## Friday, December 13, 2013

### Proof that the square roots of all non-square numbers are irrational

In a previous post I described the classical way to prove that the square root of 2 is irrational. But that proof doesn't generalize (at least not easily) to the square root of other numbers being irrational. This proof can be applied to any number and any root (cube root, etc.).

This proof uses the Fundamental Theorem of Arithmetic which states that there is only one way to factor a whole number greater than 1 into a product of prime numbers raised to some power. For example 12 = 2^2 x 3^1. There is no other way to factorize 12 into powers of prime numbers other than 2^2 and 3^1. Another example, 15 = 2^0 x 3^1 x 5^1. So the exponent of the first prime number must be 0, the second must be 1 and the third must be 1. If we had to include the fourth prime number we'd give it an exponent of 0 too, that is, 2^0 x 3^1 x 5^1 x 7^0. In other words, for any number "n" greater than 1,
n = 2^i1 x 3^i2 x 5^i3 x ... x Pn^in
where "n" is the number of prime factors used, P1 is the first prime number, P2 is the second, etc. and i1 is the exponent of the first prime factor, i2 of the second, etc.

Proof for the square root of 2
Onto the proof. We'll start from the proof for the square root of 2 and then generalize it.

We start exactly the same way we started in the other proof.
Assume sqrt(2) = a/b
2 = (a/b)^2
2 = a^2 / b^2
2 b^2 = a^2


Now we use the fundamental theorem of arithmetic on "a" and "b".

Let a = 2^i1 x 3^i2 x 5^i3 x ... x Pn^in
Let b = 2^j1 x 3^j2 x 5^j3 x ... x Pm^jm

Therefore,
2 (2^j1 x 3^j2 x 5^j3 x ... x Pm^jm)^2 = (2^i1 x 3^i2 x 5^i3 x ... x Pn^in)^2
2 (2^2j1 x 3^2j2 x 5^2j3 x ... x Pm^2jm) = 2^2i1 x 3^2i2 x 5^2i3 x ... x Pn^2in
2 x 2^2j1 x 3^2j2 x 5^2j3 x ... x Pm^2jm = 2^2i1 x 3^2i2 x 5^2i3 x ... x Pn^2in
2^(2j1+1) x 3^2j2 x 5^2j3 x ... x Pm^2jm = 2^2i1 x 3^2i2 x 5^2i3 x ... x Pn^2in


Now since the number on the left and the number on the right are both equal, then by the fundamental theorem of arithmetic, apart from the number of prime factors being equal, the exponents of their prime factors must also be equal. Therefore,

m = n

and

2j1+1 = 2i1
2j2 = 2i2
2j3 = 2i3
...
2jm = 2in


So then 2j1+1 = 2i1. But one is an odd number whilst the other is an even number. This cannot be the case, therefore "a^2" and "2b^2" cannot be equal, therefore "a/b" cannot equal "sqrt(2)" and therefore "a" and "b" cannot exist.

What happens if the square root is actually rational?
Let's try that again with the square root of 4 this time.

Assume that sqrt(4) = a/b
4 b^2 = a^2
(2^2) (2^j1 x 3^j2 x 5^j3 x ... x Pm^jm)^2 = (2^i1 x 3^i2 x 5^i3 x ... x Pn^in)^2
(2^2)(2^2j1 x 3^2j2 x 5^2j3 x ... x Pm^2jm) = 2^2i1 x 3^2i2 x 5^2i3 x ... x Pn^2in
2^(2j1+2) x 3^2j2 x 5^2j3 x ... x Pm^2jm = 2^2i1 x 3^2i2 x 5^2i3 x ... x Pn^2in

By the fundamental theorem of arithmetic,
m = n

and

2j1+2 = 2i1
2j2 = 2i2
2j3 = 2i3
2jm = 2in


As you can see, all exponents can be equal this time. 2j1+2 = 2i1 can be true because this time both numbers are even numbers. In fact j1 = 0.

You can also notice that since 4 is a composite number, we had to factor it as well in order to assimilate it into the factors of b.

Square root of 12?
Again, we use the proof on a composite number, this time with no rational square root.

Assume that sqrt(12) = a/b
12 b^2 = a^2
(2^2 x 3^1) (2^j1 x 3^j2 x 5^j3 x ... x Pm^jm)^2 = (2^i1 x 3^i2 x 5^i3 x ... x Pn^in)^2
(2^2 x 3^1)(2^2j1 x 3^2j2 x 5^2j3 x ... x Pm^2jm) = 2^2i1 x 3^2i2 x 5^2i3 x ... x Pn^2in
2^(2j1+2) x 3^(2j2+1) x 5^2j3 x ... x Pm^2jm = 2^2i1 x 3^2i2 x 5^2i3 x ... x Pn^2in

By the fundamental theorem of arithmetic,
m = n

and

2j1+2 = 2i1
2j2+1 = 2i2
2j3 = 2i3
2jm = 2in


Looks like this time it although 2j1+2 = 2i1 can be true, 2j2+1 = 2i2 cannot be.

Cube root of 2?
Let's try that again with a cube root instead.

Assume that 2^(1/3) = a/b
2 b^3 = a^3
2 (2^j1 x 3^j2 x 5^j3 x ... x Pm^jm)^3 = (2^i1 x 3^i2 x 5^i3 x ... x Pn^in)^3
2 (2^3j1 x 3^3j2 x 5^3j3 x ... x Pm^3jm) = 2^3i1 x 3^3i2 x 5^3i3 x ... x Pn^3in
2^(3j1+1) x 3^3j2 x 5^3j3 x ... x Pm^3jm = 2^3i1 x 3^3i2 x 5^3i3 x ... x Pn^3in

By the fundamental theorem of arithmetic,
m = n

and

3j1+1 = 3i1
3j2 = 3i2
3j3 = 3i3
3jm = 3in


Can 3j1+1 = 3i1? No, because no multiple of 3 can equal another multiple of 3 plus 1, since any multiple of 3 plus 1 cannot be a multiple of 3.

In general
Now let's try to generalize this to any number "q" and any root "r".

Assume that q^(1/r) = a/b
q b^r = a^r
(2^k1 x 3^k2 x 5^k3 x ... x Pl^kl) (2^j1 x 3^j2 x 5^j3 x ... x Pm^jm)^3 = (2^i1 x 3^i2 x 5^i3 x ... x Pn^in)^3
(2^k1 x 3^k2 x 5^k3 x ... x Pl^kl) (2^rj1 x 3^rj2 x 5^rj3 x ... x Pm^rjm) = 2^ri1 x 3^ri2 x 5^ri3 x ... x Pn^rin

Now since we don't know whether "q" has more factors than "b" or vice versa, the last factor in the assimilated factors will simply be called "Pz^y" where z=max(l,m) and "y" is the exponent of this last prime number.
2^(rj1+k1) x 3^(rj2+k2) x 5^(rj3+k3) x ... x Pz^y = 2^ri1 x 3^ri2 x 5^ri3 x ... x Pn^rin

By the fundamental theorem of arithmetic,
z = n

and

rj1+k1 = ri1
rj2+k2 = ri2
rj3+k3 = ri3
...
y = rin


For rj1+k1 = ri1 to be true, "k1" must be a multiple of "r", otherwise rj1+k1 cannot be a multiple of "r". The same goes for all the other exponents. So this means that every exponent of the prime factors of "q" must be a multiple of "r", that is
q = 2^rh1 x 3^rh2 x 5^rh3 x ... x Pl^rhl

But then this means
q = (2^h1 x 3^h2 x 5^h3 x ... x Pl^hl)^r

which further means that "q" is a number raised to the power of "r". Therefore, for any whole number to have its nth root be a rational number, that number must be equal to another whole number raised to the power of "n". If there aren't any whole numbers which when raised to the nth power equal the number in question, then there cannot be any fractions either.