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I would like to confirm a proof I made:

For every $a,x,y \in\mathbb{R}$ such that $x≤a≤y$, we have $|a|≤\max\{|x|,|y|\}$.

I assumed that $|x|≤|y|$, so: $-|y|≤a≤|y|$, and that means: $a≤|y|$. Then, I assumed: $|y|≤|x|$, so $-|x|≤a≤|x|$, and that means: $a≤|x|$

From here, I let $b\in\mathbb R$, such that $b=\max\{|x|,|y|\}$ and that means $a≤b$.

From here I don't really know what to do or if all I did is true or not.

Thank you for your help!

Integrand
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  • https://math.stackexchange.com/questions/3918165/prove-that-a-leq-max-b-c-if-b-leq-a-leq-c – Neat Math Nov 25 '20 at 14:08
  • Your proof is definitely going in the good direction and is almost done. I would reword it as follows: If $|x| \leq |y|$ then .... so $|a|\leq |y|\leq \max{|x|,|y|}$. Otherwise, $|y|\leq |x|$ then .... so $|a|\leq |x|\leq \max{|x|,|y|}$. In every cases, we have $|a|\leq \max{|x|,|y|}$. – Surb Nov 25 '20 at 14:21
  • Thank you! So in general, if I have -|x| ≤ a ≤ |x|, am I able to assume that |a|≤|x| ? –  Nov 25 '20 at 15:12

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