3 ms·

I might be missing something, but as I see it here: by assumption, √2 is rational. Then k(√2-1) √2 is a positive integer. This does not mean that k(√2 - 1) be

I might be missing something, but as I see it here: by assumption, √2 is rational. Then k(√2-1) √2 is a positive integer. This does not mean that k(√2 - 1) belongs to K, with K the set of positive integers {n: n√2 ∈ ℤ}, as k(√2 - 1) is not necessarily an integer. The proof can be fixed I think with K the set of rational numbers {q: q√2 ∈ ℤ}

johnnny 4y ago
k * √2 is an integer, and k is an integer, by construction of K and the initial assumption that K is not empty. Therefore their difference k * √2 - k = k * (√2 - 1) is an integer.
- janm31415 4y ago
  correct, I missed that.
  - johnnny 4y ago
    As fro your further edit: assuming √2 is rational, then the set of ~~rational~~ integers {q: q√2 ∈ ℤ} does not admit a minimum, indeed it contains arbitrarily small numbers, so we can't pick k the smallest member of that set.