# If $| X | \leq K$ and $| Y | \leq K$, then how can I show that $| X-Y | \leq 2K$?

If $|X| \leq K$ and $|Y| \leq K$, then how can I show that $|X-Y| \leq 2K$? It seems very obvious intuitively but I am not able to use the triangle inequality here. Does anyone have any ideas? thanks!

• Why cann't you use the triangle inequality? – Hetebrij Nov 28 '15 at 21:24
• @Hetebrij Yes he can! – user60589 Nov 28 '15 at 21:25
• If you cannot (are not allowed to) use the triangle inequality, you will in effect have to prove the triangle inequality. – André Nicolas Nov 28 '15 at 21:27
• @user60589 No, we can, but I want to understand why user136503 thinks he is not able to use the triangle inequality. Or if it is forbidden by some power. – Hetebrij Nov 28 '15 at 21:29
• try to use the triangle inequality with $|X+(-Y)|$ – Mirko Nov 28 '15 at 21:37

First observe that $|x-y| \leq |x| + |y|$ since $|x-y| = |x + (-y)| \leq |x| + |(-y)| = |x| + |y|$ according to the triangle inequality.

Therefore, if we suppose that for arbitrary real numbers $x, y,$ and $k$ the inequalities $|x| \leq k$ and $|y| \leq k$ are true, then

$$|x-y| \leq |x| + |y| \leq 2\cdot k \implies |x-y| \leq 2 \cdot k$$

as desired.

$\blacksquare$

$$|X - Y| \le |X| + |Y| \le K + K = 2K$$

• Reproducing (after faulty versions) @Benedict's answer. – Did Nov 29 '15 at 8:16
• If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. - From Review – Ian Miller Nov 29 '15 at 8:52
• @IanMiller I don't have a new question. Me saying 'I think' was because I found this a little too easy. It seemed like a simple application of the triangle inequality. I found the question kind of strange consider user136503 has posted a lot of advanced probability problems – BCLC Nov 29 '15 at 8:55