# 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

Updated on November 16, 2020

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