rudin数学分析原理答案
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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 a complete solution
if turned in; varying amounts of detail would need to be filled in.
Problem 1.1: 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 1.2: 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
=12n
2
.
Since 3 divides 12n
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
=12n
2
.
Since 3
2
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
n, contradicting the assumption that the fraction
m
n
is in lowest terms.
Alternate solution (Sketch): If x ∈ Q satisfies x
2
= 12, then
x
2
is in Q and satisfies
¡
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 1.5: 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 ab ove. 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) e xists. We show that −b = inf(A). That −b is
a lower bound for A is immediate from the fact that b is an upper bound for −A.
To show that −b is the greatest lower bound, we let c>−b and prove that c is not
a lower bound for A.Now−c<b,so−c is not an upper bound for −A. So there
exists x ∈−A such that x>−c. Then −x ∈ A and −x<c.Soc isnotalower
bound for A.
Problem 1.6: Let b ∈ R with b>1, fixed throughout the problem.
Comment: We will assume known that the function n 7→ b
n
,fromZ to R,is
strictly increasing, that is, that for m, n ∈ Z,wehaveb
m
<b
n
if and only if
m<n. Similarly, we take as known that x 7→ x
n
is strictly increasing when n is
Date: 1 October 2001.
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