Proof explanation of $``\exists x\in\mathbb{R}$ with $x^2=2"$

Question

Can someone please help me break down the proof below from $(*)$ onwards. I'm lost at what is going on and where the proceeding steps are coming from. Is this a proof by contradiction? Why are we assuming $M^2\lt 2 , M^2\gt 2$, and choosing $\delta$ to be the minimum of $\frac{2-M^2}{5}$ and $1$?

Theorem 2.13 (Square root 2 exists). There exists $x\in\mathbb{R}$ with $x^2=2.$

Proof.

Define $$A=\{y\in\mathbb{R}|y^2 \leq{2}\}.$$

As $0^2\leq{2}$, we have $0\in{A}$, so $A$ is non-empty. Now suppose $y\in\mathbb{R}$ has $y\geq{2}$, then $y^2\geq{4}$, so $y\notin{A}$. Thus, $2$ is an upper bound for A and hence A is bounded above. Therefore, by the completeness axiom, $M=\sup(A)$ exists. Note that $M\geq{1}$ as $1\in{A}$ and $M\leq{2}$ is an upper bound for A.

$(*)$

Suppose $M^2\lt{2}$. Choose $\delta\gt{0}$ with $\delta\lt $ min$(\frac{2-M^2}{5},1)$, and then define $ y=M+\delta$. As $\delta\lt{1}$, we have $\delta^2 \lt{\delta}$ and so

$$y^2=(M+\delta)^2=M^2+2M\delta+\delta^2\leq M^2+4\delta+\delta=M^2+5\delta\lt 2.$$

Thus $y\in A$, but this is a contradiction as $y\gt M$. Therefore $M^2\geq 2$.

Suppose $M^2\gt 2$. Choose $\delta \gt 0$ with $\delta\lt $min$(M,\frac{M^2-2}{2M})$ so that $M^2-2M\delta \gt 2$ and $M-\delta\gt 0$. Then,

$$(M-\delta)^2=M^2-2M\delta+\delta^2\geq M^2-2M\delta\gt 2.$$

Now if $y\geq (M-\delta)$, then $y^2\geq (M-\delta)^2 \gt 2$ so $y\notin A$. Thus $M-\delta$ is an upper bound for $A$, contradicting the fact that $M$ is the supremum of $A$

Since $M^2 \lt 2$ and $M^2\gt 2$ both lead to contradictions, we conclude that $M^2=2$, as required.

Answer 1

The last part is a proof by contradiction, using the fact that $M^2$ and $2$ are related either by $M^2 < 2$ or $2 < M^2$ or $M^2 = 2$, exclusively. In both cases, we contradict the supremum property of $M$; that $M$ is greater than or equal to every number in $A$, and that it is the least such number with that property.

First, assuming $M^2 < 2$, we construct a $y > M$ but $y \in A$, which is impossible if $M$ is the supremum. It's a "little" more than $M$, so add a positive $\delta$ "fudge factor." We have complete control of how little or big that $\delta$ is; i.e. $y = M + \delta$. So, if we want $y\in A, y^2 < 2$, we want $(M + \delta)^2 < 2$. We will work towards this inequality "backwards", although the engineering of the different number choices are much clearer in this direction.

Start with $(M + \delta)^2$, and expand it. Two things happen in $$M^2 + 2M\delta + \delta^2 \leq M^2 + 5\delta$$

First, we use $M \leq 2$ to get $2M\delta \leq 4\delta$. Then, we can choose for $\delta < 1$ so that $\delta^2 < \delta$.

To get this inequality: $$M^2 + 5\delta < 2$$

We can actually work backwards, and choose $\delta < \frac{2 - M^2}{5}$. If we choose $\delta < \min(\frac{2 - M^2}{5}, 1)$, then both the inequalities we need will be true. It's a lot more clear where they come from working backwards, though.

I hope this helps you understand the $M^2 > 2$ part, as well, which works on a similar reasoning scheme.

Answer 2

Do you know the trichotomy law of inequality? It says that for any two numbers $a,b$, either $a < b$, or $a > b$, or $a=b$.

So now if you can prove that $a<b$ leads to a contradiction, and that $a > b$ leads to a contradiction, the only possibility left must be true, namely $a=b$. Apply this with $a=M^2$ and $b=2$.

Answer 3

Yes, it is a proof by contradicion. The author proves the supremum $M$ satisfies the equality $M^2=2$ by deducing a contradiction in each of the other two cases, $M^2<2$ and $M^2>2$.

In the first case, he/she obtains a contradiction by finding an element $y=M+\delta$ which is both in $A$ but greater than $sup A$.

In the second case, he/she finds an upper bound for $A$ which is smaller than the least upper bound.

$\delta<\min\left(\frac{2-M^2}{5},1\right)$ simply means that we have both $\delta< \frac{2-M^2}{5}$ and $\delta<1$. These conditions are technically necessary (i.e. $\delta$ must be small enough) to deduce the contradiction.

Answer 4

Indeed, the proof is a bit confusing as it sprawls out with variable definitions and claims. Here we want to calmly prove part of the argument in theorem 2.13, highlighting some of the thinking behind the mechanics.

Proposition: Let $r \gt 0$ be a real number satisfying $r^2 \gt 2$.
Then there exists another real number $s$ with $0 \lt s \lt r$ satisfying $s^2 \gt 2$.
Proof
We can write any such number $s$ as $r - \delta$ with $\delta \gt 0$. So we are looking for any $\delta$ 'that works', leading us to analyze

$$\tag 1 s^2 = (r - \delta)^2= r^2 - 2\delta r + \delta^2$$

Now certainly if $r^2 - 2\delta r = 2$ then $s^2 = 2 + \delta^2 \gt 2$. Solving for $\delta$,

$$\tag 2 \delta = \frac{r^2 - 2}{2r}$$

So $\delta$ is positive and $s = r - \delta \lt r$ is guaranteed. We also have

$\quad s \gt 0 \, \text{ iff } \, 2r^2 - (r^2 -2) \gt 0 \, \text{ iff } \, r^2 + 2 \gt 0$

and so defining

$$\tag 3 s = r - \frac{r^2 - 2}{2r}$$

satisfies the requirements of proposition, which completes the proof. $\quad \blacksquare$

There are any number of ways of putting together the proof of the OP's theorem, one way being to incorporate the above proposition.

Note: Using and extending the above logic, an algorithm can be implemented to calculate square roots - the Babylonian method.

Answer 5

I'm having a hard time following some of these proof attemps, but what about this:

Let

$f(x) = x^2 - 2; \tag 1$

then

$f(0) = -2, \; f(2) = 2, \tag 2$

and $f(x)$ is clearly continuous (easy to prove). Thus by the itermediate value theorem,

$\exists x_0 \in (0, 2), \; f(x_0) = 0 \Longrightarrow x_0^2 - 2 = 0 \Longrightarrow x_0 = 2. \tag 3$