# Irrational

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• Jan 14th 2006, 06:35 PM
Irrational
Here's an interesting one:

Prove that $\displaystyle \sqrt{2}$ is irrational by induction
• Jan 16th 2006, 04:09 AM
TD!
Using induction on what?
• Jan 16th 2006, 04:18 AM
dud
I don't get this, how can you use induction if all there is is 1 constant?
Induction needs to have a variable n doesn't it?
• Jan 16th 2006, 05:16 AM
CaptainBlack
Quote:

Originally Posted by TD!
Using induction on what?

The convergents of the continued fraction expansion of $\displaystyle \sqrt2$,
possibly?

RonL
• Jan 16th 2006, 11:32 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
The convergents of the continued fraction expansion of $\displaystyle \sqrt2$,
possibly?

RonL

I don't have the time to do this in detail now, but further thought and
sketches of a construction make me fairly sure it can be done using the
continued fraction expansion of $\displaystyle \sqrt2$.

The key idea is that a rational number has a terminating simple cf expansion,
but we can show by induction that the simple cf expansion of $\displaystyle \sqrt 2$ does not
terminate.

RonL
• Jan 16th 2006, 02:19 PM
ThePerfectHacker
Interesting, I see what you are saying CaptainBlack,
$\displaystyle \sqrt{2}=[1;\=2]$ But how do you show it does not terminate?
• Jan 16th 2006, 02:41 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Interesting, I see what you are saying CaptainBlack,
$\displaystyle \sqrt{2}=[1;\=2]$ But how do you show it does not terminate?

Take a finite part of the expansion and the non-integer r-term, extend the
expansion by one term which will be a 2 and the r-term will remain the same.

Hence by induction the cf expansion does not terminate.

RonL
• Jan 17th 2006, 07:33 AM
Indeed, that is one way of inducing the proof. :)
• Jan 18th 2006, 12:35 AM
CaptainBlack
Quote:

Indeed, that is one way of inducing the proof. :)

Method #2:

Let $\displaystyle P_n$ be the proposition that:

For all $\displaystyle 1\le m \le n$ there is no $\displaystyle k\epsilon \mathbb{Z}_+$, such that $\displaystyle \sqrt2 = \frac{k}{m}$.

This is clearly true for $\displaystyle n=1$ as $\displaystyle \sqrt2$ is not an integer.

The induction step consists of showing that there is no $\displaystyle k\epsilon \mathbb{Z}_+$,
such that $\displaystyle \sqrt2 = \frac{k}{n+1}$, which together with $\displaystyle P_n$ will prove $\displaystyle P_{n+1}$.

This last step I can do, but will not reproduce here.

RonL
• Jan 18th 2006, 03:37 AM
CaptainBlack
Quote:

Indeed, that is one way of inducing the proof. :)

Method 3

I'm not going to do much on this other than suggest it might be made to work.

Let $\displaystyle P_n$ be something like:

The decimal expansion of $\displaystyle \sqrt2$ does not have a peiodic
tail of any period $\displaystyle \le n$.

RonL
• Jan 18th 2006, 11:04 AM
Method 2 is what I have written down originally. It goes slightly differently than what you wrote.

$\displaystyle \sqrt{2}\neq 1$

Suppose $\displaystyle \sqrt{2}\neq \frac{n}{m}$ for some general $\displaystyle n,m\in \mathbb{Z}$ Details are omitted here, but it can be shown that $\displaystyle \sqrt{2}\neq\frac{n+1}{m}$ and $\displaystyle \sqrt{2}\neq\frac{n}{m+1}$ By induction, it does not equal to any rational number.
• Jan 18th 2006, 11:07 AM
Quote:

Originally Posted by CaptainBlack
Let $\displaystyle P_n$ be something like:

The decimal expansion of $\displaystyle \sqrt2$ does not have a peiodic
tail of any period $\displaystyle \le n$.

RonL

This is the very fact that you have to prove, because it is equivalent to the definition of an irrational number. You will actually have to compute the decimals of root 2, which comes down to something similar to what you did with the continued fractions.
• Jan 18th 2006, 01:17 PM
CaptainBlack
Quote:

This is the very fact that you have to prove, because it is equivalent to the definition of an irrational number.

But aren't they all.

Quote:

You will actually have to compute the decimals of root 2, which comes down to something similar to what you did with the continued fractions.
I was thinking of doing something slightly different. I had no intention of
computing the decimal expansion of $\displaystyle \sqrt 2$, but supposing
that the tail were periodic of period $\displaystyle n$, then writing the sum of
the tail in terms of the replicand (if that is the right word) which we don't
know, and the sum of the geometric series:

$\displaystyle \sum_1^{\infty} \frac{1}{10^{n+1}}$,

and show in some way not yet determined that this lead to a contradiction.

RonL
• Jan 18th 2006, 02:13 PM
ThePerfectHacker
Speaking about continued fractions, can someone prove Euler's interesting fraction:
$\displaystyle e=[2;1,2,1,1,4,1,1,6,1,1,8,...]$
and
$\displaystyle \frac{e-1}{e+1}=[0;2,6,10,14...]$
one more,
$\displaystyle \frac{e^2-1}{e^2+1}=[0;1,3,5,7,...]$
• Jan 19th 2006, 09:02 AM
SkyWatcher
Quote:

Originally Posted by ThePerfectHacker
Speaking about continued fractions, can someone prove Euler's interesting fraction:
$\displaystyle e=[2;1,2,1,1,4,1,1,6,1,1,8,...]$
and
$\displaystyle \frac{e-1}{e+1}=[0;2,6,10,14...]$
one more,
$\displaystyle \frac{e^2-1}{e^2+1}=[0;1,3,5,7,...]$