How to prove this absolute value implication?

Asked Feb 20 '22 at 19:59

Active Feb 20 '22 at 20:26

Viewed 28 times

Prove:

$$(|x| < 1)\wedge(|y| < 1) \Rightarrow \frac{|x-y|}{|1 - xy|} < 1$$

for all real $x$ and $y$.

I've tried approaching it via proof by contradiction:

$(\exists x,y\in\mathbb{R})(|x| < 1 \wedge|y| < 1)\wedge\left(\dfrac{|x-y|}{|1-xy|} < 1\right)$

But I can't seem to get anywhere useful. I run into the same issues with contrapositive.

Many thanks!

edited Feb 20 '22 at 20:26

Átila Correia

asked Feb 20 '22 at 19:59

Thomas Hegarty

1

This has been asked and answered several times, you can search for it with Approach0 – Martin R Feb 20 '22 at 20:07
2

For real numbers: https://math.stackexchange.com/q/1535052/42969, https://math.stackexchange.com/q/2568386/42969 – Martin R Feb 20 '22 at 20:09
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For complex numbers: https://math.stackexchange.com/q/198582/42969, https://math.stackexchange.com/q/506058/42969 – Martin R Feb 20 '22 at 20:10
Could I convert |x-y| < |1-xy| to |x+y|<1+|xy|? Or somehow get to addition from subtraction? – Thomas Hegarty Feb 20 '22 at 20:16
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Just set $x = a$ and $y = -b$. – Martin R Feb 20 '22 at 20:21
Well doesn't that make things easier – Thomas Hegarty Feb 20 '22 at 20:23
I think it does: $\frac{|x-y|}{|1-xy|} = \frac{|a+b|}{|1+ab|}$ – Martin R Feb 20 '22 at 20:24
Yeah and it works directly or by contradiction ( unnecessary in hindsight ) , (x-y)^2-(1-xy)^2 >=1 yields 0>=1. – Thomas Hegarty Feb 20 '22 at 20:26

0 Answers0