{array}{l}a*\left(boc\right)=a*b=\left|a-b\right|\\ \left(a*b\right)o\left(a*c\right)=\left(\left|a-b\right|\right)o\left(\left|a-c\right|\right)=\left|a-b\right|\\ Hence,a*\left(boc\right)=\left(a*b\right)o\left(a*c\right).\\ Now,\\ 1o\left(2o3\right)=1o\left(\left|2-3\right|\right)=1o1=1\\ \left(1o2\right)*\left(1o3\right)=1*1=\left|1-1\right|=0\\ \therefore 1o\left(2o3\right)\ne \left(1o2\right)*\left(1o3\right)\left(where,1,2,3\in R\right)\end{array}$

$\therefore$ The operation o does not distribute over*.

S = {a, b, c}, T = {1, 2, 3}

(i) F: S → T is defined as:

F = { (a, 3), (b, 2), (c, 1)}

⇒ F (a) = 3, F (b) = 2, F (c) = 1 

Therefore, F⁻¹ : T → S is given by

F⁻¹  = { (3, a), (2, b), (1, c)}.

(ii) F: S → T is defined as:

F = { (a, 2), (b, 1), (c, 1)}

Since F (b) = F (c) = 1, F is not one-one.

Hence, F is not invertible i.e., F⁻¹  does not exist.

Onto functions from the set {1, 2, 3, …, n} to itself is simply a permutation on n symbols 1, 2, …, n.

Thus, the total number of onto maps from {1, 2, …, n} to itself is the same as the total number of permutations on n symbols 1, 2, …, n, which is n!

Let S be a non-empty set and P (S) be its power set. Let any two subsets A and B of S.

It is given that: $P\left(X\right)xP\left(X\right)\to P\left(X\right)$ is defined as $A.B=A\cap B\forall A,B\in P\left(X\right)$

We know that $A\cap X=A=X\cap A\forall A\in P\left(X\right)$

$\Rightarrow A.X=A=X.A\forall A\in P\left(X\right)$

Thus, X is the identity element for the given binary operation*.

Now, an element is  $A\in P\left(X\right)$ invertible if there exists $B\in P\left(X\right)$ such that

$A*B=X=B*A$  (As X is the identity element)

i.e.

$A\cap B=X=B\cap A$

This case is possible only when $A=X=B.$

Thus, X is the only invertible element in P (X) with respect to the given operation*.

Hence, the given result is proved.

Since every set is a subset of itself, ARA for all A ∈ P (X).

∴R is reflexive.

Let ARB ⇒ A ⊂ B.

This cannot be implied to B ⊂ A.

For instance, if A = {1, 2} and B = {1, 2, 3}, then it cannot be implied that B is related to A.

∴ R is not symmetric.

Further, if ARB and BRC, then A ⊂ B and B ⊂ C.

⇒ A ⊂ C

⇒ ARC 

∴ R is transitive.

Hence, R is not an equivalence relation since it is not symmetric.

Define $f:{\rm N}\to {\rm N}$ by

$f\left(x\right)=x+1$

And,  $g:{\rm N}\to {\rm N}$ by,

$g\left(x\right)\left\{\begin{array}{cc}\hfill x-1\hfill & \hfill if,x>1\hfill \\ \hfill 1\hfill & \hfill f,x=1\hfill \end{array}\right\}$

We first show that g is not onto.

For this, consider element 1 in co-domain N. it is clear that this element is not an image of any of the elements in domain.

$\therefore f$ is not onto.

Now,  $gof:{\rm N}\to {\rm N}$ is defined by,

$\begin{array}{l}gof\left(x\right)=g\left(f\left(x\right)\right)=g\left(x+1\right)=\left(x+1\right)-1\left[x,in{\rm N}=>\left(x+1\right)>1\right]\\ =x\end{array}$

Then, it is clear that for $y\in {\rm N}$ , there exists $x=y\in {\rm N}$ such that $gof\left(x\right)=gof\left(y\right)$

Hence, gof is onto.

Define $f:N\to Z$ as $f\left(x\right)$ and $g:Z\to Z$ as $g\left(x\right)=\left|x\right|$

We first show that g is not injective.

It can be observed that:

$\begin{array}{l}g\left(-1\right)=\left|-1\right|=1\\ g\left(1\right)=\left|1\right|=1\\ \therefore g\left(-1\right)=g\left(1\right),but,-1\ne 1\end{array}$

$\therefore g$ is not injective.

Now, $gof:N\to Z$ is defined as

$gof\left(x\right)=y\left(f\left(x\right)\right)=y\left(x\right)=\left|x\right|$

Let $x,y\in N$ such that $gof\left(x\right)-gof\left(y\right)$

$\Rightarrow \left|x\right|=\left|y\right|$

Since $x,y\in N$ , both are positive.

$\therefore \left|x\right|=\left|y\right|\Rightarrow x=y$

Hence, gof is injective

f: R → R is given as f (x) = x³.

Suppose f (x) = f (y), where x, y ∈ R.

⇒ x³ = y³  … (1)

Now, we need to show that x = y.

Suppose x ≠ y, their cubes will also not be equal.

⇒ x³ ≠ y³

However, this will be a contradiction to (1).

∴ x = y

Hence, f is injective.