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Let f: ZZ->ZZ\ a n d\ g: ZZ->ZZ\ be fun...

Let `f: ZZ->ZZ\ a n d\ g: ZZ->ZZ\ ` be functions defined by `f={(n , n^2): n in ZZ}` and `g={(n ,|n|^2): n in ZZ}` . Show that. `f=g` .

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Domain of f = Domain of g = Z and,
Co-domain of f = Co-domain of g = Z
We have, `f(n)=n^2` and
g(n)=`(∣n∣)^2=n^2`
:. f(n)=g(n) for all n in Z.
Hence, f=g.
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