Can the square root of 2 be constructed to show its irrationality? Very poorly chosen words on my part...what I am getting at is that instead of using proof by contradiction is it possible to build an algorithm that returns the digits of the square root of 2 or some other irrational number in such a way as to demonstrate the digits will never be repeating themselves?
I realize that proof by contradiction may well be the only way but in that case must one also show the law of excluded middle will apply in order to use the contradiction?  And I was trying to avoid this possibility because I did not know how to do it.  Thank you 
 A: This may not be the kind of proof the OP is looking for, but it might be of interest anyway.
Let's show that $\sqrt2$ cannot have an eventually repeating binary expansion. (I'm taking for granted the theorem that rational numbers are eventually repeating in any base and irrational numbers are not.)  That is, $\sqrt2=1.0110101\ldots$ does not settle into any $n$-bit repetition after any finite number of binary bits, say $m$.
Suppose it did.  Then, since ${1\over2^n}+{1\over2^{2n}}+\cdots={1\over2^n-1}$, we would have $2^m\sqrt2=M+{N\over2^n-1}$ for some (positive) integers $M$ and $N$ (with $N\lt2^n$).  We can assume we've taken $m$ to be minimal.  We'll first show that $m=0$ (and hence $M=1$, since $1\lt\sqrt2\lt2$).  This is because squaring both sides of $(2^n-1)2^m\sqrt2=2^nM+(N-M)$ gives an even number of the left hand side, which implies $M$ and $N$ must have the same parity. This means they end in the same bit, either $0$ or $1$.  So if $m$ were greater than $0$, we could move the (binary) decimal point to the left and still have an $n$-bit repeating string.
So now we need only consider the possibility $(2^n-1)\sqrt2=2^n+(N-1)$ for some positive integers $n$ and $N$ -- i.e., that the binary string $.0110101\ldots$ is periodic with some period $n$, which must clearly be at least $5$.  As before, $N$ must be odd, so the right hand side is even, which means its square is congruent to either $0$ or $4$ mod $8$, since $n\gt3$. But the square of the left hand side is congruent to $2$ mod $8$, which gives a contradiction.
Surely this proof is already in the literature somewhere.  Can anyone supply a reference?
