Is the Square Root of 2 Irrational?

[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?

 

[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.

 

[mathwonk]

boy you had a good 6th grade teacher!

 

[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

 

[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.

 

[matt grime]

I just don't like using fractions unnecessarily.

 

[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.

 

[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.

 

[ComputerGeek]

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.

would it not be:

p^2 = nq^2

?

 

[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.

 

[bombadillo]

I've heard that the nth root of 2 (n an integer, greater than or equal to 2) is irrational. How would one prove that?

 

[Hurkyl]

Have you tried applying the same technique?