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 filled 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 field properties of both R and Q,wehavex =(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 field
properties of R and Q,weuser =6 0.) Thus x 6∈ Q implies rx 6∈ Q.
Problem : Prove that there is no x ∈ Q such that x2 = 12.
Solution: We prove this by contradiction. Suppose there is x ∈ Q such that
2 m 2 2 2
x = 12. Write x = n in lowest terms. Then x = 12 implies that m =12n .
Since 3 divides 12n2, it follows that 3 divides m2. Since 3 is prime (and by unique
factorization in Z), it follows that 3 divides m. Therefore 32 divides m2 =12n2.
Since 32 does not divide 12, using again unique factorization in Z and the fact that
3 is prime, it follows that 3 divides n. We have proved that 3 divides both m and
m
n, contradicting the assumption that the fraction n is in lowest terms.
x ∈ Q x2 x Q
¡Alternate¢ solution (Sketch): If satisfies = 12, then 2 is in and satisfies
x 2 y ∈ Q y2
2 = 3.√ Now prove that there is no such that = 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 −A is nonempty and bounded above. Indeed, A contains
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 immediate from