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Proving the Square Root of a Prime is Irrational

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Date: 07/15/98 at 18:20:51
From: April Smothers
Subject: (Proof) square root of p is irrational if p is prime

of p is irrational? Is there a way to do this using the Fundamental
Theorem of Arithmetic?

Thank you!
April
```

```
Date: 07/24/98 at 10:50:05
From: Doctor Anke
Subject: Re: (Proof) square root of p is irrational if p is prime

Hi April,

Well, in my opinion, the best way to prove this is by contradiction.
There are definitely other ways as well. I haven't checked to see if
there is one using the Fundamental Theorem of Arithmetic.

Let's assume that p is prime and the square root of p is rational.
This means there are (positive) integers a, b such that sqrt(p) = a/b.
Therefore:

p = (a/b)*(a/b) = (a^2)/(b^2)

This shows that a/b already has to be a (positive) integer, so that
(a^2)/(b^2) is also one. (If a/b is not an integer, multiplying it by
itself wouldn't create one, since no elements would come in that you
could cancel the numerator and the denominator with.) So we have shown
that (a^2)/(b^2) = (a/b)^2 = p. But this means that p isn't prime,
because it has a/b (an integer) as a divisor, so we have a
contradiction of the given fact that p is prime.

This makes our assumption that sqrt(p) is rational false, and therefore
proves that if p is prime, sqrt(p) is irrational.

I hope this helps. If have further questions, please feel free to
write again.

- Doctor Anke, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
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Associated Topics:
High School Number Theory

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