Irrational

**Treadstone 71** · January 14th 2006, 06:35 PM

Here's an interesting one:

Prove that $\sqrt{2}$ is irrational by induction

**TD!** · January 16th 2006, 04:09 AM

Using induction on what?

**dud** · January 16th 2006, 04:18 AM

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?

**CaptainBlack** · January 16th 2006, 05:16 AM

Originally Posted by TD!

Using induction on what?

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

RonL

**CaptainBlack** · January 16th 2006, 11:32 AM

Originally Posted by CaptainBlack

The convergents of the continued fraction expansion of $\sqrt2$ ,
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 $\sqrt2$ .

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 $\sqrt 2$ does not
terminate.

RonL

**ThePerfectHacker** · January 16th 2006, 02:19 PM

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

**CaptainBlack** · January 16th 2006, 02:41 PM

Originally Posted by ThePerfectHacker

Interesting, I see what you are saying CaptainBlack,
$\sqrt{2}=[1;\=2]$ 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

**Treadstone 71** · January 17th 2006, 07:33 AM

Indeed, that is one way of inducing the proof.

**CaptainBlack** · January 18th 2006, 12:35 AM

Originally Posted by Treadstone 71

Indeed, that is one way of inducing the proof.

Method #2:

Let $P_n$ be the proposition that:

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

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

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

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

RonL

**CaptainBlack** · January 18th 2006, 03:37 AM

Originally Posted by Treadstone 71

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 $P_n$ be something like:

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

RonL

**Treadstone 71** · January 18th 2006, 11:04 AM

Method 2 is what I have written down originally. It goes slightly differently than what you wrote.

$\sqrt{2}\neq 1$

Suppose $\sqrt{2}\neq \frac{n}{m}$ for some general $n,m\in \mathbb{Z}$ Details are omitted here, but it can be shown that $\sqrt{2}\neq\frac{n+1}{m}$ and $\sqrt{2}\neq\frac{n}{m+1}$ By induction, it does not equal to any rational number.

**Treadstone 71** · January 18th 2006, 11:07 AM

Originally Posted by CaptainBlack

Let $P_n$ be something like:

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

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.

**CaptainBlack** · January 18th 2006, 01:17 PM

Originally Posted by Treadstone 71

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.

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 $\sqrt 2$ , but supposing
that the tail were periodic of period $n$ , 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:

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

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

RonL

**ThePerfectHacker** · January 18th 2006, 02:13 PM

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

**SkyWatcher** · January 19th 2006, 09:02 AM

Originally Posted by ThePerfectHacker

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

I dont understand your notation mister Hacker please could you explain?

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 !!!

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Irrational

could you please explain your notation i dont undestand them

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