Proof that Sqrt(2) is Irrational

Date: 8/29/96 at 12:49:58
From: Tim
Subject: Proof that Sqrt(2) is Irrational

Dr. Math -

For years I've tried to recall the proof that the square root of 2 is
irrational.  The proof I recall begins with the assumption that the
square root of 2 is rational and therefore = a/b. The proof then shows
this is impossible. I remember the proof being clever and elegant.

Date: 8/30/96 at 19:17:39
From: Doctor Tom
Subject: Re: Proof that Sqrt(2) is Irrational

Yup.  Assume that it is rational, of the form a/b, where a/b is
reduced to lowest terms.  So (a/b)*(a/b) = 2, or a^2 = 2b^2.  
Since a^2 is even, a must be even, say a = 2c  so (2c)^2 = 2b^2, or
4c^2 = 2b^2 or 2c^2 = b^2, so b must also be even.  So in a/b,
both a and b are even, but we assumed we'd reduced the fraction
to lowest terms.  We've got a contradiction, so sqrt(2) must be
irrational.

-Doctor Tom,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/

Date: 10/29/2005 at 20:41:41
From: Mitch
Subject: Your proof of sqrt(2)

Dear Dr. Math, 

I saw the proof on your website for the square root of 2 being 
irrational.  I can follow all the steps, but the proof doesn't seem 
valid to me.  I looked around the web and saw very similar 
descriptions for this proof and they too all seem invalid to me.  
All the proofs start with the idea that a rational number can be 
written as the ratio of two integers, say a/b, and that for any 
ratio there exists exactly one fully reduced fraction (where no 
integer greater than 1 exists that can be evenly divided into both
the numerator and denominator.)

What I see is there may exist a fraction that is not fully reduced.  
Even if I assumed a non-fully reduced fraction did exist, this does 
not imply to me there does not exist a non-fully reduced fraction.  
I would want to see that no fully reduced fraction of integers 
exists anywhere to see the conclusion.  

Thanks in advance for any help.

Date: 11/01/2005 at 21:15:47
From: Doctor Rick
Subject: Re: Thank you (Your proof of sqrt(2))

Hi, Mitch. 

The version shown above is very brief, more an outline of the proof, 
and shouldn't be judged as if it were a formal proof.  Here's a 
little more formal version that addresses your concerns about the 
ratio a/b being in lowest terms.

=======================================================
To prove: The square root of 2 is irrational. In other words, there 
is no rational number whose square is 2.

Proof by contradiction: Begin by assuming that the thesis is false, 
that is, that there does exist a rational number whose square is 2.

By definition of a rational number, that number can be expressed in 
the form c/d, where c and d are integers, and d is not zero. 
Moreover, those integers, c and d, have a greatest common divisor, 
and by dividing each by that GCD, we obtain an equivalent fraction 
a/b that is in lowest terms: a and b are integers, b is not zero, 
and a and b are relatively prime (their GCD is 1).

Now we have

[1]  (a/b)^2 = 2

Multiplying both sides by b^2, we have

[2]  a^2 = 2b^2

The right side is even (2 times an integer), therefore a^2 is even. 
But in order for the square of a number to be even, the number 
itself must be even. Therefore we can write

[3]  a = 2f

Using this to replace a in [2], we obtain

[4]  (2f)^2 = 2b^2
[5]  4f^2 = 2b^2
[6]  2f^2 = b^2

The left side is even, therefore b^2 must be even, and by the same 
reasoning as before, b must be even. But now we have found that both 
a and b are even, contradicting the assumption that a and b are 
relatively prime. Therefore the assumption is incorrect, and there 
must NOT be a rational number whose square is 2.
=======================================================

That's my proof. Admittedly it still leaves out some steps, for 
instance a proof that if the square of a number is even then the 
number itself must be even. But I filled in the part that had you 
confused.

-Doctor Rick,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/