# Thread: proving (x^3)=2 using least upper bound

1. ## proving (x^3)=2 using least upper bound

Okay, my homework is "Prove that there exists a positive real number x such that (x^3)=2."
and I have no clue how I can solve it. sigh.

Is there anyone who can me to solve it using least upper bound property??

Thank you !!

2. I'm pretty sure I sure the exact same problem on a physics forum, here's what I put:

Define $\displaystyle f(x):=x^3-2$

$\displaystyle f(1)=1-2=-1<0$
$\displaystyle f(2)=2^3-2=8-2=6>0$

$\displaystyle f(x)$ is also continous over $\displaystyle \mathbb{R}$ since it's a polynomial.

Hence, by the intermediate value theorem $\displaystyle \exists \ x_0 \in \mathbb{R} \ s.t \ f(x_0)=0$.

Hence $\displaystyle x_0^3-2=0 \Rightarrow \ x_0^3=2$ as required.

3. You should probably add that since $\displaystyle f(x)=x^3-2$ is increasing on [1,2] the point is unique. The intermediate vlaue theorem tells you that there is at least one point satisfying $\displaystyle f(x)=0$ on [0,1].

4. That's quite true, except I thought it was unnecessary since the question wants proof that the "number exists" not that the "number exists and is unique".

However, if i'm going to do something I might as well do it right!

Claim: $\displaystyle f(x)=x^3-2$ is continuous over $\displaystyle \mathbb{R}$.

Let $\displaystyle g(x)=x$.

Let $\displaystyle \epsilon>0$ and define $\displaystyle \delta:=\epsilon$. When $\displaystyle 0<|x-x_0|< \delta$:

$\displaystyle |g(x)-g(x_0)|=|x-x_0|<\epsilon$ as required.

By algebra of continuity $\displaystyle f(x)=(g(x))^3-2$ so $\displaystyle f(x)$ is continuous on $\displaystyle \mathbb{R}$.

Claim: $\displaystyle f(x)=x^3-2$ is strictly increasing.

$\displaystyle f(x)$ is continuous on $\displaystyle \mathbb{R}$ as proven above. Alternatively, $\displaystyle f(x)$ is continuous on $\displaystyle [b,c]$ where $\displaystyle b,c \in \mathbb{R}$ and $\displaystyle c>b>0$.

Subclaim: $\displaystyle f(x)$ is differentiable on $\displaystyle \mathbb{R}$.

$\displaystyle \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim_{h \rightarrow 0} \frac{(x+h)^3-2-x^3+2}{h}=\lim_{h \rightarrow 0} \frac{x^3+3hx^2+3h^2x+h^3-x^3}{h}$ $\displaystyle =\lim_{h \rightarrow 0} 3x^2+3hx+h^2=3x^2$.

Hence a limit exists so the function is differentiable on $\displaystyle \mathbb{R}$.

$\displaystyle f(x)$ is continuous on $\displaystyle [b,c]$ and differentiable on $\displaystyle (b,c)$. By the Mean Value Theorem $\displaystyle \exists \ x_0 \in (b,c) \ s.t \ f'(x_0)=\frac{f(c)-f(b)}{c-b}$.

Since $\displaystyle x_0 \in (b,c)$, $\displaystyle b>c>0$ and $\displaystyle f'(x)=3x^2 \Rightarrow f'(x)>0 \ \forall \ x \in (b,c)$ we get that $\displaystyle \frac{f(c)-f(b)}{c-b}>0 \ \Rightarrow \ f(c)>f(b)$ as required.

If you let (b,c) be subsets of [1,2] then f(x) is injective over [1,2]. Hence if $\displaystyle x_0^3=2$, $\displaystyle x_0$ has to be unique.