Timeline for Is there a proof for triangle inequality in $\mathbb{R}$ by contradiction/absurd?
Current License: CC BY-SA 4.0
9 events
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| Nov 26, 2021 at 19:00 | comment | added | Ted Shifrin | OK, you've mostly taken care of my original issue. However, you should make it clear that you are assuming $|a|+|b|<|a+b|$ holds for some $a,b$, not all. You're missing the case with $a=-b$. Admittedly it's an easy case. ... The upshot, though, is that this proof by contradiction is probably more torture than a direct proof, since we need not proceed by cases. | |
| Nov 26, 2021 at 16:52 | history | edited | soupless | CC BY-SA 4.0 |
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| Nov 26, 2021 at 16:34 | comment | added | soupless | I'll edit it now. | |
| Nov 26, 2021 at 16:33 | comment | added | Jakob Streipel | Yes. Since you know nothing about the assumed $a$ and $b$, you would need to show that whatever they are, they lead to a contradiction. | |
| Nov 26, 2021 at 16:32 | comment | added | soupless | @prets This means that I should: 1) state which values of $a$ and $b$ make the equation true, and/or; 2) state all the possible cases and prove them one by one, right? | |
| Nov 26, 2021 at 16:27 | comment | added | Jakob Streipel | Sure, but all you're assuming in your original assumption is that for some $a$ and $b$, we have $|a| + |b| < |a + b|$. You've no idea which, and how their sizes compare, etc. | |
| Nov 26, 2021 at 16:26 | comment | added | soupless | @prets Isn't it that one case is enough to prove the claim by contradiction? | |
| Nov 26, 2021 at 16:16 | comment | added | Jakob Streipel | There's also no reason an inequality (involving negatives and positives!) without absolute value signs remains true with absolute value signs, so be careful about this, too. Some cases are required. | |
| Nov 26, 2021 at 16:11 | history | answered | soupless | CC BY-SA 4.0 |