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Re:Math Bite: Irrationality of √m (1999) by contravariant on Wednesday December 23, 2015 @06:19PM (#10786245)
During the proof it is assumed, this assumption leads to a contradiction, hence the assumption 'sqrt(m) is in Q' is false.

The reason this is not a circular argument is that they don't assume the thing they're trying to prove, but rather it's negation.

Re:Math Bite: Irrationality of √m (1999) by vog on Thursday December 24, 2015 @07:50AM (#10787999)
You are right. My bad.