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

**psb01080** · March 21st 2009, 08:01 AM

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 !!

**Showcase_22** · March 21st 2009, 10:28 AM

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

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

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

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

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

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

**matheagle** · March 21st 2009, 10:56 PM

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

**Showcase_22** · March 22nd 2009, 04:46 AM

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

Let $g(x)=x$ .

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

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

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

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

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

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

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

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

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

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

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

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