Square root of 2 is Irrational

I was looking at the in Internet now and found in this site a different proof of this fact:

"Another elegant proof:
In 1952, Robert Gauntt, then a freshman at Purdue University, gave the following proof[2].
a ² = 2b² cannot have a non-zero solution in integers because the last non-zero digit of a square, written in base 3, must be 1, whereas the last non-zero digit of twice a square is 2."

Interesting! note the "last non-zero digit" part.

This proof would be obvious if we were used to base 3.

Here at cut the knot there is a list of different proofs. Proofs number 4 and 3 (in that order) are the ones I like most. The proofs at cut the knot are also reproduced below:

Proof 4

Following is yet another illuminating proof. As in the standard proof, assume p and q are mutually prime (numbers with no common factors). Their squares are still mutually prime for they are built from the same factors. Therefore, the fraction p ² /q² cannot cancel out. In particular, p² /q² cannot cancel down to equal 2. Therefore, p² /q ² .

Proof 3

Following is another intuitive one [Lasckovich, p. 4]. This one is based on the assertion that every number is uniquely (up to the order of factors) representable as a product of primes. (A prime is a number divisible only by itself and 1.) The fact which is known as The Fundamental Theorem of Arithmetic. Assuming it as an axiom (or a given fact), let (p/q)² = 2 for some integers p and q. Then p² = 2q² . Factor both p and q into a product of primes. p² is factored into a product of the very same primes as p each taken twice. Therefore, p² has an even number of prime factors. So does q² . Therefore, 2q² has an odd number of prime factors. Contradiction.