Prove there is a real number whose cube is 3

**ardam** · April 15th 2009, 11:48 AM

considering x is an element of the real numbers and x^3 < 3
prove that there exists a real number 'alpha' such that
'alpha'^3 = 3

**Showcase_22** · April 15th 2009, 01:20 PM

What does the first part ave to do with the proof?

Here's what I did:

Define $f(x)=x^3-3$ .

$f(1)=1-3=-2$
$f(2)=8-3=5$

By algebra of continuity, f(x) is continuous. You can prove this using the delta-epsilon definition.

By the intermediate value theorem $\exists \ \alpha \in \ \mathbb{R}$ s.t $f(\alpha)=\alpha^3-3=0 \Rightarrow \alpha^3=3$ .

You can prove that $\alpha$ is unique using the mean value theorem. It does not mention in the question that this is the case so I have no included it in my answer.

Hope this helps!

**Jhevon** · April 15th 2009, 01:24 PM

Originally Posted by Showcase_22

What does the first part ave to do with the proof?

i believe, based on the way the question was presented, that they want you to use the concept of "supremum". you have a set of real numbers that is bounded above. so by the completeness axiom....

**Showcase_22** · April 15th 2009, 01:33 PM

$\exists \ \alpha: \ 3^{\frac{1}{3}}- \epsilon < \alpha< 3^{\frac{1}{3}}$

$3^{\frac{1}{3}}-\epsilon<\alpha<3^{\frac{1}{3}}<3^{\frac{1}{3}}+\e psilon$

$-\epsilon<\alpha -3^{\frac{1}{3}}< \epsilon \Rightarrow \ |\alpha-3^{\frac{1}{3}}|< \epsilon$

I've never done it this way before. Is it like that?

**Jhevon** · April 15th 2009, 01:48 PM

Originally Posted by Showcase_22

$\exists \ \alpha: \ 3^{\frac{1}{3}}- \epsilon < \alpha< 3^{\frac{1}{3}}$

$3^{\frac{1}{3}}-\epsilon<\alpha<3^{\frac{1}{3}}<3^{\frac{1}{3}}+\e psilon$

$-\epsilon<\alpha -3^{\frac{1}{3}}< \epsilon \Rightarrow \ |\alpha-3^{\frac{1}{3}}|< \epsilon$

I've never done it this way before. Is it like that?

yes. just say that the completeness axiom tells us we have a supremum, call it $\alpha$ and claim that $\alpha = \sqrt[3]{3}$ so that $\alpha^3 = 3$ , as desired. then you go on to show, as you did (stating that $\epsilon > 0$ ), that $\alpha \to \sqrt[3]{3}$ , proving the claim

indeed, for this type of problem, what you did (using the intermediate value theorem) is more conventional. but it can also be done by the above method, in the first part of a beginning analysis class, before IVT even comes on the scene

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