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Math Help - 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 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.
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  3. #3
    MHF Contributor matheagle's Avatar
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    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].
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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: 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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