Re: trouble starting a proof

Re: issue starting a prrof

Note that x^{3} + 4x = x(x^{2} + 4).

For any real number (and thus rational number or integer) x, x^{2} ≥ 0. Thus x^{2} + 4 ≥ 0 + 4 = 4 > 0.

so x^{3} + 4x > 0 implies x > 0 (we can divide both sides by the positive number x^{2} + 4).

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

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?

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 x^{3} + 4x = x(x^2 + 4) > 0 and x^{2}+4 > 0, then x > 0 (using a = x, b = x^{2}+4).

Basically, if number*number > 0, then either we have positive*positive, or negative*negative. But x^{2}+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.