Is 1 Included in the Mathematical Set S?

[Bachelier]

In one of my old notes, I was reviewing this proof. It started with:

Consider $S = \{ x \in \mathbb{R} \ | \ x^2 = 2, \ x > 0\}$, then S is not empty because $1 \in S$

Why is 1 in the set?

Thanks.

 

[lavinia]

In one of my old notes, I was reviewing this proof. It started with:

Consider $S = \{ x \in \mathbb{R} \ | \ x^2 = 2, \ x > 0\}$, then S is not empty because $1 \in S$

Why is 1 in the set?

Thanks.

This isn't true. The square of 1 is not 2

 

[flatmaster]

x^2 = 2 doesn't seem like a particular interesting thing to look at. I can't help but notice that there are two "2's" in your expression. Perhaps you were looking at a specific case of the set x^n = n. The number 1 would be in this set for the case that n=1.

 

[micromass]

In one of my old notes, I was reviewing this proof. It started with:

Consider $S = \{ x \in \mathbb{R} \ | \ x^2 = 2, \ x > 0\}$, then S is not empty because $1 \in S$

Why is 1 in the set?

Thanks.

It's probably a typo and it should be

$$S=\{x\in \mathbb{R}~\vert~x^2\leq 2,~x>0\}$$

I bet they end up taking the supremum of S.

 

[Bachelier]

It's probably a typo and it should be

$$S=\{x\in \mathbb{R}~\vert~x^2\leq 2,~x>0\}$$

I bet they end up taking the supremum of S.

Indeed. Please see the attached file for the complete proof.