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
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!
$\displaystyle \exists \ \alpha: \ 3^{\frac{1}{3}}- \epsilon < \alpha< 3^{\frac{1}{3}}$
$\displaystyle 3^{\frac{1}{3}}-\epsilon<\alpha<3^{\frac{1}{3}}<3^{\frac{1}{3}}+\e psilon$
$\displaystyle -\epsilon<\alpha -3^{\frac{1}{3}}< \epsilon \Rightarrow \ |\alpha-3^{\frac{1}{3}}|< \epsilon$
I've never done it this way before. Is it like that?
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