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Prove: If $|x+y|<|x|+|y|$, then $x<0$ or $y<0$

This looks as though it's true from the start. Take $x=-4, y=4$.

$|-4+4|<|-4|+|4|$

$0<8$ is true.


The question is asking for a proof by contradiction or contrapositive. Which means I am going to negate some part of the ending in order to find a contradiction in the hypothesis.

This is of form: If $ P\implies Q$

So for a proof by contradiction I need:

$P \implies \lnot Q $

If pr the contrapositive: $ \lnot Q \implies \lnot P$


Will the following proof work? Also, is my proof formal enough? What can be done to improve it's form?

PF. (by contradiction)

If $|x+y|<|x|+|y|, \implies x \geq 0 \lor y \geq 0$

$x \geq 0, y \geq0$

since $x \geq 0$

$|x+y|<|x|+|y|$ is false proof by contradiction

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  • $\begingroup$ No, for a proof by contradiction, you need $P\land \neg Q\implies \text{contradiction}$. A proof by contrapositive would be a proof that $\neg Q\implies\neg P$. You cannot prove $P\implies Q$ by proving $\neg P\implies \neg Q$. $\endgroup$ – Arturo Magidin Oct 9 '11 at 21:38
  • $\begingroup$ Looks like that was a typo. Would my following proof suffice, since it clearly is not true $\endgroup$ – user17366 Oct 9 '11 at 21:56
  • $\begingroup$ The first step ($x\geq 0, y\geq 0$) is fine. After that, it just falls off the rails. You cannot simply assume $x=y$; and if you do, then you only establish the result for the case when $x=y$. What about the case where $x\neq y$? So, no, the following proof will not suffice. $\endgroup$ – Arturo Magidin Oct 9 '11 at 22:04
  • $\begingroup$ @arete: Why does $x\ge 0$ imply the contradiction? Also, you misquoted the claim you want to prove. $x<0\lor y<0$ would negate to $x\ge 0\land y\ge 0$. $\endgroup$ – Asaf Karagila Oct 9 '11 at 22:21
  • $\begingroup$ For an alternative kind of proof: note that using the alternative definition $\;\left|x\right| = x\mathbin\max -x\;$, using the properties of $\;\mathbin\max\;$ (like the fact that $\;+\;$ distributes over it, and $\;x<y\mathbin\max z\;\equiv\;x<y\;\lor\;x<z\;$), it is straightforward to reduce $\;\left|x+y\right|<\left|x\right|+\left|y\right|\;$ to $\;(x>0 \land y<0) \lor (x<0 \land y>0)\;$. Which incidentally is equivalent to $\;xy>0\;$. $\endgroup$ – Marnix Klooster Jan 27 '18 at 18:45
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When coming to prove $P\Rightarrow Q$, we can either:

  1. Prove directly, that is assume $P$ and show $Q$;
  2. Prove by contradiction, that is assume $P$ and $\lnot Q$ and derive contradcition; or
  3. Prove the contrapositive, that is assume $\lnot Q$ and show $\lnot P$.

So to prove by contrapositive is to assume that $\lnot(x<0\lor y<0)\equiv (x\ge 0\land y\ge 0)$ and show that $|x+y|\nless|x|+|y|$;

and to prove by contradiction is to assume that $|x+y|<|x|+|y|$ but $x>0$ and $y>0$, and thus derive contradiction.


As for the added proof attempt:

This proof is unclear, when I was grading papers in a similar introductory course I would probably mark this question with a nice little X mark and move on.

  • The statement of the assumption by contradiction is unclear. You should say something such as "assume by contradiction $x\ge 0, y\ge 0$." to make it clear where is the contradiction is going to come from.

  • Now you are trying to prove, not to falsify. The next step is to say $x+y\ge 0$, therefore $|x+y|=x+y$ and since $|x|=x, |y|=y$ we have that $x+y=|x+y|<|x|+|y|=x+y$.

Now we have derived a contradiction, since $x+y< x+y$ is clearly a false theorem in the context above.

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  • $\begingroup$ I'm confused about negating inequalities. If you negate the original statement's conclusion $x<0$ or $y<0$ then is the negation $x \geq 0$ and $y\geq 0$ ? $\endgroup$ – user17366 Oct 9 '11 at 21:25
  • $\begingroup$ @areteornothing: Yes, I had a mistake there which is now corrected. $\endgroup$ – Asaf Karagila Oct 9 '11 at 21:25
  • $\begingroup$ Thank you for your input, I arrived at a similar conclusion before I saw this comment. I will post what I have and see if the approach is enough to not get that little x of yours. $\endgroup$ – user17366 Oct 9 '11 at 22:08
  • $\begingroup$ @arete: Fear not, I have realized that one of the greatest difficulties when starting with formal mathematics is to understand how a proof should look like; and how formal it should be. Many students either underdo it, or overdo it. Only a few get it right from the start. Just don't give up, and things will start fall into place. $\endgroup$ – Asaf Karagila Oct 9 '11 at 22:10

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