# Prove root 3 is irrational

1. Aug 19, 2012

### gottfried

1. The problem statement, all variables and given/known data

Prove that there is no rational x such that x2=3

2. The attempt at a solution

Suppose that there is a rational x=$\frac{a}{b}$=$\sqrt{3}$ and that the fraction is fully simplified. (ie. a and b have no common factor)

Then a2/b2=3 which means a2=b2.3 and it follows that a is a factor of 3 and be written a=3.k (k is an integer). Therefore a2/b2 = 9.k2/b2=3. Therefore b2=3.k2.

This means both a and b are multiples of 3 and this contradicts our original assumption that the fraction is fully simplified.

3. Question

As far as I can tell the proof is sound but if you replace 3 with 4 the same logic holds and this means there is no rational number x such that x2=4 which is obviously wrong since 2 does. So what is wrong with this proof.

2. Aug 19, 2012

### HallsofIvy

No, the "same logic" does not work with 4 because 3 is a prime number and 4 is not. 6 is not a multiple of 4 but $6^2= 36= 4(9)$ is. That's because 6(6)= (2(3))(2(3))= (2)(2)(3)(3)= 4(9). You can't do that with 3.

More formally, every integer is of the form 3k or 3k+1 or 3k+ 2 for some integer k. $(3k)^2= 9k^2= 3(3k^2)$ is a multiple of 3. $(3k+1)^2= 9k^2+ 6k+ 2= 3(3k^2+ 2k)+ 1$ is not a multiple of 3. $(3k+2)^2= 9k^2+ 12k+ 4= 3(3k^2+ 4k+ 1)+ 1$ is not a multiple of 3. That is, for any integer, n, $n^2$ is a multiple of 3 if and only if n is a multiple of 3.

That does NOT work with 4. Every integer is of the form 4k, or 4k+1, or 4k+ 2, or 4k+ 3. BOTH $(4k)^2= 16k^2= 4(4k^2)$ and $(4k+ 2)= 16k^2+ 16k+ 4= 4(4k^2+ 4k+ 1)$ are multiples of 4.

Last edited by a moderator: Aug 19, 2012
3. Aug 19, 2012

### voko

Your proof uses your previous result, that if $p^2$ is a multiple of 3, then $p$ also is. This is only true for prime numbers.

4. Aug 19, 2012

### gottfried

Thanks for clearing it up that makes perfect sense