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

  1. #1
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    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 !!
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  2. #2
    Super Member Showcase_22's Avatar
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    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.
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  3. #3
    MHF Contributor matheagle's Avatar
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    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].
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  4. #4
    Super Member Showcase_22's Avatar
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    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.
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