# Proofs involving real numbers

1. Oct 16, 2012

### tehdiddulator

1. The problem statement, all variables and given/known data
Let x $\in$ ℝ
Prove that if 3x$^{4}$+1≤x$^{7}$+x$^{3}$, then x > 0

2. Relevant equations

None

3. The attempt at a solution
Assume
3x$^{4}$+1≤x$^{7}$+x$^{3}$
then 0 ≤ -3x$^{4}$-1≤x$^{7}$+x$^{3}$
Then I assumed that each was greater than or equal to 0, which I thought gave the desired result. No examples in the book to really guide me...any help would be greatly appreciated.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 16, 2012

### clamtrox

That can't be right, no matter what value x has.

You might want to think it like this instead: If x≤0, then x7+x3 is always smaller than 1+3x4. Why is this?

3. Oct 16, 2012

### tehdiddulator

Would it be because both powers of x are odd, which will mean it is always negative? So proving the contapositive would be an easier route than a direct proof?

4. Oct 16, 2012

### Staff: Mentor

To be clearer, say what "it" refers to.

5. Oct 16, 2012

### tehdiddulator

Still not entirely sure if I did it correctly....this class is giving me such a headache. Not sure if its the book or just my lack of understanding but some of these topics covered are going straight over my head...

Let x$\in$ℝ. If 3x$^{4}$+1 ≤ x$^{7}$ + x$^{3}$, then x > 0.
Proof
Contrapositive.
If x≤ 0, then 3x$^{4}$+1> x$^{7}$ + x$^{3}$
Assume x ≤ 0 for x $\in$ ℝ.
Since for all values of x less than or equal to zero would produce a true statement.
Since 3x$^{4}$+1 < 0 for x ≤ 0 and 0 > x$^{7}$ + x$^{3}$ for x ≤ 0.

Would this suffice as a proof? It seems like I don't know the tools to know how to get there, it always seems obvious when I look at the answer key, but have really no idea how to get there.

6. Oct 16, 2012

### tehdiddulator

Oh...and I would just like to point out how the book does this type of problem. Seems incredibly non-intuitive to me.

Let x $\in$ ℝ. If x$^{5}$ -3x$^{4}$+2x$^{3}$-x$^{2}$ +4x - 1 ≥ 0, then x ≥ 0.
Proof.
Assume that x < 0. Then x$^{5}$ < 0, 2x$^{3}$ < 0, and 4x < 0. In addition, -3x$^{4}$ < 0 and -x$^{2}$ < 0.

Thus x$^{5}$ -3x$^{4}$+2x$^{3}$-x$^{2}$ +4x - 1 < 0 - 1 < 0.
as desired.

This is what I have to go on for an example...there has got to be a better way to do these proofs. I'm just regurgitating the same problem, except with different numbers and yet I have no idea what I'm doing.

7. Oct 16, 2012

### Staff: Mentor

You can omit the sentence above. Instead, show using inequalities why this is so.

For any real x, 3x4 + 1 ≥ 1 > 0
For any x ≤ 0, x7 ≤ 0 and x3 ≤ 0, hence x7 +x3 ≤ 0
Then for any x ≤ 0, 3x4 + 1 > x7 +x3.

8. Oct 17, 2012

### clamtrox

Yes I think this is one of the most obvious cases where indirect proof is easier than direct proof. For a direct proof, you'd need to solve the equation 1+3x4-x3-x7=0.