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Given examples of two functions `f:" "N ->N" "a n d""""""g:" "N->N` such that of is onto but f is not onto. (Hint: Consider `f(x)" "=" "x" "a n d""""""""g(x)" "=" "|x|` ) .

Text Solution

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Let f:N→N
be
f(x)=x+1
and
g:N→N be
g(x)={x−1,x>11,x=1}
We will first show that f is not onto.
Checking f is not onto.
...
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