Prove that the set S = { q in Q : sqrt(2) < q } has no least member.

I will show that given any rational q < sqrt(2), I can construct another rational number, r, such that q < r < sqrt(2). I just have to add a little bit, call it e, to q such that

q + e < sqrt(2)
e < sqrt(2) - q
e (sqrt(2) + q) < (sqrt(2) - q)(sqrt(2) + q)
e (sqrt(2) + q) < 2 - q^2
e < (2 - q^2)/(sqrt(2) + q)

but e = (2 - q^2)/4 works. so r = q + e is the desired number.

I accidentally showed that T has no greatest member rather than that S has no least member, but just turn the problem upside down. they are exactly equivalent.

By the way, it is easy to show with a similar procedure that there is a rational number between any two real numbers.