# Irrationaltiy of sqrt(2)

1. Mar 5, 2005

### mathwonk

Have you seen this argument? If sqrt(2) = p/q is in lowest terms, then also 2/1 = p^2/q^2 is in lowest terms. Since lowest terms is unique, p^2 = 2 and q^2 = 1. Thus sqrt(2) is the integer p. But 1^2 is too small, 2^2 is too big and all the resta re even bigger, so this is false. So sqrt(2) is not rational.

does anyone have a shorter one?

2. Mar 5, 2005

### dextercioby

Nope.I've known this proof from 6-th grade.Learnt it in school.It's much nicer (though roughly equivalent) than the one with prime factors...

Daniel.

3. Mar 6, 2005

4. Mar 6, 2005

### matt grime

A better proof, perhaps, and one that works for all non-squares.

let sqrt(x) = p/q, then xq^2=p^2, counting primes with multiplicity, unless x is a perfect square there will be an odd number of prime factors of the left hand side and an even number of the right handside

5. Mar 6, 2005

### mathwonk

thats a nice proof, but when i first saw that as a student i didn't understand it, so i came up with the one above as using fewer tools, or at least apparently so.

moreover, doesn't the first proof above also work on all non squares? I.e. if p/q = sqrt(n) is in lowest terms, then so also is p^2/q^2 = n/1, so p^2 = n, and hence n is a perfect square.

of course both proofs rest on exactly the same fact, uniqueness of factorization into primes.

6. Mar 6, 2005

### matt grime

I just don't like using fractions unnecessarily.

7. Mar 6, 2005

### mathwonk

chaque a son gout. but you do not mind using prime factorization.

To me it seems rather hard to avoid fractions in discussing rational numbers, but i agree the correct first step in a problem involving fractions is normally to remove them.

Last edited: Mar 6, 2005
8. Mar 7, 2005

### hedlund

This may be the same argument as you're using, don't know really. Assume that sqrt(2) is rational, then we have p and q such as sqrt(2) = p/q. This gived 2 = p^2/q^2. But since gcd(p,q)=1 then p^2/q^2 isn't an integer unless q=1. This gives the equation 2 = p^2 with integers, no integer can solve this equation, and hence sqrt(2) cannot be rational.

9. Mar 11, 2005

### ComputerGeek

would it not be:

p^2 = nq^2

?

10. Mar 11, 2005

### mathwonk

well if p^2 /q^2 = n/1, and the left side is known to be in lowest terms, and if the lowest terms form of a fraction is unique, then since the right side n/1 is obviously also in lowest terms, then the two fractions must be equal, i.e;. they must have the same top and same bottom.

Since the tops are equal, p^2 = n.

so of course you are right, that p^2 = nq^2, but in fact also q^2 = 1.

11. Mar 12, 2005