Download Math 500 – Intermediate Analysis Homework 2 – Solutions!

Math 500 – Intermediate Analysis
Homework 2 – Solutions!
Fall 2014
for all n ∈ N.
1.1 Prove that 12 + 22 + · · · + n2 = n(n+1)(2n+1)
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Solution: To prove the base case, we must verify the formula for n = 1. In
= 1 so the case holds when n = 1. For the
this case, 12 = 1 and n(n+1)(2n+1)
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induction step, we assume the formula holds for some n ∈ N and show this
implies it holds for n + 1. To this end, notice that
n(n + 1)(2n + 1)
+ (n + 1)2
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n(n + 1)(2n + 1) + 6(n + 1)2
=
6
(n + 1)(n + 2)(2(n + 1) + 1)
=
6
12 + 22 + · · · + n2 + (n + 1)2 =
by induction hypothesis
by tedious algebra.
Thus, the induction step holds and hence the proof is complete by mathematical
induction.
1.2 Prove that 3 + 11 + · · · + (8n − 5) = 4n2 − n for all n ∈ N.
Solution: When n = 1, we have 3 = 4 − 1 so the base case holds. Now, assume
the formula holds for some n ∈ N, i.e. suppose that for some n ∈ N we have
3 + 11 + · · · + (8n − 5) = 4n2 − n.
Then
3 + 11 + · · · + (8n − 5) + (8(n + 1) − 5) = 4n2 − n + (8(n + 1) − 5) by induction hypothesis
= 4n2 − n + 8n + 3
= 4(n + 1)2 − (n + 1) by algebra.
Thus, the induction step holds and hence the proof is complete by mathematical
induction.
1.11 For each n ∈ N, let Pn denote the assertion “n2 + 5n + 1 is an even integer.”
(a) Prove that Pn ⇒ Pn+1 , i.e. show that Pn+1 is true whenever Pn is true.
Solution: Suppose Pn is true for some n ∈ N. Then
(n + 1)2 + 5(n + 1) + 1 = (n2 + 5n + 1) + 2n + 6.
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Since we are assuming Pn is true, it follows that Pn is true since it is the sum
of two even integers (CHECK: Show that the sum of even numbers is even!).
Thus, Pn+1 is true whenever Pn is true.
(b) For which n ∈ N is Pn actually true? What is the moral of this exercise?
Solution: Pn is actually FALSE for all n ∈ N. Indeed, when n is even it follows
that n2 + 5n is even (CHECK!) and hence n2 + 5n + 1 is odd and when n is
odd the quantity n2 + 5n + 1 is the sum of three odd integers and is hence
odd (CHECK!). Thus, although Pn ⇒ Pn+1 , none of the statements Pn are
actually true. This shows the importance of establishing the base step when
using mathematical induction. Moreover this exercise should emphasize the
important fact that when proving a theorem by mathematical induction: when
showing Pn ⇒ Pn+1 in the induction step one never actually proves that Pn+1
is true, only that if Pn is true then Pn+1 is true.
2.2 Show that 21/3 , 51/7 , and 131/4 do not represent rational numbers.
Solution: We begin by showing that 21/3 is not
√ rational. This could either be
done by mimicking my proof from class that 2 is not in Q, or else one could
use the Rational Zeros Theorem from the text. Here, we will use the latter
approach. First, notice that the number 21/3 is a solution of the polynomial
equation
x3 − 2 = 0.
By the Rational Zeros Theorem, the only rational roots are of the form r = pq
where p and q are integers with no common factors such that p divides 1 and
q divides 2. Thus, the only possible rational roots are r = ±1 or r = ±2, none
of which satisfy the required polynomial equation. Thus, 21/3 is not in Q as
claimed. The other cases are handled similarly.
√
2.3 Show that (2 + 2)1/2 does not represent a rational number.
Solution: We begin by finding a polynomial equation this number satisfies. To
this end, notice that
√
x = (2 + 2)1/2
√
⇒ x2 = 2 + 2
√
⇒ x2 − 2 = 2
⇒ (x2 − 2)2 = 2
and hence the number (2 +
√ 1/2
2) is a root of the polynomial equation
x4 − 4x2 + 2 = 0.
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By the Rational Zeros Theorem the only possible rational roots are √
±1 and ±2,
none of which satisfy the desired polynomial equation. Thus, (2 + 2)1/2 does
not belong to Q.
3.5 (a) Show that |b| ≤ a if and only if −a ≤ b ≤ a.
Solution: See text.
(b) Prove that ||a| − |b|| ≤ |a − b| for all a, b ∈ R.
Solution: Notice that |a| = |a − b + b| ≤ |a − b| + |b| and hence |a| − |b| ≤ |a − b|
for all a, b ∈ R. Since the argument is symmetric in a and b, we also have that
|b| − |a| ≤ |a − b| and hence the proof is complete by part (a).
3.6 (b) Use induction to prove
|a1 + a2 + · · · + an | ≤ |a1 | + |a2 | + · · · + |an |
whenever n ∈ N and a1 , . . . , an ∈ R.
Solution: In the case n = 1, we clearly have |a1 | ≤ |a1 | for any a1 ∈ R so the
base case holds. Now, suppose the inequality holds for some collection {aj }nj=1
of n real numbers. If an+1 ∈ R, it follows that
|a1 + a2 + · · · + an + an+1 | ≤ |a1 + a2 + · · · + an | + |an+1 | by Triangle ineq.
≤ |a1 | + |a2 | + · · · + |an | + |an+1 | by induction hypothesis.
Thus, the induction step holds and hence the proof is complete by mathematical
induction.
3.8 Let a, b ∈ R. Show that if a ≤ b1 for every b1 > b, then a ≤ b.
Solution: We will prove this by contrapositive. This could also be proven by
contradiction, but seems a bit more clear this way. Recall that to prove an
implication P ⇒ Q by contrapositive, one assumes that Q is false and tries to
show that P is also false. In the notation of mathematical logic, one is hence
trying to show that P ⇒ Q by showing ¬Q ⇒ ¬P .
To this end, suppose that a > b. By the denseness of Q in R, there exists a
be the
r ∈ Q such that b < r < a (alternatively, you could have just let r = a+b
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arithmetic average). Thus, under the assumption that a > b we have proven
the existence of a real number r > b for which the inequality a ≤ r does not
hold. It follows that if it is true that a ≤ b1 for every b1 > b, it must be that
a ≤ b as claimed.
Note: The difference between proofs by contradiction and proofs by contrapositive is quite subtle. When proving the statement P ⇒ Q by contradiction,
one assumes that P is true and that Q is false and tries to derive some sort
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of contradiction. Thus, proofs by contradiction can be quite cumbersome and
awkward to follow. However, when using the method of contrapositive to prove
P ⇒ Q, one assumes that Q is false and tries to prove that P is also false. As
a result, one has a very clear defined goal in mind: mainly to start from the
assumption that Q is false and to show this implies that P is false.
4.3 See text for statement of problem and answers for left column. The answers for
the
√ right column are as follows: (b) 1, (d) π, (f) 0, (h) NO sup, (j) 1, (l) 2, (n)
2, (p) 10, (r) 1, (t) 2, (v) 1.
4.4 See text for statement of problem. The answers are as follows: (a) 0, (b) 0, (c)2,
√
(d) e, (e) 0, (f) 0, (g) 0, (h) 2, (i) 0, (j) 23 , (k) 0, (l) NO inf, (m) -2, (n) - 2,
(o) No inf, (p) 1, (q) 0, (r) 1, (s) 0 (Since
there are infinitely many primes....
√
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CHECK!), (t) No inf, (u) 0, (v) -1, (w) - 2 .
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