Supremum in a set of rational numbers?

Question

Suppose E={x|x is a rational number and x^2<2}
then prove that supE=sqrt2.

Answer 1

E has as an upper bound the number sqrt(2)(why?)
In order to show supE=sqrt2, you have to prove that sqrt(2) is the lowest such upper bound
So suppose by reduction to absurd that there exists an upper bound A such that
A< sqrt(2).
Since A is real there exist a rational number c between A and sqrt(2): A an upper bound(why?). Contradiction

Answer 2

To prove that sup E = √2, we need to prove two things. First, that √2 is an upper bound for E, and second, that no lower number is an upper bound for E.

The first property is easy -- for suppose x>√2. Then x² > 2, so x∉E. Taking the contrapositive, we see that x∈E ⇒ x≤√2, so √2 is an upper bound.

To prove the second property, suppose u<√2. By the density of the rational numbers, there exists r∈ℚ such that uu and r²<2, so r∈E and thus u fails to be an upper bound of E. Therefore, √2 is the smallest upper bound for E and so sup E = √2. Q.E.D.

Answer 3

keep in innovations sup(S) = min{y in R: for all x in S, y >= x}, dually for inf. so for a) inf = 0, sup = a million for b) inf = 0, sup = a million for c) inf= 0, sup undefined. for d) inf = 2, sup undefined. enable y = ax. then (x^2+y^2)/xy = (x^2 + a^2x^2)/ax^2 = (a^2 + a million)/a = a + a million/a it is better than a and can want to hence be arbitrarily great by technique of choosing small x and great y. hence there's no sup. a + a million/a is minimum at the same time as a = a million. those are the factors on the effective 1/2-line x=y, which elements inf 2.