# Absolute value in rational numbers

We define the absolute value in $$\mathbb{Q}$$ as an application $$||\, \cdot \, || : \mathbb{Q}\rightarrow [0,\infty )$$ that fulfills the properties:

• $$||x||=0$$ if and only if $$x=0$$.
• $$||xy||=||x||\, ||y||$$.
• $$||x+y||\leq ||x||+||y||.$$

We have to show that if for $$n\in \mathbb{Z}$$ we have $$||n||\leq 1$$, then it is fulfilled that for all $$x,y\in \mathbb{Q}$$ we have $$||x+y||\leq \max (||x||, ||y||).$$

My attempt is as follows:

$$||x+y||^N =\prod_{i=1}^N ||x+y|| =||(x+y)^N||=|| \sum_{k=0}^N {N\choose k}x^{N-k}y^k|| \leq \sum_{k=0}^N ||{N\choose k}x^{N-k}y^k||$$

Can you help me doing this proof?

Thanks

• That's not true: $||1+1||=2 > 1=||1||.$ – Dzoooks Sep 15 '19 at 21:00
• You'll get some useful ideas here. – J.G. Sep 15 '19 at 21:01
• Something seems jumbled about your problem statement. In particular when you reach the part "We have to show...", you put a condition on all (?) integers $n$ that $||n||\le 1$. So this is not satisfied by the usual "absolute value" on rational numbers. It might be clearer if you gave an explicit function satisfying your properties. – hardmath Sep 15 '19 at 21:02
• It isn't the usual absolute value. The absolute value is defined with the three properties given and nothing more. – Guillemus Callelus Sep 15 '19 at 21:07

Let $$(x,y)\in\mathbb{Q}^2$$, you have $$\|x+y\|^n=\|(x+y)^n|\|=\left\|\sum_{k=0}^n{\binom{n}{k}x^k y^{n-k}} \right\|\leqslant\sum_{k=0}^n{\underbrace{\left\|\binom{n}{k}\right\|}_{\leqslant 1}\|x^k y^{n-k}\|}\leqslant\sum_{k=0}^n{\|x\|^k\|y\|^{n-k}}$$ Since $$\|x\|\leqslant\|\max(x,y)\|$$ and $$\|y\|\leqslant \|\max(x,y)\|$$ you then have $$\|x+y\|^n\leqslant\sum_{k=0}^n{\|\max(x,y)\|^k\|\max(x,y)\|^{n-k}}=(n+1)\|\max(x,y)\|^n$$ Hence $$\|x+y\|\leqslant (n+1)^{\frac{1}{n}}\|\max(x,y)\|\underset{n\rightarrow +\infty}{\longrightarrow}\|\max(x,y)\|$$
• Because $\binom{n}{k}\in\mathbb{Z}$, thus $\left\|\binom{n}{k}\right\|\leqslant 1$. – Tuvasbien Sep 24 '19 at 3:07