Q. 125.0( 2 Votes )

Give examples of

Answer :

Define f: N Z as f(x) = x and g: N N as g(x)=|x|.


We first show that g is not injective.


It can be observed that:


g(– 1)=| – 1| = 1


g(1) =|1| = 1


Therefore, g(– 1) = g(1), but —1 1.


Therefore, g is not injective.


Now, gof: N Z is defined as gof(x) = g(f(x)) =g(x)=|x|.


Let x, y N such that gof(x) = gof(y).


|x|=|y|


Since x and y N both are positive.


|x|=|y| x=y


Hence, gof is injective


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