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I already have proved by induction that the equalities are true $\forall r,s\in\mathbb{Z}$

Because induction only works for statements that uses natural numbers. We first of all have to find a bijection from $\mathbb{Q}\rightarrow\mathbb{N}$ and reframe the statement.

$f:\mathbb{Q}\rightarrow\mathbb{Z}$

We are assuming that we have already reduced the rational number which we put in the function.

Then $f(\frac{x}{z}):=$

$f(\frac{1}{1}) = 1$

$f(\frac{0}{1}) = 0$

case a $\frac{x}{z}\geq 0 \rightarrow$ case 1

case b $\frac{x}{z}< 0 \rightarrow -f(-\frac{x}{z})$

case 1 $\exists k\in\mathbb{N}:\frac{x/k}{z/k}=\frac{x}{z}$ then $f(\frac{x}{z})=f(\frac{x-1}{z})$

for example: $f(\frac{3}{4})=f(\frac{2}{4})+1 =f(\frac{1}{4})+1$

case 2 $x-1 = z$ then $f(\frac{x}{z})=f(\frac{z}{x})+1$

for example: $f(\frac{5}{4})=f(\frac{4}{5})+1$

case 3 $x>z$ then $f(\frac{x}{z})=f(\frac{x}{z+1})+1$

for example: $f(\frac{7}{3})=f(\frac{7}{4})+1$

case 4 $x=1$ then $f(\frac{x}{z})=f(\frac{z-1}{x})+1$

for example: $f(\frac{1}{6})=f(\frac{5}{1})+1$

case 5 $z>x$ then $f(\frac{x}{z})=f(\frac{x-1}{z})+1$

for example: $f(\frac{3}{4})=f(\frac{2}{4})+1 =f(\frac{1}{4})+1$

This is the list of the first shortened rational numbers if we would look at the preimages:

1, 1/2, 2, 1/3, 2/3, 3/2, 3, 1/4, 3/4, 4/3, 4, 1/5, 2/5, 3/5, 4/5, 5/4, 5/3, 5/2, 5, 1/6, 5/6, 6/5, 6, 1/7, 2/7, 3/7, 4/7, 5/7, 6/7, 7/6, 7/5, 7/4, 7/3, 7/2, 7

We define

$g:\mathbb{Z}\rightarrow\mathbb{N}$

with $g(x):= 2x$ if $x$ is positve otherwise $2|x|+1$

I substitute $r$ with $(f$ o $g)^{-1}(n)$= $f^{-1}$o $g^{-1}$(n) and choose a random $s$

Inductionbase n=1

Because $g(0)=1$ and $f(\frac{0}{1}) = 0$ we have $(f$ o $g)^{-1}(1) = 0$ and

because $x^0 = 1$ the inductionbase is proved.

Can somebody help me with the inductionstep?

RM777
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  • What definition of $a^b$ are we using here? – SmileyCraft Nov 14 '18 at 17:17
  • I know that for a equation $x^n - c = 0, c\geq 0$ the unambigious solution is defined as $x=: \sqrt[n]{c} = c^{1/n}$ and $x^{p/q}:=(x^{1/q})^p=(x^p)^{1/q}$ and $a^n =\prod_{i=1}^n a,\forall n\in\mathbb{N}$ – RM777 Nov 14 '18 at 17:26
  • In that case this is much easier to prove by simply using the fact that you already proved the equations hold for natural numbers. Do you have to come up with an inductive proof for some reason, or are you ok with a direct proof using the fact that you already proved it for natural numbers? – SmileyCraft Nov 14 '18 at 17:37
  • Yes you are Right that would make more sense, they wrote only it can be shown by induction but not that is already sufficient if it is proved for natural numbers. But theoreticaly, for the inductionstep: I have to show for example $x^{r+s} = x^r x^s$ I Substitute the $r$ with my function, which I abridge to $h$, do I have to show $x^{h(n+1)+s}=x^{h(n+1)}x^s$ or could I say $x^{(r+1)+s}=x^{h(n)+1+s}=x^{h(n)+1}x^s=x^{r+1}x^s$? – RM777 Nov 14 '18 at 17:51

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