question_answer

Show that the relation R in the S at \[A=\{x:x\in W,0\le x\le 12\}\] given by \[R=\{(a,\,\,b):|a-b|\] is multiple of 4} equivalence relation. Also, find the set of all elements related to 2.

OR

Show that the function

\[f:R\to \{x\in R:-1<x<1\}\] defined by

\[f(x)=\frac{x}{1+|x|},\] \[x\in R\] is one-one and onto function.

Answer:

\[\therefore \]\[A=\{x:x\in W,\,\,0\le x\le 12\}\]

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

\[R=\{(a,\,\,b):|a-b|\]is multiple of 4}

For any element \[a\in A,\] we have \[(a,\,\,a)\in R\]  as \[|a-a|\,\,=0\]is a multiple of 4.

\[\therefore \]R is reflexive.

Now, let \[(a,\,\,b)\in R\Rightarrow |a-b|\] is a multiple of 4.

\[\Rightarrow \]\[|-\,(a-b)|\,\,\Rightarrow \,\,|b-a|\] is a multiple of 4.

\[\Rightarrow \]               \[(b,\,\,a)\in R\]

\[\therefore \]R is symmetric

Let \[(a,\,\,b),\,\,(b,\,\,c)\in R\]

\[\Rightarrow \]\[|a-b|\] is a multiple of 4and \[|b-c|\]is a multiple of 4

\[\Rightarrow \]\[(a-b)\]is a multiple of 4 and \[(b-c)\]is a multiple of 4.

\[\Rightarrow \]\[(a-c)=(a-b)+(b-c)\]is a multiple of 4

\[\Rightarrow \]\[|a-c|\]is a multiple 4

\[\Rightarrow \]\[(a,\,\,c)\,\,\in R\]

\[\therefore \]R is transitive

Hence, R is an equivalence relation.

The set of elements related to 2 is {2, 6, 10}

\[\therefore \]      \[|2-2|\,\,=0\]is a multiple of 4

\[|6-2|\,\,=4\] is a multiple of 4

\[|9-1|\,\,=8\]is a multiple of 4

OR

It is given that \[f:R\to \{x\in R:-\,1<x<1\}\] is defined as

\[f\,(x)=\frac{x}{1+|x|},\,\,x\in R\].

Suppose, \[f\,(x)=f\,(y),\]where

\[x,\,\,y\in R\Rightarrow \frac{x}{1+|x|}=\frac{y}{1+|y|}\]

It can be observed that if x is positive and y is negative, then we have

\[\frac{x}{1+x}=\frac{y}{1-y}\Rightarrow 2xy=x-y\]

Since, x is positive and y is negative, then

\[x>y\Rightarrow x-y>0\]

But \[2xy\] is negative. Then, \[2xy\ne x-y\].

Thus, the case of x being positive and y being negative can be ruled out.

Under a similar argument, x being negative and y being positive can also be ruled out. Therefore, x and y have to be either positive or negative.

When x and y are both positive, then we have

\[f\,(x)=f\,(y)\Rightarrow \frac{x}{1+x}=\frac{y}{1+y}\]

\[\Rightarrow x+xy=y+xy\Rightarrow x=y\]

When x and y are both negative, then we have

\[f\,(x)=f\,(y)\Rightarrow \frac{x}{1-x}=\frac{y}{1-y}\]

\[\Rightarrow x-xy=y-yx\Rightarrow x=y\]

Therefore, f is one-one. Now, let \[y\in R\]such that \[-\,1<y<1\].

If y is negative, then there exists \[x=\frac{y}{1+y}\in R\]such that

\[f\,(x)=f\left( \frac{y}{1+y} \right)=\frac{\left( \frac{y}{1+y} \right)}{1+\left| \frac{y}{1+y} \right|}=\frac{\frac{y}{1+y}}{1+\left( \frac{-\,y}{1+y} \right)}=\frac{y}{1+y-y}=y\]

If y is positive, then there exists \[x=\frac{y}{1-y}\in R\]such that

\[f\,(x)=f\left( \frac{y}{1-y} \right)=\frac{\left( \frac{y}{1-y} \right)}{1+\left| \left( \frac{y}{1-y} \right) \right|}=\frac{\frac{y}{1-y}}{1+\frac{y}{1-y}}=\frac{y}{1-y+y}=y\]Therefore, f is onto.

Hence, f is one-one and onto.

Hence proved.