# Thread: Do not quite understand this simple proof

1. ## Do not quite understand this simple proof

Claim: There exists a number $\displaystyle \alpha$ such that $\displaystyle \alpha ^2 = 2$

Proof:
So, we look at the set $\displaystyle S = \{x \in R | x^2 < 2\}$. We want to consider $\displaystyle \alpha = sup(S)$. Claim that $\displaystyle \alpha ^2 = 2$, and prove by contradiction.

There are two possibilites: $\displaystyle \alpha ^2 < 2$ or $\displaystyle \alpha ^2 > 2$.

For $\displaystyle \alpha ^2 < 2$: Let $\displaystyle n \in R$ be large enough so that $\displaystyle n > \frac{2\alpha + 1}{2 - \alpha ^2}$ (<---Do not understand choice of n here). Then $\displaystyle \frac{1}{n} < \frac{2 - \alpha^2}{2\alpha + 1}$. Then

$\displaystyle (\alpha + \frac{1}{n})^2 = \alpha^2 + \frac{2\alpha}{n} + \frac{1}{n^2} < \alpha^2 + \frac{2\alpha}{n} + \frac{1}{n} = \alpha^2 + \frac{2\alpha + 1}{n}$

So $\displaystyle \alpha^2 + \frac{2\alpha}{n} < \alpha^2 + (2\alpha + 1) \frac{2 - \alpha^2}{2\alpha + 1} = 2$

But then $\displaystyle \alpha + \frac{1}{n} \in S$, and $\displaystyle \alpha + \frac{1}{n} > \alpha$, contradicting the fact that $\displaystyle \alpha$ is an upper bound for S.

This isn't the complete proof. You then have to use P.B.C. for the other case, $\displaystyle \alpha > 2$. But, my problem is that I don't understand the choice of n here. Of course it makes sense looking at the proof, but how does one know how to choose n for any other problem, say, for $\displaystyle \alpha^3 = 4$?

2. I rather suspect that this choice of n was done by "trial and error"! (Many proofs are created by trial and error, then written out with showing the "errors".) Of course, it is not really the "n" that is important- it is $\displaystyle \frac{1}{n}= \frac{2n+1}{2- \alpha^2}$ that we want. We want to prove that if $\displaystyle \alpha^2< 2$, then $\displaystyle \alpha$ is not an upper bound for the set given. $\displaystyle 2- \alpha^2$ is the amount by which $\displaystyle \alpha^2$ is less than 2 and the denominator was chosen so that $\displaystyle (\alpha+ \frac{2-\alpha^2}{2\alpha+1})^2$ is still less than 2.

You say "it makes sense looking at the proof" and I suspect that they just tried $\displaystyle 2- \alpha^2$ divided by something to get a fraction of the interval and tried different divisors until they got something that worked.

Here's how I would do that, with a slightly different "estimate".

Suppose $\displaystyle \alpha^2$ is less than 2. Then we must have $\displaystyle \alpha^2= 2- h$ for some h. Use that h to find some number larger than $\displaystyle \alpha$ whose square is still less than 2.

First note that $\displaystyle 1.4^2= 1.96< 2$ and $\displaystyle 1.5^2= 2.25> 2$ so we can assume that $\displaystyle 1.4< \alpha< 1.5$ and that $\displaystyle h< 2- 1.96= 0.04$.

What about the number half way between? $\displaystyle (\alpha+ h/2)^2= \alpha^2+ 2(h/2)\alpha+ h^2= \alpha^2+ h\alpha+ h^2= \alpha^2+ h(\alpha+ h)$. No, that won't work- $\displaystyle \alpha+ h> 1.4> 1$, $\displaystyle (\alpha+ h)h> h$ so $\displaystyle (\alpha+ h)^2> \alpha^2+ h= 2$.

What about $\displaystyle \alpha+ h/3$? $\displaystyle (\alpha+ h/3)^2= \alpha^2+ 2(h/3)\alpha+ h^2/9= \alpha^2+ h(\frac{2}{3}\alpha+ h/9)$. $\displaystyle \frac{2}{3}\alpha> \frac{2}{3}1.4= .933$ and $\displaystyle \frac{2}{3}\alpha< \frac{2}{3}(1.5)= 1.0$. Since we have to add $\displaystyle h^2/9$, that still isn't good enough- we might have $\displaystyle \frac{2}{3}\alpha+ h^2/9$ larger than 1 so that $\displaystyle (\alpha+ \frac{h}{3})^2= \alpha^2+ h(\frac{2}{3}\alpha+ \frac{h^2}{9})> \alpha+ h= 2$.

Okay, what about $\displaystyle \alpha+ h/4$? $\displaystyle (\alpha+ h/4)^2= \alpha^2+ \frac{1}{2}h\alpha+ \frac{h^2}{16}$$\displaystyle = \alpha^2+ h(\frac{\alpha}{2}+ \frac{h}{16}$.

Since $\displaystyle 1.4< \alpha< 1.5$, $\displaystyle .7< \alpha< .75$ and since $\displaystyle h< .04$, $\displaystyle \frac{h}{16}<0.0025$, $\displaystyle \frac{\alpha}{2}+ \frac{h}{16}< 0.7525< 1$ so $\displaystyle \alpha^2+ h(\frac{\alpha}{2}+ \frac{h}{16})< \alpha^2+ h= 2$. Yes, we have a number, larger than $\displaystyle \alpha$ whose square is still less than 2 contradicting the fact that \alpha is the least upper bound of that set.

3. People usually find these proofs in different way and then when they put them in this form, they make less sense.

The important parts of this proof are that you are choosing $\displaystyle \alpha = \sup(S)$. You know that $\displaystyle \alpha^2$ should equal 2, but you want to assume on the contrary. Once you do that, you want to pick a very small number, say $\displaystyle \epsilon$ and add it to $\displaystyle \alpha$. You want to show that this is also in the set S and thus this contradicts your choice of $\displaystyle \alpha$.

Given $\displaystyle \alpha^2<2[/Math], you want [Math](\alpha + \epsilon)^2 = \alpha^2+2\alpha\epsilon +\epsilon^2 < 2$. Since $\displaystyle \epsilon$ is supposedly very small, it is definitely $\displaystyle < 1$, so $\displaystyle \epsilon^2 < \epsilon$. So you can sub this to get

$\displaystyle \alpha^2+2\alpha\epsilon +\epsilon^2 < \alpha^2 + (2\alpha+1)\epsilon.$

You want to make this $\displaystyle <2$ so you might as well try to get it to look like $\displaystyle \alpha^2 +(2 - \alpha^2)$. comparing the two expressions, you want to make $\displaystyle (2\alpha+1)\epsilon < (2 - \alpha^2)$.

Well, just divide out to get $\displaystyle \epsilon < \frac{2-\alpha^2}{2\alpha+1}$. when you sub this in for epsilon, you will get what you want. All this shows that you didn't have to choose and n and stuff, which makes it somewhat better to understand I think.

4. The choice of $\displaystyle n$ looks in first instance a bit arbitrary, but it's chosen such that the little trick works...

We can pull the same trick for all $\displaystyle c> 0$ and showing the existence of $\displaystyle \alpha$ s.t $\displaystyle \alpha^3= c$

We'll decide what we want $\displaystyle n$ to be afterwards...

Let $\displaystyle S = \left\{x\in R: x^3< c\right\}$ and $\displaystyle \alpha = \sup(S)$. And let's assume $\displaystyle \alpha^3< c$.

First observe that $\displaystyle \alpha\geq 0$ and ...

$\displaystyle (\alpha+\frac{1}{n})^3< \alpha^3+\frac{1}{n}(3\alpha^2+3\alpha+1)$

So that if we want our little trick to work we need $\displaystyle \frac{1}{n}< \frac{c-\alpha^3}{3\alpha^2+3\alpha+1}$

And since $\displaystyle \alpha \geq 0$, this number is positive and it's obvious we can find such $\displaystyle n$.

Now we have $\displaystyle (\alpha+\frac{1}{n})^3< \alpha^3+\frac{1}{n}(3\alpha^2+3\alpha+1) < \alpha^3+\frac{c-\alpha^3}{3\alpha^2+3\alpha+1}(3\alpha^2+3\alpha+1 )=c$

since $\displaystyle \alpha+\frac{1}{n}> \alpha$ this contradicts the fact that $\displaystyle \alpha= \sup(S)$

We may do something likewise in the case $\displaystyle \alpha^3> c$

5. More general case. Let $\displaystyle \alpha \in \mathbb{R}$, you want to show $\displaystyle \alpha^n = c$ for some real and positive c and let it be >1.

You let the set be $\displaystyle S = \{x \in \mathbb{R}\mid x^n<c\}$. Let $\displaystyle \alpha = \sup(S)$.

Then we want $\displaystyle (\alpha+\epsilon)^n < c.$
We need a little result before that. We have $\displaystyle \epsilon<1$. Then We have $\displaystyle \epsilon^2<\epsilon$. So by induction you will have that $\displaystyle \epsilon^k<\epsilon$ for all $\displaystyle k \in \mathbb{N}$.

$\displaystyle \displaystyle{(\alpha+\epsilon)^n = \alpha^n + \sum_{k=1}^n \binom{n}{k}\alpha^{n-k}\epsilon^k < \alpha^n+\sum_{k=1}^n \binom{n}{k}\alpha^{n-k}\epsilon = \alpha^n+\epsilon\sum_{k=1}^n \binom{n}{k}\alpha^{n-k} = \alpha^n +\epsilon D}$
where D is the sum. You want $\displaystyle \epsilon D < c - \alpha^n$, so you choose
$\displaystyle \displaystyle{\epsilon < \frac{c-\alpha^n}{D}}$