Here's an interesting one:

Prove that $\displaystyle \sqrt{2}$ is irrational byinduction

Printable View

- Jan 14th 2006, 06:35 PMTreadstone 71Irrational
Here's an interesting one:

Prove that $\displaystyle \sqrt{2}$ is irrational by*induction* - Jan 16th 2006, 04:09 AMTD!
Using induction on what?

- Jan 16th 2006, 04:18 AMdud
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 AMCaptainBlackQuote:

Originally Posted by**TD!**

possibly?

RonL - Jan 16th 2006, 11:32 AMCaptainBlackQuote:

Originally Posted by**CaptainBlack**

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 PMThePerfectHacker
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 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

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 AMTreadstone 71
Indeed, that is one way of inducing the proof. :)

- Jan 18th 2006, 12:35 AMCaptainBlackQuote:

Originally Posted by**Treadstone 71**

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 AMCaptainBlackQuote:

Originally Posted by**Treadstone 71**

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 AMTreadstone 71
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 AMTreadstone 71Quote:

Originally Posted by**CaptainBlack**

- Jan 18th 2006, 01:17 PMCaptainBlackQuote:

Originally Posted by**Treadstone 71**

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.

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 PMThePerfectHacker
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 AMSkyWatchercould you please explain your notation i dont undestand themQuote:

Originally Posted by**ThePerfectHacker**

In fact i have been teached a very basic and simple method to proove that any integer that is not a square have an irational square root. I am in doubt that any steps of the tree induction method, not developed and using weird notation for some of them (this is the math puzzle section not the advanced question section), so any steps would not use the same or a more complicated argument that the basic method

People that dont know it are invited to search in <<square root>>thread !!!