Correctness of proof that every positive rational with square $>2$ is an upper bound for those with square $<2$

Question

I would like to know whether my proof makes sense or not, and if not where should it be corrected.

Let $E=\{x \text{ is rational }: x>0 \text{ and } x^2<2\}.$

Claim: Every member of $F=\{x \text{ is rational }: x>0 \text{ and } x^2>2\}$ is an upper bound for set $E$.

Proof:

$F$ is an upper bound of $E$ if all elements of $F$ are greater than or equal to all elements in $E$.

I drew a number line outlining sets $F$ and $E$ in relative positions. My number line shows $E$ being the set greater than $0$ and up to $\sqrt2$ noninclusive and F being the set excluding $\sqrt2$ all the way to infinity.

By contradiction let's assume there is at least one member of $F$ that is not an upper bound for set $E$. This means there exists a rational $y$ s.t. $0<y<\sqrt2$. This means $y$ is now part of set $E$ and the only way it can be an upper bound is if $y=\sqrt2$, which it cannot, based on the rules of set $E$. So we have a contradiction and $y$ has to be an upper bound.

Answer 1

You can prove this directly without contradiction by using the fact that $x^{2}$ is an increasing function if $x > 0$. This means if $t \in F$, we have $t^{2} > 2$. So for any $s \in E$, since $s^{2} < 2$, this gives $s^{2} < t^{2}$. But since $s$ and $t$ are both positive and $x^{2}$ is an increasing function, this implies $s < t$. Since this is true for all $s \in E$, $t$ is an upper bound for $E$, as desired.

Having said that, I think your argument could be tweaked a little bit in the following way: As you said, suppose there is at least one member $y$ of $F$ that is not an upper bound for the set $E$, then this means there is some $g \in E$ such that $y < g$. But $g^{2} < 2$, and $x^{2}$ is an increasing function if $x > 0$, which implies $y^{2} < 2$. But this is a contradiction to the fact that $y \in F$ since $y \in F$ implies $y^{2} > 2$, and we can't have both $y^{2} > 2$ and $y^{2} < 2$ be true.

Answer 2

Your proof is close, but not quite correct. You are correct in observing that $y \neq \sqrt{2}$ implies $y$ cannot be an upper bound of $E$. But this is not yet the desired contradiction because you have already assumed that $y$ is not an upper bound. So far, your hypothesized world is consistent.

You are, however, very close, and one way to obtain a contradiction does depend upon the observation that $y < \sqrt{2}$. Give that a little bit of thought, I'm sure you can figure it out having gotten this far! :D

Answer 3

By contradiction let's assume there is at least one member of $F$ that is not an upper bound for set $E$.

fine

This means there exists a rational $y$ s.t. $0<y<\sqrt2$.

No, it means that that there is a rational $y$ such that $y^2>2$ (because it is from $F$) and there is an $x$ from $E$ such that $y \le x$ is not valid.

The order $\lt$ on $\mathbb{Q}$ is a total order, so that $a \le b$ or $b \le a$ for a pair $(a,b)$ from $\mathbb{Q}$. Because $y \le x$ is not valid we have $x \le y$.

$\mathbb{Q}$ is and ordered field so we have $0\le(y-x)(y+x)$ and therefore $x^2 \le y^2$. From this and $2 \lt x^2$ and transitivity we conclude $2 \le y^2$. Using antisymmetry we get $y^2=2$. That contradicts $y^2 \lt 2$

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A direct proof:

If $e \in E$ and $f \in F$ then we have

$$(f+e)(f-e)=f^2-e^2\ge 0$$ and therefore $f-e\ge 0$ and this is equivalent to $f \ge e$. (we have to apply all the already mentioned axioms to get this result). So $f$ is an upper bound to $E$.