Prove that for every three non-zero integers, a,b and c, at least one of the three products ab,ac,bc is positive

Solution 1:

The product of three negative numbers is negative. So if $ab$, $ac$, and $bc$ are all negative, then $(ab)(ac)(bc)\lt0$. But $(ab)(ac)(bc)=a^2b^2c^2$ is the product of three squares, which are all positive.

Solution 2:

HINT: One of the following is true about the integers: (all positive), (all negative), (one positive and two negative), or (two positive and one negative).

Solution 3:

Assume to the contrary, there exist $a$, $b$ and $c$ that are non-zero integers and none of the three products $ab$, $ac$ and $bc$ are positive.

Things generally look OK up to this point. In particular, the statement above is a good assumption to make for the desired proof by contradiction.

Now, I pick a = 1, b = -1 , c = 2

This is not OK: the assumption did not say "for any $a$, $b$ and $c$ that are non-zero integers" (which would allow us to choose any non-zero $a$, $b$, and $c$ for a counterexample), it said merely "there exist". As it turns out, there are some choices of $a$, $b$, and $c$ that do not satisfy the "none of the products" condition, but so what? All it takes to justify a "there exist" statement is to find one set of numbers that do satisfy the condition.

Here's an example of why a counterexample does not contradict a "there exists" condition. Let's try to prove this statement:

Every integer is even.

Proof by contradiction:

Assume the contrary, that there exists an integer $n$ that is not even.

Now pick $n = 2$.

But $2$ is even, therefore the assumption (that it is not even) is contradicted.

Do you see why this does not work? Can you see how the logic progresses just like the logic in your proof? (Assume there exists ____ such that ____; choose some values for the variables in the first blank; then show that for these values, the statement in the second blank is false.)

Solution 4:

ab and ac negative => sign(b) = sign(c) => bc positive