文档介绍：MATH 413 [513] (PHILLIPS) SOLUTIONS TO HOMEWORK 1 Generally, a “solution” is something that would be acceptable if turned in in the form presented here, although the solutions given are often close to minimal in this respect. A “solution (sketch)” is too sketchy to be considered plete solution if turned in; varying amounts of detail would need to be ?lled in. Problem : If r ∈ Q \{ 0 } and x ∈ R \ Q, prove that r +x, rx 6∈ Q . Solution: We prove this by contradiction. Let r ∈ Q \{ 0 }, and suppose that r + x ∈ Q. Then, using the ?eld properties of both R and Q,wehave x =( r + x ) ? r ∈ Q . Thus x 6∈ Q implies r + x 6∈ Q . Similarly, if rx ∈ Q, then x =( rx ) /r ∈ Q. (Here, in addition to the ?eld properties of R and Q,weuse r 6= 0.) Thus x 6∈ Q implies rx 6∈ Q . Problem : Prove that there is no x ∈ Q such that x 2= 12. Solution: We prove this by contradiction. Suppose there is x ∈ Q such that x 2= 12. Write x = m n in lowest terms. Then x 2= 12 implies that m 2=12 n 2 . Since 3 divides 12 n 2, it follows that 3 divides m 2. Since 3 is prime (and by unique factorization in Z), it follows that 3 divides m. Therefore 3 2 divides m 2=12 n 2 . Since 3 2does not divide 12, using again unique factorization in Zand the fact that 3 is prime, it follows that 3 divides n. We have proved that 3 divides both m and n, contradicting the assumption that the fraction m nis in lowest terms. Alternate solution (Sketch): If x ∈ Q satis?es x 2= 12, then x 2is in Q and satis?es ? x 2 ￠ 2= 3. Now prove that there is no y ∈ Q such that y 2= 3 by repeating the proof that √ 2 6∈ Q . Problem : Let A ? R be nonempty and bounded below. Set ? A = {? a : a ∈ A }. Prove that inf( A )= ? sup( ? A ). Solution: First note that ? Ais nonempty and bounded above. Indeed, Acontains some element x, and then ? x ∈ A; moreover, A has a lower bound m,and ? m is an upper bound for ? A . We now know that b= sup( ? A) exists. We show that ? b= inf( A). That ? b is a lower bound for A is