Proof $\sqrt2$ is supremum of $x^2 <2 $

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The wording is a bit imprecise, but the reasoning is correct.

You start with: "Let $u$ be another upper bound for $M$.

Assumption: $\sqrt{2}>u$"

Right now I could set $u=10$ and the proof ends with "and so $u$ can't be another upper bound of $M$", which is obviously wrong.

So start the proof with "Let $u$ be another upper bound for $M$ that fulfills $\sqrt{2}>u$." and the rest is fine.

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franz3
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franz3

Updated on November 16, 2020

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  • franz3
    franz3 almost 3 years

    Let $M = \{x \in \mathbb{R}| x^2 <2\}$. Prove that $\sqrt2$ is the supremum of M.

    Proof:

    $\forall x \in M: x^2 <2 \Leftrightarrow |x| <\sqrt 2 \Leftrightarrow -\sqrt2 < x < \sqrt 2 \Rightarrow \sqrt 2 $ is an upper bound for $M$.

    Hypothesis: $\sup(M) = \sqrt 2$.

    Let $u$ be another upper bound for $M$.

    Assumption: $\sqrt2 >u$.

    Let $\epsilon:= \sqrt2 - u >0$. By the theorem of Eudoxos: $\forall \epsilon > 0$ $ \exists m \in \mathbb{N}: \dfrac{1}{m} < \epsilon$, that means: $\dfrac{1}{m} < \sqrt2 - u \Leftrightarrow u< \sqrt2 - \dfrac{1}{m}$. Since $m>0 \Rightarrow \dfrac{1}{m} >0 \Rightarrow \sqrt2 - \dfrac{1}{m} \in M$ and so $u$ can't be another upper bound of $M$. Then $\sqrt2 \leq u \Rightarrow \sqrt2 = \sup(M)$.

    Is this proof correct?

  • TonyK
    TonyK almost 4 years
    I think your instances of $2>\sqrt u$ should be $\sqrt 2>u$ (or $2>u^2$).
  • Nurator
    Nurator almost 4 years
    Oh yeah sorry, I copied incorrectly. Thank you!