Download Advanced Calculus I MAA4226 Homework 1 Solutions

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Advanced Calculus I MAA4226
Homework 1 Solutions
I. Do problems 2, 4, and 6 from page 22 of Rudin.
2. You are allowed to use the unique prime factorization property of integers in these problems.
Assume p/q is a rational number in lowest terms such that (p/q)2 = 12. Then
p2 = 12q 2 ⇒ 3|p2 ⇒ 3|p.
So, write p = 3k; hence
9k 2 = 12q 2 ⇒ 3k 2 = 4q 2 ⇒ 3|q.
Since p and q are both divisible by 3, this contradicts the fact that p/q was in lowest terms,
completing the proof.
4. Since E 6= φ, choose x ∈ E. The definition of lower and upper bounds implies that α ≤ x ≤ β,
so α ≤ β.
6. Fix a real number b > 0. In this problem, you are allowed to use the standard facts about
integer exponents.
(a) If m, n, p, q are integers, with n, q > 0, such that m/n = p/q, we need to prove that (bm )1/n =
(bp )1/q . The assumption implies that mq = np. Let K = mq = np. Now, let x = (bm )1/n and let
y = (bp )1/q . Compute as follows:
xnq = (((bm )1/n )n )q = bmq = bK
y nq = (((bp )1/q )q )n = bpn = bK .
(In the above computations, we used that fact that (L1/n )n = L; this is true by definition of L1/n .)
Hence x = (bK )1/nq = y, finishing part (a).
(b) Need to prove that for if r and s are rational, then br bs = br+s . To do this, just give r and s a
common denominator, writing r = m/n and s = p/n. (Note these are not in lowest terms, but it
doesn’t matter by (a)). Then calculate
br bs = (bm )1/n (bp )1/n = (bm bp )1/n = (bm+p )1/n = b(m+p)/n = br+s .
For the first = sign, we used the definition of bm/n from (a); for the second = sign, we used the
Corollary on page 11; the other = signs should be clear.
(c) For this part, let’s start with a couple of Lemmas:
Lemma 1: For all positive integers m, n, where n > 0, (bm )1/n = (b1/n )m .
Proof of Lemma 1: We know (b1/n )m is positive. Now compute:
((b1/n )m )n = (b1/n )mn = ((b1/n )n )m = bm .
Hence, by definition, (b1/n )m = (bm )1/n .
Lemma 2. If r and s are rational, then r ≤ s if and only if br ≤ bs .
Poof of Lemma 2: The fact that this lemma is true for r and s integers follows from the basic laws
of ordered fields. Letting r and s be general rational numbers, give them a common denominator,
so r = m/n and s = p/n, where m, n, p are integers, and n > 0. Then
r ≤ s ⇔ m ≤ p ⇔ (b1/n )m ≤ (b1/n )p ⇔ br ≤ bs
proving Lemma 2.
Now, we want to prove that if x is rational, bx = sup B(x). This is immediate, since the elements of
B(x) are just the real numbers of the form bt , where t is rational and satisfies t ≤ x. From Lemma
2, it follows immediately that bx is the largest of these numbers, and hence the least upper bound
of B(x).
(d) We are now supposed to prove that bx+y = bx by for all real x and y. This is difficult! Let’s
start by making the following definition: Given two sets A and B of real numbers, define the set
AB = {ab : a ∈ A and b ∈ B}.
Lemma X: If A and B are nonempty sets of positive real numbers and are bounded above, then
sup AB = (sup A)(sup B).
Proof of Lemma X: Let α = sup A, β = sup B, and γ = sup C. Given x ∈ AB, we can write x = ab
with a ∈ A and b ∈ B. Since a ≤ α and b ≤ β, we know that x = ab ≤ αβ. This proves that αβ is
an upper bound for AB, so γ ≤ αβ. To prove that γ = αβ, suppose to the contrary that γ < αβ.
Let = αβ − γ, a positive real number. Let δ = /(α + β).
Since α = sup A, we know that α − δ is not an upper bound for A, so there exists an a ∈ A such
that α − δ < a. Likewise, there exists a b ∈ B such that β − δ < b. It follows that
γ = αβ − < αβ − + (/(α + β))2 = αβ − δ(α + β) + δ 2 = (α − δ)(β − δ) < ab.
But since ab ∈ AB, this contradicts the fact that γ = sup AB. We conclude that γ = αβ, proving
Lemma X.
To complete the proof of part (d) is, for the most part an application of Lemma X, but we need to
consider some different cases.
Case 1. If x and y are both rational, we already proved the required fact in part (b).
Case 2. If x is rational and y is not rational, then B(x + y) = B(x)B(y), so the result follows
immediately from Lemma X.
Case 3. If x, y, and x + y are all irrational, then, just as in Case 2, B(x + y) = B(x)B(y) so Lemma
X applies here too.
Case 4. If x and y are irrational but x + y is rational, then it’s not quite true that B(x + y) =
B(x)B(y) (because bx+y ∈ B(x + y), while bx+y 6∈ B(x)B(y)), so we cannot apply Lemma X
directly. There is a little more work required here, but I am temporarily out of energy!
II. A. Prove the following statement: Assume F is an ordered field and x, y, z, t ∈ F are all positive.
If x < y and z < t, then xz < yt. You should be able to do this using only Definition 1.17 and
Proposition 1.18.
Proof: By Proposition 1.18(b), we know that xz < yz, and by the same Proposition, yz < yt. Then
from the definition of an ordering, it follows that xz < yt.
II. B. Let p be a prime integer, and let Z/pZ = {[0], [1], . . . , [p − 1]} denote the field of integers
modulo p, which is a field where the usual operations are defined using “clock arithmetic,” as
discussed in class. Is it possible to put an ordering on Z/pZ to make it into an ordered field? If
so, give the ordering. If not, prove it.
Proof: Suppose we had an ordering <, making Z/pZ into an ordered field. Clearly, [0] is the
additive identity element and [1] is the multiplicative identity element. Hence, by Proposition
1.18(d), [1] > [0]. By adding this inequality to itself (applying II.A.), we see that [2] > 0. Adding
these last two inequalities gives [3] > [0]. Continuing, gives this gives [p − 1] > [0], and after that,
[0] > [0]!. But clearly this violates the definition of an ordered set (since [0] = [0]). Hence our
assumption that an ordering exists must be false.