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Consider three real numbers $a$, $b$, and $c$. Prove that we can pick two of them such that their product is non-negative.

My Proof:

Using proof by cases:

Case 1:
$a >0,\space b>0,\space c>0$
Above, we can see that if we pick ANY two pairs that their product is alway positive.
Ex: $$a\cdot b = ab \qquad (ab>0)$$

Case 2:
Lets say $a <0, b<0,c>0$
There is only one way to get a positive product out of this group. You need to pick two variables that have the same "sign"(The variables must both be positive or negative). If the two variables are not the same sign then their product will always be negative.
Ex: $$a \cdot b = ab \qquad (ab>0, a<0, b<0)$$ $$a \cdot c = ac \qquad (ac <0, a<0, b >0)$$

$\therefore$ You can pick two variables with like signs from this group and their producgt will always be positive. $\square$

My Question:

Is this the correct way to go about this proof? I don't see another way with the small set of proof methods we have. I don't know if this is enough to prove this though. It seems too simple... Any thoughts?

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    $\begingroup$ You missed the case that exactly one number is negative , the case that all numbers are negative and the case that at least one number is $0$, but it is no problem to include those cases. You can shorten the proof by splitting in the case that at least one number is $0$ (trivial) and the case that there are two positive or two negative numbers. $\endgroup$ – Peter Sep 19 at 20:09
  • $\begingroup$ So basically this can be solved by: case 1, two variables are negative. Case 2, two variables are negative. Case 3, one variable is 0? This covers all the cases I really need to think about? $\endgroup$ – E__ Sep 19 at 20:20
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    $\begingroup$ You mentioned one case twice. If one of them is replaced by "two variables are positive" , then in fact you covered all possible cases. $\endgroup$ – Peter Sep 19 at 20:21
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    $\begingroup$ In fact, the "$0$-case" is utterly trivial since any number multiplied with $0$ is $0$ , which is non-negative. $\endgroup$ – Peter Sep 19 at 20:23
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    $\begingroup$ "positive (non-negative)" is a confusing formulation, since "positive" is not the same as "non-negative" $\endgroup$ – Peter Sep 19 at 20:30
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Proof by contradiction.

We want to prove that given any triple of numbers $a,b,c\in\mathbb{R}$ we can extract two such that their product is non negative.

Suppose that there exists a triple $(x,y,z)$ such that any of the three possible products $$xy,\;xz,\;yz$$ is negative. So their product should be negative as well. But $$(xy)(xz)(yz)=(xyz)^2$$ Contradiction.

Thanks to @Servaes for the collaboration.

Edit.

This property can be easily generalized to any set $S \subseteq\mathbb{R}$ containing more than two elements.

We want to prove that from any of those subsets it can be chosen a pair of elements $(x,y)\in S\times S$ such that $xy\ge 0$. Indeed suppose all pairs $(a,b)\in S\times S$ are such that $ab<0$ we could chose $(p,q)$ and $(p,r)$ such that $pq<0,\;pr<0$. So $(pq)(pr)>0$ which means $p^2(qr)>0$ and as $p^2>0$ we would have $qr>0$ in contradiction with the assumpt that all pair gave a negative product.

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    $\begingroup$ How about finding a contradiction by noting that if $xy$, $xz$ and $yz$ are negative, then so is their product $$(xy)(xz)(yz)=(xyz)^2.$$ $\endgroup$ – Servaes Sep 19 at 20:48
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    $\begingroup$ @Servaes Maybe, this is at last the most elegant proof. $\endgroup$ – Peter Sep 19 at 20:50
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I'll assume that here that $a$, $b$ and $c$ must be nonzero, as otherwise the statement to prove is false.

You've treated two cases, but there are many more cases. As each of $a$, $b$ and $c$ is either positive or negative (you seem to exclude the possibility that the variables are $0$), there are $2^3=8$ cases to consider. Of course this is cumbersome, and it is not the way to go about things.

One way to reduce the number of cases, is to consider the number of negative variables among $a$, $b$ and $c$. This leaves just $4$ cases to consider. Another approach is the following:

The product of two nonzero real numbers is positive if and only if their signs agree. By the pigeonhole principle, given three nonzero real numbers, there must be two with the same sign. So you can pick two of them so that their product is positive.

Edit: Another way to keep the statement to prove from being false is to interpret positive as meaning nonnegative. Then the numbers $a$, $b$ and $c$ may equal $0$, but of course in this case the product is also $0$.

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    $\begingroup$ The statement is also true , if there is a $0$ (see comments above). I guess you overlooked "non-negative" because in the body we have "positive" and only in parantheses "non-negative". $\endgroup$ – Peter Sep 19 at 20:28
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    $\begingroup$ @Peter I understood positive to mean greater than $0$, but interpreting it as nonnegative indeed also makes the statement true. I have updated my answer to include this interpretation. $\endgroup$ – Servaes Sep 19 at 20:36
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    $\begingroup$ It is true that "positve" menas "greater than $0$" , but I think the author means actually "non-negative". $\endgroup$ – Peter Sep 19 at 20:38
  • $\begingroup$ Mentioning the pigeonhole-principle is useful since similar questions could be solved with this technique (+1) $\endgroup$ – Peter Sep 19 at 20:41
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Have you ever heard the puzzle if you have a drawer full of black and blue socks and you are blindfolded and need to pull out a pair of matching socks, how many do you need to pull?

Answer: $3$.

If you only have $2$ categories for things to be (blue or blue) and you pull out $3$ off them, as you can't have all $3$ items be different because you have more items then categories so at least $2$ must be in the same category. And if you want to spell it out; you can have all three socks black (and you have more than two matching socks); you can have two black and one blue (and you have a pair of matching black socks); you can have one black and two blue (and you have one pair of matching blue socks); or you can have three blue socks (and you have more than one pair of matching socks).

This is the exact same thing. All real numbers are either 1) negative or 2) not negative. So if you draw three real numbers at random the at least two of the them will be the same type.

If you have at least two negatives, then their product is the product of two negatives. That is positive, and so the product is non negatives.

If you don't have at least two negatives, then you have at most one negative and you have at least two numbers that are not negative. If you multiply two non-negative numbers together you can not get a negative number. (Either they are both positive; or one is zero; or both are zero-- in any case the product can not be negative.)

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You can prove that the product of two of the numbers is nonnegative, otherwise the choice of $0,1,-1$ would be a counterexample.

If one of the numbers is $0$, we're done. So we can assume that the numbers are nonzero.

Take two of them. If their product is positive, we're done. Otherwise one is positive and the other one is negative. Now the third number is either positive or negative: combine it with the one with the same sign among the first two.

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