How to prove square root 2 is irrational?

• Jun 21st 2007, 01:20 AM
Benno_11
How to prove square root 2 is irrational?
I have the first bit to this problem but I do not know how to prove that x^2 is even and then y is even. I am using the contradiction method. Here is what I have so far :

Assume sqrt2 is rational

therefore sqrt2=(x/y)^2
2y^2=x^2

thanks for any help.

p.s how do you write with apprioate symbols? I have never done maths on the computer.
• Jun 21st 2007, 03:30 AM
topsquark
Quote:

Originally Posted by Benno_11
I have the first bit to this problem but I do not know how to prove that x^2 is even and then y is even. I am using the contradiction method. Here is what I have so far :

Assume sqrt2 is rational

therefore sqrt2=(x/y)^2
2y^2=x^2

thanks for any help.

p.s how do you write with apprioate symbols? I have never done maths on the computer.

We have to state something slightly stronger than $\frac{x}{y}$ is rational, we have to state that it has been reduced to lowest terms. We may freely make this assumption. The point of the proof is to show that if $\sqrt{2} = \frac{x}{y}$ then the fraction cannot possibly be in reduced form.

So assume we can write
$\sqrt{2} = \frac{x}{y}$
where x and y are integers ( $y \neq 0$) and the fraction is in reduced form. Then by squaring both sides we see that:
$\frac{x^2}{y^2} = 2$

$x^2 = 2y^2 = 2(\text{integer})$

Thus $x^2$ is an even number.

So if x is an even number it is of the form $x = 2n$ where n is an integer. Thus $x^2 = 4n^2$ and so
$2y^2 = 4n^2$

$y^2 = 2n^2 = 2(\text{integer})$

Thus y^2 is also an even number.

Thus y is of the form $y = 2m$ where m is a (non-zero) integer.

Thus
$\frac{x}{y} = \frac{2n}{2m}$
which can be reduced, contrary to assumption.

Thus $\sqrt{2}$ is not rational.

-Dan
• Jun 21st 2007, 05:22 PM
Benno_11
thanks

You could also use this for sqrt 3, couldn't you?

Matt

P.S
• Jun 21st 2007, 06:45 PM
ThePerfectHacker
Quote:

Originally Posted by Benno_11
You could also use this for sqrt 3, couldn't you?

Yes
• Jun 21st 2007, 09:13 PM
CaptainBlack
Quote:

Originally Posted by Benno_11

You could also use this for sqrt 3, couldn't you?

Matt

P.S

It works for the square root of every positive integer that is not a
perfect square.

RonL
• Jun 21st 2007, 09:22 PM
Benno_11
How would you ...
How do you prove that e and pi are irrational? I do not need to know it, but I was just wondering how complex it would be. For e, would you have to use the maclaurin series?
• Jun 22nd 2007, 12:04 AM
CaptainBlack
Quote:

Originally Posted by Benno_11
How do you prove that e and pi are irrational? I do not need to know it, but I was just wondering how complex it would be. For e, would you have to use the maclaurin series?

There is a neat, short proof of the irrationality of $e$ in Hardy
and Wright.

It gose like this:

Let's suppose that $e$ is rational, so that $e=a/b$ and $a$ and $b$ are integers.

If $k \ge b$, and:

........... $
\alpha=k!\left(e-1-\frac{1}{1!}-\frac{1}{2!}- \cdots -\frac{1}{k!}\right)
$

then $b|k!$ and $\alpha$ is an integer. But (replacing $e$ by its series expansion)

........... $
0<\alpha=\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}+ \cdots < \frac{1}{k+1}+\frac{1}{(k+1)^2}+\cdots = \frac{1}{k}
$

(the last part of the above uses the sum of the geomentric series)

RonL
• Jun 22nd 2007, 06:13 AM
ThePerfectHacker
Quote:

Originally Posted by Benno_11
How do you prove that e and pi are irrational? I do not need to know it, but I was just wondering how complex it would be. For e, would you have to use the maclaurin series?

The proof that $\pi$ is irrational is much more complicated. The first proof was published by the mathematician Lambert.

Here.

Here is a good link devoted to Real Analysis. It does a better job explanaing the proof.
Here.
• Jun 24th 2007, 07:40 AM
chandrasekhar_mallisetty
squareroot-of-2
Quote:

Originally Posted by Benno_11
I have the first bit to this problem but I do not know how to prove that x^2 is even and then y is even. I am using the contradiction method. Here is what I have so far :

Assume sqrt2 is rational

therefore sqrt2=(x/y)^2
2y^2=x^2

thanks for any help.

p.s how do you write with apprioate symbols? I have never done maths on the computer.

Here both x and y are relatively prime nos(i.e they don"t have a common factor.)then their squares are also relatively prime. Now we have 2y^2=x^2
therefore y^2=x^2/2 which is a contradiction that x^2 and y^2 are realtivelyprime.I hope this may clear your doubt.