Prove that for every three non-zero integers, a,b and c, at least one of the three products ab,ac,bc is positive The question is:
Prove that for every three non-zero integers, a,b and c, at least one of the three products ab,ac,bc is positive. Use proof by contradiction.
My general approach to doing contradiction is as follow:
I always like to turn the statements into propositional logic, with an implication. In this case it would be like:
[For all a,b,c in the domain of non-zero integer If a,b and c are three non-zero integers], then at least one of the three products ab,ac,bc is positive.
Then I take the negation of it: (P ^ ~Q)
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.
Now, I pick a = 1,  b = -1 , c = 2
a.b = -1
a.c = 2
b.c = -2
Since a.c is positive, our assumption is false and we have a contradiction. Hence, the original statement is true.
Please feel free to share any other alternatives...(contradiction ones)
 A: 
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.)
A: 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.  
A: You can't prove this with just an example.
By contradiction, suppose $ab<0$, $ac<0$ and $bc<0$.
Since $ab<0$, then either $a<0$ and $b>0$, or $a>0$ and $b<0$.
In the first case, the condition $ac<0$ implies $c>0$, but then $bc>0$.
In the second case $bc<0$ implies $c>0$, but then $ac>0$.
Both cases give a contradiction.
A: Map each of $a,b,c$ to its sign:
$$
sign\colon x \mapsto \frac x {\lvert x \rvert}\colon\{a,b,c\}\to \{-1,1\}.
$$
If the image of $sign$ is just $\{-1\}$ or just $\{1\}$, then all of $a,b,c$ have the same sign, and the product of any two of them is positive. If the image of $sign$ is all of $\{-1,1\}$, then by the pigeonhole principle some two distinct $x,y\in \{a,b,c\}$ have the same sign, and then $xy$ is positive.
A: ab and ac negative
=> sign(b) = sign(c) => bc positive
A: This is an attempt to answer the question about the logic here implicit in misheekoh's comment. What we are trying to prove is:
$$
\forall a, b, c \in \Bbb{Z}_{\neq0}(ab > 0 \lor ac > 0 \lor bc > 0)
$$
To do this by contradiction, we negate the above to give:
$$
\exists a, b, c \in \Bbb{Z}_{\neq0}(ab \le 0 \land ac \le 0 \land bc \le 0)
$$
and try to derive a contradiction from that. So we suppose we have $a, b, c \in \Bbb{Z}_{\neq0}$ such that 
$$
ab \le 0 \land ac \le 0 \land bc \le 0
$$
and we observe that none of the products can be zero so we have
$$
ab < 0 \land ac < 0 \land bc < 0
$$
from which we can conclude that $a$ and $b$ have opposite signs and that $a$ and $c$ have opposite signs, so that $b$ and $c$ must have the same sign contradicting $bc < 0$.
However, what we did in the end game of this proof is a direct proof of:
$$
\forall a, b, c \in \Bbb{Z}_{\neq0}(ab > 0 \lor ac > 0 \lor bc > 0)
$$
thought of as:
$$
\forall a, b, c \in \Bbb{Z}_{\neq0}(\lnot(ab > 0 \lor ac > 0) \Rightarrow bc > 0)
$$
without any need for a proof by contradiction.
Why do teachers encourage illusory non-constructive proofs?
A: So simple to answer but also so easy to trick those who want it to be difficult.
The Simple Answer is:-
Regardless of the sign of a, b, or c, (Oxford comma included to negate the possibility of some smart alec trying to imply boolean logic in this answer) the product will always be positive when like signed elements are multiplied.
As there are only 3 elements, at least one of the products is therefore positive.
A: 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). 
A: Assume we have some nonzero numbers such that each pairwise product is negative. Therefore we don't have two positive numbers (their product would be positive), and we don't have two negative numbers (their product would be positive), so we have at most one positive and one negative number, for a total of two numbers. Contradiction to "we have three numbers". 
