1. existanse of real number

Hi, I got a question that I'm having trouble with.

Prove that there exists a real number a such that $a^3=5$

I tried to prove it by contradiction,so I assume a is a complex number (x+yi), then expend a^3, but I still cannot see any partern to show the contradiction.
Can anyone please give some suggestions?
Thanks a lot.

2. Since $\mathbb{R} \subset \mathbb{C}$, I don't think you'll have a lot of luck proving that it's not a complex number. In particular, for $z = a + bi$, taking $b = 0$ is a valid choice for the complex number $z$.

Try reading Construction of the real numbers - Wikipedia, the free encyclopedia

Edit: Proving that a number is real requires a definition for a real number. Of course, you could always rely on your natural intuition for what a real number is, but that seems to defeat the purpose a bit in what you're trying to do . There are a few ways to define (or construct) the real number system, and the method by Dedekind cuts doesn't require a substantial amount of mathematical background to understand (ie, it doesn't rely on a precise definition of, say, Cauchy sequences).

Anyway, what I'm trying to get at is that defining a number as real is a bit more involved than defining rational numbers (which seems to be the analogous proof in this case).

3. Continuity of f(x) = x^3 plus the intermediate value theorem?

4. Thank you, I'm still a little bit struggling with it. But you are giving me some idea, I will keep trying, if I still can't get it, I will get back to you.
Thanks again.

5. How to do this depends, as Math Major suggests, on exactly how you define the real numbers and what basic properties of the real numbers you can use. I would suggest showing that the set of all real numbers, x, such that $x^3< 5$, is non-empty and has, say, 2 as an upper bound. By the "least upper bound" property (which is easily proved by using the Dedekind cut definition of the real numbers), that set must have a least upper bound. Now prove that the least upper bound has the property that its cube is equal to 5.

6. Originally Posted by HallsofIvy
How to do this depends, as Math Major suggests, on exactly how you define the real numbers and what basic properties of the real numbers you can use. I would suggest showing that the set of all real numbers, x, such that $x^3< 5$, is non-empty and has, say, 2 as an upper bound. By the "least upper bound" property (which is easily proved by using the Dedekind cut definition of the real numbers), that set must have a least upper bound. Now prove that the least upper bound has the property that its cube is equal to 5.

Hi, thanks a lot for your help. I took your advice. However, I did a lot algebra munipulation, but I still cannot prove 5 is the "least upper bound". I certainly got your idea, great idea, but I still cannot solve it. Could you please help me more with it? Thanks a lot.

7. ??? I didn't suggest that 5 was the least upper bound of any set.

Start with the set of all numbers whose cube is less than 5, as I suggested. That set has an upper bound and so a least upper bound. Call it "x" (it is NOT 5).

One of three things must be true. $x^3> 5$, $x^3< 5$, or $x^3= 5$. You show that $x^3= 5$ by showing that the other two are NOT true.

Suppose $x^3> 5$. Let $d= 5- x^3> 0$. Now show that $(x- \epsilon d)^3> 5$ for some sufficiently small $\epsilon$. That's the hard part- choosing $\epsilon$. If you can do that you will have proved that $x- \epsilon d$ is also an upper bound for the set contrdicting the fact that x is the least upper bound.

Do something similar assuming that $x^3< 5$.

8. I think OP question can be proved also with the intermediate value theorem.

Edit: Sorry JG89!!! I didn't see your post!