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1.1 Prove that there is no rational number whose square is 12. Rudin’s Ex. 2 Proof Suppose there is a rational number whose square is 12. Hence there are two 2 integers m and n, being co-prime, such that m = 12. This gives m2 = 12n2 , n which implies that m is of multiple of 3. Put m = 3k. Then we have 4n2 = 3k 2 . This implies that n is also of multiple of 3. Both m and n being of multiple of 3 contradicts to the assumption that they are co-prime. 1.2 Let E be a nonempty subset of an ordered set; suppose α is a lower bound of E and β is an upper bound of E. prove that α ≤ β. Rudin’s Ex. 4 Proof Since α a lower bound of E, and E be nonempty, there is x ∈ E such that α ≤ x. Since β is an upper bound of E, then for this element x, we have x ≤ β. By the transitivity, we have α ≤ β. 1.3 Let A be a nonempty set of real numbers which is bounded below. Let −A be the set of all numbers −x, where x ∈ A. Prove that inf A = − sup(−A). Proof Since A is nonempty and bounded below, −A is nonempty and bounded above. By the least-upper-bound-property of R, b = sup(−A) exists. We shall show that −b = inf(A). First, for any x ∈ A, since b = sup(−A) is an upper bound of −A, we have −x ≤ b, or −b ≤ x. This means that −b is a lower bound of A. To show that −b is the greatest lower bound of A, for any c satisfying −b < c, we only need to show that c cannot be a lower bound of A. Indeed, Since −c < b, so −c is not an upper bound of −A. Hence there is x ∈ −A such that −c < x. In other words, we have −x ∈ A such that −x < c. So c is not a lower bound of A. 1 Rudin’s Ex. 5