2 has a square root in $\mathbb{R}$ - proof explanation

Question

Can someone please help me to understand the steps of the proof below.

What do the assumptions become when we use the proof by contradiction in the cases where $s^2\gt{2}$ and $s^2\lt{2}$? Can you please state the theorem in the form of an "if..., then... . " statement and negate it. I think not knowing this is giving me the following confusions.

I don't understand how $t^2\gt{2}$ leads to contradicting that $s$ is a least upper bound. If we first assume that $s^2\gt{2}$ why can't it also be that $t$ has $t^2\gt{2}$. Couldn't it be possible that $t^2\gt{2}$ and $s^2\gt{2}$ with $t\lt{s}$. Im also having trouble with the algebra and where everything is coming from, which again, I feel like this is because of the above confusions. I've been having trouble with this proof for quite some time and can't seem to understand anybody's proof of it!

Theorem: The number $2$ has a square root in $\mathbb{R}$.

Proof:

Let $A = \{x ∈ R : x^2\leq{2}\}$ and note that $A$ is bounded above, for example $u = 5$ is an upper bound. By the least upper bound axiom, there exists a least upper bound for $A$. Set $s$ to be the least upper bound of $A$. We will show that $s^2 = 2$. The proof will be by contradiction.

Suppose first that $s^2 > 2$. Let $\epsilon = \frac{s^2−2}{2s} > 0$ and set $t = s−\epsilon < s$. Then $$t^2 = s^2−2s\epsilon + \epsilon^2 > s^2−2s\epsilon = 2,$$

showing that $t$ is an upper bound for $A$ and contradicting the fact that $s$ is the least upper bound.

On the other hand if $s^2 < 2$ put $\epsilon = \text{min}\{\frac{2−s^2}{ 2s},1\}$ , and set $t = s+ \epsilon > s$. Then $$t^2 = s^2 +2s\epsilon + \epsilon^2 \leq s^2 + (2s+1)\epsilon \leq{2},$$ contradicting the fact that s is an upper bound for A.

Answer 1

The goal is to prove the statement: if $s$ is the least upper bound of $A=\{x \in \mathbb{R} \mid x^2 \leq 2\}$, then $s^2=2$. The proof by contradiction goes by assuming $s$ is the least upper bound of $A$, yet $s^2 \neq 2$, and from these two assumptions derives a contradiction. The assumption $s^2\neq 2$ splits into two possible cases: $s^2 >2$ or $s^2<2$.

In the first case, $s^2>2$, you construct a $t<s$ such that $t^2>2$. The fact that $t^2>2$ implies $t$ is also an upper bound for $A$. This is because if $x \in A$, then $x^2 \leq 2<t^2$ and so $x \leq t$. However, $s$ was assumed to be the least upper bound of $A$, and we have just found another upper bound that is smaller than $s$. This is a contradiction.

Answer 2

The assumption that $s$, the least upper bound of $A$ has the property $s^2>2$ is brought to a contradiction by showing that $s$ cannot be the least upper bound if $s^2>2$. Unfortunately, the proof as written has a bit of a gap there: Starting from $s^2>2$, we construct $t=s-\epsilon$ with $\epsilon$ as given and thereby have

• $t<s$
• $t^2>2$

Immediately, the latter only means $t\notin A$, and not that $t$ is an upper bound for $A$. However, one readily shows that $\epsilon<s$, hence $t>0$, hence if $x>t$ for some $x\in A$, then also $x^2>t^2>2$.