question_answer

Let \[A=\{x\in R:0\le x\le 1\}.\] If \[f:A\to A\] is defined by

\[f(x)=\left\{ \begin{matrix}    x, & i\text{f}\,\,x\in Q  \\    1-x & \text{if}\,\,x\notin Q  \\ \end{matrix} \right.\]

Then prove that \[fof(x)=x\] for all \[x\in A.\]

OR

Let \[A=N\times N\] and * be the binary operation on A defined by (a, b) * (c, d) = \[(a+c,\,\,b+d)\]

Show that * is commutative and associative.

Find the identity element for * on A, if any.

Answer:

Let \[x\in A\] Then, either x is rational or x is irrational.  there are two cases arise

Case I When \[x\in Q\]

In this case, we have f(x) = x

\[\therefore \]      fof(x) = f(f(x))                 [\[\because \] f(x) = x]

= f(x)

= x

Case II When \[x\notin Q\]

In this case, we have \[f(x)=1-x\]

\[\therefore \]      fof(x) = f(f(x))

\[=f(1-x)\]

\[=1-(1-x)\]

\[[\because \,\,\,x\notin Q\,\,\,\,\Rightarrow \,\,\,\,1-x\notin Q]\]

Thus, fof(x) = x whether \[x\in Q\,\,\,\text{or}\,\,\,x\notin Q.\]

Hence, fof(x) = x for all \[x\in A.\]     Hence proved.

OR

Given, (a, b)\[*\](c, d) \[=(a+c,\text{ }b+d)\]for (a, b),\[(c,\,\,d)\in A.\]

For commutative

Let (a, b), \[(c,\,\,d)\in A.\]

Then, (a, b)\[*\] (c, d) \[=(a+c,\text{ }b+d)\]

\[=(c+a,\,\,d+b)=(c,\,\,d)*(a,\,\,b)\]

[\[\because \] Commutative property of N]

\[\Rightarrow \] (a, b)\[*\](c, d) = (c, d)\[*\](a, b)

Hence, \[*\] is commutative.

For associative

Let (a, b),(c, d),\[(e,\,\,f)\in A.\]

Then,{(a, b)\[*\](c, d)}\[*\](e, f) \[=(a+c,\text{ }b+d)\]\[*\](e, f)       \[=\{(a+c)+e,\text{ }(b+d)+f\}\]

\[=\{a+(c+e),\text{ }b+(d+f)\}\]

[\[\because \] associative property of N]

\[=(a,\,\,b)*(c+e,\,d+f)\]

= (a, b)\[*\]{(c, d)\[*\](e, f)}

Hence, \[*\] is associative.

For \[(a,\,\,b)\in A,\,\,if\,\,(c,\,\,d)\in A\] is identity element, then

\[=(c+a,\,\,d+b)=(c,\,\,d)*(a,\,\,b)\]

\[\Rightarrow \]   \[(a+c,\,b+d)(c+a,\,b+d)=(a,\,b)\]

\[\Rightarrow \]   \[a+c=a\]and \[b+d=b\]

\[\Rightarrow \]   c = 0, d = 0

\[\Rightarrow \] \[(0,\,\,0)\in A\] is identity element.

But \[(0,\,\,0)\notin A\]

Hence, no identity element.