Problem
ALG-B3-M04-P003 Iteration Gives a Bijection
#3
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Let \(f:\mathbb Q\to\mathbb Q\) satisfy \(f(f(x))=x+1\) for all \(x\in\mathbb Q\). Prove that \(f\) is bijective.
For injectivity compare \(f(f(a))\) and \(f(f(b))\). For surjectivity represent an arbitrary \(T\) as a value of the function.
If \(f(a)=f(b)\), then \(f(f(a))=f(f(b))\). By the condition, \(a+1=b+1\), so \(a=b\). Thus \(f\) is injective.
Now let \(T\in\mathbb Q\). Then \(f(f(T-1))=T\). Hence \(T\) lies in the image of \(f\). Since \(T\) was arbitrary, \(f\) is surjective. Therefore \(f\) is bijective.
This is not a function-finding problem; it extracts properties from an iteration.