Square Root of 3

Date: 05/18/99 at 02:35:50
From: sheila
Subject: Pure Maths

What is the appropriate way to prove that the square root of 3 is an 
irrational number?

Date: 05/18/99 at 13:28:50
From: Doctor Nick
Subject: Re: Pure Maths

Hi Sheila -

Here's one way to prove this.

Suppose that the square root of 3 is r/s, where r and s are relatively 
prime positive integers (if the square root of 3 is rational, then we 
can do this).

Squaring both sides of the equation yields 3 = (r^2)/(s^2). That is, 
3 s^2 = r^2.

This shows that r^2 must be divisible by 3, from which it follows that 
r is divisible by 3, and r^2 is divisible by 9. This implies that s^2 
must be divisible by 3, and so s must be divisible by 3. Hence, r and 
s are divisible by 3. This is a contradiction to our assumption that 
r and s are relatively prime. Thus, there must not exist integers r 
and s so that the square root of 3 is r/s. In other words, the square 
root of three is irrational.

You can use an almost identical proof to show that the square root of 
any prime is irrational. With a little modification, you can use it to 
show that the square root of an integer is either an integer or an 
irrational number.

Have fun,

- Doctor Nick, The Math Forum
  http://mathforum.org/dr.math/