# Thread: issue starting a prrof

1. ## trouble starting a proof

So the claim is: If X^3+4X>0, then 5x+9>6

We are supposed to show our scratch work which generally means he wants us to figure it out algebraicly and then use our work backwards to form the proof. However, i'm not sure how to start. We don't have any theorems that apply so it has to be something simple i'm just not thinking of.

Any ideas?

2. ## Re: trouble starting a proof

Originally Posted by shmiks
So the claim is: If X^3+4X>0, then 5x+9>6

We are supposed to show our scratch work which generally means he wants us to figure it out algebraicly and then use our work backwards to form the proof. However, i'm not sure how to start. We don't have any theorems that apply so it has to be something simple i'm just not thinking of.

Any ideas?
Well, observe that $\displaystyle 5x+9>6$ holds if and only if $\displaystyle x>-1$, so if we can prove that $\displaystyle x^3+4x>0$ implies that $\displaystyle x>-1$ we are done.

Proceed by contradiction, assume $\displaystyle x^3+4x>0$ and suppose that $\displaystyle x\le -1$, then $\displaystyle x^3+4x\le -5$ which is a contradiction. hence if $\displaystyle x^3+4x>0$ then $\displaystyle x > -1$ ...

.

3. ## Re: issue starting a prrof

Note that x3 + 4x = x(x2 + 4).

For any real number (and thus rational number or integer) x, x2 ≥ 0. Thus x2 + 4 ≥ 0 + 4 = 4 > 0.

so x3 + 4x > 0 implies x > 0 (we can divide both sides by the positive number x2 + 4).

Since x > 0 > -3/5 (and 5 > 0), 5x > -3. Add 9 to both sides, and you're done.

4. ## Re: issue starting a prrof

@ zzephod i actually completely get that and i really wish i could do it that way but he said not to use contradictions bc that is the next section

@deveno ok that helps a bit, i like the x^2 part quite a lot. My question is first how did you get rid of the x^2 im not quite following through that part. And second what about the second phrase of the claim? How does that tie in?

5. ## Re: issue starting a prrof

I'm using the following fact:

If ab > 0 and b > 0, then a > 0. Because:

b > 0 implies 1/b > 0, which implies a = (1/b)(ab) > (1/b)(0) = 0.

So if x3 + 4x = x(x^2 + 4) > 0 and x2+4 > 0, then x > 0 (using a = x, b = x2+4).

Basically, if number*number > 0, then either we have positive*positive, or negative*negative. But x2+4 is NEVER negative, its the sum of two POSITIVE numbers, so it is positive. So we must have positive*positive, which means "the other factor" (x, in this case) is positive, too.

Finally, at the end: x > 0 (we just went over that), and 0 > -3/5 (which is equivalent to 5 > -3). So 5x > 5*0 > -3 (or: x > 0 > -3/5). This uses the TRANSITIVE property of ">" namely:

If a > b, and b > c, then a > c. From 5x > -3, we quickly arrive at:

5x + 9 > -3 + 9 = 6, and we're done.