Show that the function f: N rightarrow N,give | vedclass

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(N/A) To check if $f$ is one-one:$A$ function $f$ is one-one if $f(x_1) = f(x_2) \implies x_1 = x_2$.Here,$f(1) = 1$ and $f(2) = 1$.Since $f(1) = f(2)

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(N/A) To check if $f$ is one-one:$A$ function $f$ is one-one if $f(x_1) = f(x_2) \implies x_1 = x_2$.Here,$f(1) = 1$ and $f(2) = 1$.Since $f(1) = f(2)$ but $1 
eq 2$,the function $f$ is not one-one.To check if $f$ is onto:$A$ function $f: N ightarrow N$ is onto if for every $y \in N$,there exists an $x \in N$ such that $f(x) = y$.Case $1$: If $y = 1$,we have $f(1) = 1$. Thus,$1$ has a pre-image.Case $2$: If $y > 1$,then $y+1 > 2$. We can choose $x = y+1$. Then $f(x) = f(y+1) = (y+1) - 1 = y$.Since every $y \in N$ has a pre-image in $N$,the function $f$ is onto.

Show that the function $f: N \rightarrow N$,given by $f(1)=f(2)=1$ and $f(x)=x-1$ for every $x>2$,is onto but not one-one.

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