I think that the reasoning is the same:

* sqrt(-1) = a/b

* a^2 = -1 * b^2

Then either a^2 or b^2 are negative but a square can't be negative, so contradiction and sqrt(-1) is not rational.

The main "problem" with this proof (and the original with sqrt(2)) is "how to prove that a^2 >= 0" (or that "if a^2 is even, then a is even")

The first one is easy to prove:

* a^2 = sign(a)^2 * abs(a)^2

* abs(a) >= 0 for any a

* sign(a) = 1 or -1 for any a so sign(a)^2 = 1 (either 1*1 or -1*-1)

* so a^2 >= 0

The second one may be proved:

* Assume a is even, then a=2n, then a^2=4n^2=2*(2n^2) so a is even => a^2 is even

* Assume a is odd, then a=2n+1, then a^2=4n^2+4n+1=2*(2n^2+2n)+1, then a^2 is odd

* a is either even or odd

* So the only possibility for a^2 to be even is for a to be even