1

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.

kccu
  • 20,808
  • 1
  • 22
  • 41
  • 2
    The argument starts with a $t$ that's smaller than $s$ and shows it is an upper bound, so $s$ can't be the least upper bound, which is your contradiction. – Ethan Bolker Jan 23 '19 at 00:46

2 Answers2

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.

kccu
  • 20,808
  • 1
  • 22
  • 41
  • 1
    $s$ is certainly not equal to $2$. Don't understand your first sentence at all. – Matt Samuel Jan 23 '19 at 00:52
  • 1
    Sorry, typo. I meant $s^2$ of course. – kccu Jan 23 '19 at 00:54
  • Thanks for your answer. Just to clarify, from the negated statement we assume that $s$ is the least upper bound for $A$; that is $s=\sqrt{2}$. Then take $t\lt{s}$, so $t$ is not an upperbound. Therefore, we get a contradiction when we conclude that $t^2\gt{2}$; that is, $t\gt{\sqrt{2}=s}$. Is this correct? –  Jan 23 '19 at 01:38
  • If you are trying to prove $\sqrt{2}$ exists, you cannot speak of $\sqrt{2}$. It will turn out that $s=\sqrt{2}$, but the only way we have to write this is $s^2=2$. However, if you simply define $s$ as the least upper bound of $A$, you don't yet know that $s^2=2$; that is precisely what we have to prove. Now you assume $s^2>2$ and define $t=s-\frac{s^2-2}{2s}$, which is certainly less than $s$. Then $t$ is not an upper bound, since $s$ is the least upper bound and $t$ is smaller. However, $t^2>2$ implies $t$ is an upper bound for $A$ ( but it doesn't imply $t>s$). That's the contradiction. – kccu Jan 23 '19 at 13:48
0

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 $x^2>t^2>2$.

Hagen von Eitzen
  • 374,180
  • Thank you for your answer. If the latter does not mean that $t$ is not an upper bound for $A$, then what is it? I don't understand what it is from what you've stated. –  Jan 23 '19 at 02:10
  • @James $t$ is an upper bound for $A$, but this answer is pointing out that that is not an immediate consequence of the fact $t^2>2$. To show $t$ is an upper bound for $A$, you have to show that if $x \in A$ then $x \leq t$. This is done by contradiction: suppose $x \in A$ but $x>t$; then since $t$ and $x$ are both positive, $x^2>t^2>2$ so in fact $x \notin A$. – kccu Jan 23 '19 at 13:52