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

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- Apr 15th 2009, 11:48 AMardamProve there is a real number whose cube is 3
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 - Apr 15th 2009, 01:20 PMShowcase_22
What does the first part ave to do with the proof?

Here's what I did:

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

$\displaystyle f(1)=1-3=-2$

$\displaystyle f(2)=8-3=5$

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

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

You can prove that $\displaystyle \alpha$ 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! - Apr 15th 2009, 01:24 PMJhevon
- Apr 15th 2009, 01:33 PMShowcase_22
$\displaystyle \exists \ \alpha: \ 3^{\frac{1}{3}}- \epsilon < \alpha< 3^{\frac{1}{3}}$

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

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

I've never done it this way before. Is it like that? - Apr 15th 2009, 01:48 PMJhevon
yes. just say that the completeness axiom tells us we have a supremum, call it $\displaystyle \alpha$ and claim that $\displaystyle \alpha = \sqrt[3]{3}$ so that $\displaystyle \alpha^3 = 3$, as desired. then you go on to show, as you did (stating that $\displaystyle \epsilon > 0$), that $\displaystyle \alpha \to \sqrt[3]{3}$, 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