Re: Formal proof that sqrt(2) is irrational?

fernando revilla <frej0002@xxxxxxxxxxxxxxxxxx> wrote:

The set IK = {a+bS : a,b in Q} and S a letter with:

(a) Sum. (a+bS)+(a'+b'S) = (a+a')+(b+b')S

(b) Product. (a+bS) (a'+b'S) = aa'+2bb'+(ab'+a'b)S

is a field (extension of Q), and S^2=2. This provides an algebraic
interpretation of sqrt(2) as the letter S, and S not in Q.

If you are claiming that yields a proof that sqrt(2) is irrational
then you will need to elaborate much more to complete such a proof.
However, I suspect that the proof you have in mind is incorrect
since it will end up being circular, i.e. begging the principle.

Indeed, e.g. if you intend on using the following implication

field Q[x]/f(x) -> f irred in Q[x] -> f has no roots in Q

then how do you propose to deduce that Q[x]/(xx-2) is a field
in a manner that is not essentially equivalent to proving that
xx-2 is irred. over Q? (which is equivalent to /2 irrational)

--Bill Dubuque
.

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