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.
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