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I would like to offer my own attempt, but I honestly don't have a clue. I'm completely baffled. Any help would be appreciated.

Full question: Using results from Book A, one can show that for each non-zero rational number $x$, there is a unique $m∈\mathbb{Z}$ such that:

$x=5^m\frac{a}{b}$, $a∈\mathbb{Z}-${$0$}, $b∈\mathbb{N}$, and 5 does not divide either of a and b. We use this fact to define a function $f_5:\mathbb{Q}\rightarrow \mathbb{R}$ by,

$f_5(x)= 5^{-m}$, if $x=5^m\frac{a}{b}$,

$f_5(x)= 0$, if $x=0$

(I don't know how to do a piecewise function on here)

Write down $f_5(1), f_5(5)$ and $f_5(5^n)$ for $n∈\mathbb{N}$

I'm not sure which particular results they are referenceing from the Book A, but I don't think they are needed for the question.

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    Is this all the information given from the question? – 2017-02-26
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    I'll edit to add more detail, I thought it was one that could just be answered immediately if someone knew. Gimme a few minutes. – 2017-02-26
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    Thanks for the response. I was under the impression neither $a$ nor $b$ can be multiples of 5, but your answer make me think I have misinterpreted the question. So for $f_5(1)$, $1 = 5^0\frac{1}{1}$ and so $f_5(1) = 1$? – 2017-02-26
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    Yes that's correct – 2017-02-26

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$f_{5}(1) = 1, 1 = 5^{0}*\frac{1}{1}$

$f_{5}(5) = \frac{1}{5}, 5 = 5^{1} * \frac{1}{1}$

$f_{5}(5^n) =\frac{1}{5^{n}}, 5^{n} =5^{n} *\frac{1}{1}$

Are there any further proofs or anything required? Otherwise question seems pretty short.

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    It was simply write down, so that will definitely suffice. I thought $a$ and $b$ had to be distinct numbers for some reason, which I had no reason to think. I feel pretty silly, I've spent quite a while racking my brain as to how I can represent it without $\frac{1}{1}$. Thanks for your help, I'm gonna slink off in shame now! – 2017-02-26