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Square Root of a Prime


Date: 07/14/99 at 20:23:13
From: Stephen F. Moran
Subject: Irrational Numbers

I am trying to grasp the concept of irrational numbers. Suppose p is a 
prime number. Show that sqrt(p) is irrational. (Please don't use 2 as 
an example; the textbook I have uses it and I am missing somthing.)  
Is there a way to demonstrate this graphically?

Steve Moran


Date: 07/15/99 at 10:13:07
From: Doctor Floor
Subject: Re: Irrational Numbers

Dear Steve,

Thanks for your question.

Here is one proof from the Dr. Math archives:

http://mathforum.org/dr.math/problems/smothers7.15.98.html   

Another proof:

Let p be a prime number.

To prove that sqrt(p) [sqrt means square root] is irrational number, 
we have to show that it cannot be written as a/b, where a and b are 
integers. 

Such a proof has to be done indirectly: 

1. We assume that sqrt(p) could be written as a/b for integer a and b;
2. We show that this leads to a contradiction. So something 
   impossible has to be derived from our assumption.

The fact that we derive someting impossible from the fact that sqrt(p)
= a/b, shows that sqrt(p) = a/b must be false, and we have proven the 
theorem.

We are going to try this:

So let's assume that sqrt(p) could be written as a/b for integer a 
and b. We can derive:

  sqrt(p) = a/b [sqare both sides]
        p = a^2/b^2 [multiply by b^2; ^2 means squared]
    p*b^2 = a^2
      a^2 = p*b^2

Now we have two integers in the equation, p*b^2 and a^2. We can factor 
these integers into primes, and find:

  a^2   = p^m * {product of primes unequal to p}
  p*b^2 = p^n * {product of primes unequal to p}

Because a^2 is a square, m has to be even.
Because b^2 is a square, and "p*" brings an additional factor p, n has 
to be odd.

So m and n are not equal.

But if a^2 = p*b^2, then the factorization into primes of these two 
has to be equal. So, m and n should be equal.

We have found a contradiction. And our assumption, that sqrt(p) = a/b 
for integers a and b, cannot hold. This proves that sqrt(p) is 
irrational.

I hope this helps you to understand.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Number Theory

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