# Prove by contradiction or contrapositive? If $|x+y|<|x|+|y|$, then $x<0$ or $y<0$.

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?

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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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 $\neg P\implies \neg Q$. –  Arturo Magidin Oct 9 '11 at 21:38
Looks like that was a typo. Would my following proof suffice, since it clearly is not true –  user17366 Oct 9 '11 at 21:56
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. –  Arturo Magidin Oct 9 '11 at 22:04
@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 $x\ge 0\land y\ge 0$. –  Asaf Karagila Oct 9 '11 at 22:21

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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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 $y\geq 0$ ? –  user17366 Oct 9 '11 at 21:25
@areteornothing: Yes, I had a mistake there which is now corrected. –  Asaf Karagila Oct 9 '11 at 21:25
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. –  user17366 Oct 9 '11 at 22:08
@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. –  Asaf Karagila Oct 9 '11 at 22:10