we denote the solution to to be . then clearly, for all ,

it can be shown (by examples if necessary) that . thus

(am i begging the question here? i hate doing proofs for things that seem "obvious")

well, if you accept what is in (a), this is immediate. i guess you can show it by example, or come up with some approximation to or just some number greater than 1 but less than , and then it would be in A.(b) Suppose that y is the least upper bound of A. Show

that

i) y >= 1;

assume to the contrary that y^2 < 2. then by the denseness of Q, we can find some rational x in A such that y^2 < x^2 < 2. a contradictionii)y^2 >= 2;

assume to the contrary that y^2 > 2. clearly we have that y^2 > 2 > x^2 for all x in A. but that means that there is some upper bound of the set A that is less than y, therefore, y cannot be the supremum as stated. there's our contradictioniii)y^2 <= 2;

i'm not in the mood for any fancy arguments at this point, i never liked this question to begin with. so i'd just slap on the abstract property of real numbers that says: if a and b are real numbers, with a <= b and b <= a then a = biv)y^2 = 2.