Let  f:N\rightarrow Y be a function defined as f(x)=4x+3,\: \: where\: \: Y=\left \{ y\equiv N:y=4x+3,for\: \: some\: \: x\equiv N \right \}. Show that f is invertible. Find its inverse.

 

 

 

 
 
 
 
 

Answers (1)

Let f:N \rightarrow Y be a function defined as 

f(x)=4x+3,\: \: where\: \: Y=\left \{ y\equiv N:y=4x+3,for\: \: some\: \: x\equiv N \right \}.Arbitrary Element Y 

y=4x+3

x=\frac{y-3}{4}

Define g:Y\rightarrow N\: \: by\: \: g(y)=\frac{y-3}{4}

Now gof(x)=f(f(x))=g(4x+3)=\frac{4x+3-3}{4}=x

and fog(y)=f(g(y))=f\left ( \frac{y-3}{4} \right )=\frac{4(y-3)}{4}+3=y-3+3=y

This shows that gof=I_x and fog=Iy which implies that f is invertible and g is the inverse of f.

\therefore Inverse\: \: of\: \: f'=g(y)=\frac{y-3}{4}

Hence Proved.

 

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