Proof By Contrapositive?

Question

I just started a proofs course in college and have been doing great so far with direct proofs (if x and y are both even, then x + y is even - simple stuff like that) ...until this last section hit us...I can't seem to do any of the proofs. We just learned what proof by contrapositive was. Here is one I have no idea about...

If ab > 0 and bc < 0, then ax^2 + bx + c =0 has two distinct real roots.

A couple issues arise...first, what is a distinct real root? She never mentioned it. I know real roots, but distinct...does that mean unique? or what?

Next, I cannot say whether or not a,b, or c is negative. All I know is one or more will have to be to fit both hypothesis.

If I should use proof by contrapositive, well..it simply leaves me stuck at the hypothesis ax^2 + bx + c = 0 does not have two distinct real roots. I still have no idea what to do with that knowledge or how to write it out to even get started with something...

Well I can write the contrapositive. It just doesn't seem to help.

Contrapositive: If ax^2 + bx + c =0 does not have two distinct real roots, then ab <= 0 or bc >= 0.

I think my biggest problem with that is writing out the new hypothesis in mathematical language instead of english. I can't do anything with the English statement. From what you say, it will need to be in the form

j^2 = 0 where j can be a variety of things. I guess my problem is figuring out how to write that j in terms of a, b, and c. In order to get the ax^2, I would think it needs to be in the form [sqrt(a)x + sqrt(c)]^2 = 0 where b = 2sqrt(c)sqrt(a) or something unless we are talking about the difference of squares situation in which case b will be 0...but this just seems to be getting more difficult the more I dig into it.

Hmm..never heard of Viete Relation yet. This is our transition to advanced math course, and the only prereq. is Calculus II, nothing too fancy yet and we have just started proofs in the course (first month or so was spent on formal logic).

I think I can put together a poorly written proof that somewhat makes sense if I can make this one claim...

If ax^2 + bx + c =0 does not have two distinct real roots, then we can put it in the form

[sqrt(a)x + sqrt(c)]^2 = 0.

This will produce ax^2 + 2sqrt(a)sqrt(c) + c = 0 when expanded, so let b = 2sqrt(a)sqrt(c).

From there I can say bc >= 0 since b cannot be negative and neither can c. Again, that's assuming we are dealing with real numbers.

This only works if we assume we have real numbers, but not DISTINCT real numbers. Of course, if we bring nonreal numbers into the picture, I'm all lost again as I haven't dealt with those since 8th grade probably.

Answer 1

Two distinct roots means that they are unique.
x^2 = x * x = 0 is an example of something that doesn't have 2 distinct roots.

Contrapositive
Suppose the statement "if A then B" is true. then the statement "if not B then not A" is also true.

Determine what the A and B are in the proof you are asked to do. Figure out what the negative of these are. Then rewrite what you are asked to prove as the contrapositive.
This will be a good exercise even if it doesn't help you solve the proof. Now you have two ways of looking at the problem. You can either find a way to prove the original problem of the contrapositive.

EDIT:
When does ax^2+bx+c not have 2 distinct real roots?
Hint: examine the quadratic formula.

Answer 2

A quadratic has two roots, these two roots can be the same or not. If they are not the same, then they are distinct( sic!)

Now by Viete relation
x_1+x_2=-b/a and
x_1 * x_2 = c/a
ab>0 and bc<0 implies c/a <0

by your conditions it results that x_1+x_2 <0 and
x_1 *x_2 <0
Now if x_1= x_2 then x_1 *x_2 = x_1^2 >=0 contradiction
So this cannot happen, so the equation has distinctive roots.

edit: another method, you can use delta form i.e
a quadratic has only one root iff delta = 0
Now by your conditions delta=b^2-4ac >0

"This will produce ax^2 + 2sqrt(a)sqrt(c) + c = 0 when expanded, so let b = 2sqrt(a)sqrt(c).

From there I can say bc >= 0 since b cannot be negative and neither can c. Again, that's assuming we are dealing with real numbers.

This only works if we assume we have real numbers, but not DISTINCT real numbers. Of course, if we bring nonreal numbers into the picture, I'm all lost again as I haven't dealt with those since 8th grade probably."

edit: good thinking.
From b = 2sqrt(a)sqrt(c), you can square them and you get
b^2 = 4 ac
But ab>0 and bc<0 implies ac<0. This contradicts b^2 = 4ac.