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Find all functions $ f : \mathbb Z \to \mathbb Z $ that satisfy the following conditions:

(i) $ f ( 0 ) = 1 $;
(ii) $ f \big( f ( x ) \big) = x $ for all integers $ x $;
(iii) $ f \big( f ( x + 1 ) + 1 \big) = x $ for all integers $ x $.

How can I prove it by induction?

Mohsen Shahriari
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Denise
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  • As in my argument below, we know that $f(x)=f(x+1)+1$ for all integers $x$. Now let us prove that $f(n)=1-n$ and that $f(-n)=1+n$ for all positive integers $n\geq 1$.

    The induction start is easy: we know that $f(0)=f(1)+1$ and that $f(-1)=f(0)+1$. As $f(0)=1$, it follows that $f(1)=0$ and $f(-1)=2$.

    If it holds for some $n\geq 1$ that $f(n)=1-n$ and $f(-n)=1+n$, then $f(n)=f(n+1)+1$, implying $f(n+1)=f(n)-1=-n=1-(n+1)$, and $f(-(n+1))=f(-n)+1$, implying $f(-(n+1))=1+n+1=2+n=1+(n+1)$. Hence it follows that $f(n)=1-n$ and $f(-n)=1+n$ for all integers $n\geq 1$, so that $f(x)=1-x$.

     – Bryder Apr 14 '14 at 19:28

2 Answers2

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Quite straightforward. $f$ is a bijection by (ii), so (ii) and (iii) imply $f(x)=f(x+1)+1$. Now $f(n)=f(n-1)-1=f(n-2)-2=f(n-3)-3=\ldots=f(0)-n=1-n$ and $f(-n)=f(1-n)+1=\ldots=f(0)+n=1+n$ for all integers $n>0$, so that $f(x)=1-x$ for all integers $x$. Induction is of course also applicable.

Bryder
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By (ii), $$f(f(0))=f(1)=0$$ By (iii) $$f(f(-1+1)+1)=f(f(0)+1)=f(2)=-1$$ So $$f(f(2))=f(-1)=2$$ So $$f(f(-2+1)+1)=f(f(-1)+1)=f(3)=-2$$ So $$f(f(3))=f(-2)=3$$ So $$f(f(-3+1)+1))=f(4)=-3$$ So...