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Most proofs that I read start the part of proving the contradiction (that if a is the supremum of the set of numbers whose squares are less than or equal than 2, then $a^2<2$ will led to a contradiction) by saying

let $e$ by a number such that $0<e<1$ and $e<(2-a^2)/(2a+1)$

Here's my problem: where does the number on the RHS came from? Are you supposed to do lots of trial and error until you find something that "fits" the algebraic manipulations? Wouldn't be easy, for example, to take $e$ to be $(2-a^2)/2$?

Allonsy
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  • One usually works backwards from the end goal doing what is called "scratch work." In scratch work, you can violate any math rules that you want to get a formula. Then, you start the actual proof with the magic formula that you derived from the scratch work and show (using only valid mathematics) that it satisfies the desired properties. – Michael Burr Mar 09 '17 at 01:29
  • Almost exactly the same question: http://math.stackexchange.com/questions/1319580/proof-that-square-root-of-2-exists-and-is-a-real-number?rq=1. See also http://math.stackexchange.com/questions/1415235/prove-the-existence-of-the-square-root-of-2?rq=1 http://math.stackexchange.com/questions/1314123/existence-of-square-root-of-2?rq=1 – mathematician Mar 09 '17 at 01:36
  • If you look at the steps, the $e<(2-a^2)/(2a+1)$ is exactly the $e$ you need if you start from $(a+e)^2<2$. You'd start with $(a+e)^2<2$ then see that you needed a special $e$ satisfying this condition to get the proof to work. – mathematician Mar 09 '17 at 01:42
  • That e won't do what you want. Yes, you are supposed to do trial and error. Sort of. – fleablood Mar 09 '17 at 01:42
  • however the e that i've choose is actually gt a^2 and less that 2. Isn't strange that I cannot prove someting that I know that is true? – Allonsy Mar 09 '17 at 02:03

2 Answers2

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This is a small part of an overall proof. What we want to prove demonstrate is:

Lemma: the set $A = \{s\in \mathbb Q|s^2 < 2\}$ has no maximal element. In other words. For any $a \in A$ there will always another $q\in A$ so that $q > a$. (and then there will be another $q_1 \in A$ so that $a < q < a_1$ etc.

And as $1^2 = 1 < 2; 1 \in A$ so we only have to prove this for values $a\ge 1$. (If $a \in A$ and $a < 1$ then if $q=1$ we know $a < q$ and $q\in A$ and there is nothing to prove.)

If $0 < a < q$ then $0 < a^2 < q^2$ so this is what we want to prove.

Lemma: If $a > 0$ is rational and $a^2 < 2$ there exists a rational $q > a ge 1$ so that $a^2 < q^2 < 2.

That is what we need to prove.

.....

To find such a $q$ in terms of $a$ we set $q = a+e$ ($e$ can be anything we want so long as i) $e > 0$ ii) $e$ is rational and iii) $a^2 < (a+e)^2 < 2$.

So we need to find an $e$ (any positive rational $e$) where:

$a^2 < (a+e)^2 = a^2 + 2ae + e^2 < 2$

so we want $0 < 2ae+ e^2 = e(e+2a) < 2- a^2$ so

$0 < e < \frac {2-a^2}{e+2a}$

At first glance this doesn't seem to help us as we have $e$ in terms of $e$. But $e$ can be anything we want. If we pick $e < 1$ then we can pick $e$ to be:

$0 < e < \frac {2-a^2}{1+2a} < \frac {2-a^2}{e+2a}$

(Note: we aren't trying solve $e$. We are trying to find an $e$ that works. If $0 < e < \frac{1-a^2}{1+2a}$ then the $e$ will work. There may be other $e$ that will work as well, but we know this one will.)

So if we let $e$ be any rational so that $0 < e < \frac {2-a^2}{1+2a}$ and we let $q = a + e$ then we know $a^2 < q^2 < 2$.

Pf: $1\le a < q = a+e < a +\frac {2-a^2}{1+2a}$

$a^2 < q^2 = (a+e)^2 < (a +\frac {2-a^2}{1+2a})^2$

$a^2 < q^2 < a^2 + 2a\frac {2-a^2}{1+2a}+(\frac {2-a^2}{1+2a})^2$

$a^2 < q^2 < a^2 + (2a+1)\frac {2-a^2}{1+2a}+(\frac {2-a^2}{1+2a})^2-\frac {2-a^2}{1+2a}$

$a^2 < q^2 < a^2 + (2-a^2) + \frac{(2-a^2)^2-(2-a^2)(1+2a)}{(1+2a)^2}=2+ \frac{2-a^2}{(1+2a)^2}[(2-a^2)-(1+2a)]$

$a^2 < q^2 < a^2 + (2-a^2) + \frac{(2-a^2)^2-(2-a^2)(1+2a)}{(1+2a)^2}=2+ \frac{2-a^2}{(1+2a)^2}(1-2a - a^2)$

as we are assuming $a \ge 1$ we know $1-2a - a^2 < 0$

so

$a^2 < q^2 < 2$.

we are done.

Now we couldn't just assume $e < \frac{2-a^2}2$ because then we'd have

$a < q + e< a + \frac{2-a^2}2$ so

$a^2 < q^2 < a^2 + 2a\frac{2-a^2}2 + (\frac{2-a^2}2)^2 = a^2 +\frac a{2-a^2} +(\frac{2-a^2}2)^2 =a^2 +\frac a{2-a^2} + 1 - a^2 +\frac {a^4}4= \frac a{2-a^2} + 1+\frac {a^4}4$

which doesn't help us at all.

fleablood
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  • why can we asume the 0<e<1? is that necessary? – Allonsy Mar 09 '17 at 02:10
  • also, what do you mean that there's no maximal? and why q is a rational? – Allonsy Mar 09 '17 at 02:12
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    We can't solve for e(e+2a) < K because we have a quadratic. So we want to replace e+2a. We can't replace it with something smaller. so we want something bigger. We decllare e <1 by fiat. We could have use e < 2 instead. If fact I think that is how I've seen it done. – fleablood Mar 09 '17 at 02:14
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Here's how I would do it (off the top of my head).

If $r$ is a rational such that $r^2 < 2$, we wand to find another rational $s$ such that $r^2 < s^2 < 2$.

Let's increment $r$ a little and see what we can find.

Letting $d$ be the increment, we want $(r+d)^2 < 2$.

Expanding this, $2 \gt (r+d)^2 =r^2+2rd+d^2 $ or $d^2+2rd < 2-r^2 $ or $d(d+2r) < 2-r^2$.

Since $d < 1$ once $r \ge 1$ (because $2^2 > 2$), $d+2r < 1+2r$.

Therefore, if $d(2r+1) \le 2-r^2$ then $d(d+2r) < 2-r^2$.

So choosing $d = \dfrac{2-r^2}{2r+1} $ will make $d > 0$ (since $r^2 < 2$) and $(r+d)^2 < 2$, so we can get closer to $2$.

As a check, assuming $r \ge 1$,

$\begin{array}\\ (r+d)^2 &=\left(r+\dfrac{2-r^2}{2r+1}\right)^2\\ &=r^2+2r\dfrac{2-r^2}{2r+1}+\left(\dfrac{2-r^2}{2r+1}\right)^2\\ &=\dfrac{r^2(2r+1)+2r(2-r^2)}{2r+1}+\left(\dfrac{2-r^2}{2r+1}\right)^2\\ &=\dfrac{2r^3+r^2+4r-2r^3}{2r+1}+\left(\dfrac{2-r^2}{2r+1}\right)^2\\ &=\dfrac{r^2+4r}{2r+1}+\left(\dfrac{2-r^2}{2r+1}\right)^2\\ \text{so}\\ (r+d)^2-2 &=\dfrac{r^2+4r}{2r+1}+\left(\dfrac{2-r^2}{2r+1}\right)^2-2\\ &=\dfrac{r^2+4r-2(2r+1)}{2r+1}+\left(\dfrac{2-r^2}{2r+1}\right)^2\\ &=\dfrac{r^2-2}{2r+1}+\left(\dfrac{2-r^2}{2r+1}\right)^2\\ &=\dfrac{r^2-2}{(2r+1)^2}\left(2r+1-(2-r^2)\right)\\ &=\dfrac{r^2-2}{(2r+1)^2}\left(r^2+2r-1)\right)\\ &=\dfrac{r^2-2}{(2r+1)^2}((r+1)^2-2))\\ &< 0\\ \end{array} $

since $r^2 < 1$ and $(r+1)^2 > 2$.

(That checking was harder than I thought it would be. I wanted to avoid just working backwards.)

Note that

$\begin{array}\\ \dfrac{(r+d)^2-2}{r^2-2} &=\dfrac{(r+1)^2-2}{(2r+1)^2}\\ &=\dfrac{r^2+2r-1}{4r^2+4r+1}\\ &\approx\dfrac{2+2\sqrt{2}-1}{8+4\sqrt{2}+1} \qquad\text{since } r \approx \sqrt{2}\\ &\approx\dfrac{3+2\sqrt{2}}{8+4\sqrt{2}}\\ &=\dfrac{2+\sqrt{2}}{8}\\ &\approx 0.42677...\\ \end{array} $

so the error decreases by more than a factor of 2 each iteration.

marty cohen
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