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?


  • $\begingroup$ That's not true: $||1+1||=2 > 1=||1||.$ $\endgroup$ – Dzoooks Sep 15 '19 at 21:00
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    $\begingroup$ You'll get some useful ideas here. $\endgroup$ – J.G. Sep 15 '19 at 21:01
  • $\begingroup$ 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. $\endgroup$ – hardmath Sep 15 '19 at 21:02
  • $\begingroup$ It isn't the usual absolute value. The absolute value is defined with the three properties given and nothing more. $\endgroup$ – 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 $$ \|x+y\|\leqslant (n+1)^{\frac{1}{n}}\|\max(x,y)\|\underset{n\rightarrow +\infty}{\longrightarrow}\|\max(x,y)\| $$

  • $\begingroup$ Why is the absolute value of the combinatorial number less than or equal to 1? $\endgroup$ – Guillemus Callelus Sep 23 '19 at 19:51
  • $\begingroup$ Because $\binom{n}{k}\in\mathbb{Z}$, thus $\left\|\binom{n}{k}\right\|\leqslant 1$. $\endgroup$ – Tuvasbien Sep 24 '19 at 3:07

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