Download Math 201 – Homework 5 – solutions

Math 201 – Homework 5 – solutions
Problem 7.6 Suppose x, y ∈ R. Then x3 + x2 y = y 2 + xy if and only if
y = x2 or y = −x.
Solution. Proof ⇐: Substitute y = x2 and check that you get a true
equation. Then do the same with y = −x.
Proof ⇒: Suppose x3 + x2 y = y 2 + xy. Take the difference between left and
right side of this equation, then factorize:
x3 + x2 y − y 2 − xy = 0
x2 (x + y) − y(y + x) = 0
(x2 − y)(x + y) = 0.
The last line is equivalent to the first, which is equivalent to the given equation. And from the last line, it becomes obvious that it can only be true if
at least one of the factors is zero, meaning x2 − y = 0 or x + y = 0. This
gives the two possibilities y = x2 or y = −x.
Problem 7.8 Let a, b ∈ Z. Prove that a ≡ b modulo 10 if and only if a ≡ b
(mod 2) and a ≡ b (mod 5).
Solution. Proof ⇒: Suppose a ≡ b (mod 10). So 10|(a − b). Therefore
5|(a − b) and 2|(a − b), and we obtain the two statements modulo 2 and
modulo 5.
Proof ⇐: Suppose a ≡ b (mod 2) and a ≡ b (mod 5). So 2|(a − b) and
5|(a − b). We can write
a − b = 2m = 5n.
Since 2 is a prime, it divides at least one of the factors of 5n, It does not
divide 5, so it divides n and we can write n = 2k, hence a − b = 10k and
a ≡ b (mod 10).
Problem 7.20 There exists an n ∈ N for which 11|(2n − 1.
Solution. One example is enough, eg n = 10
210 − 1 = 1023
is divisible by 11 (this is the smallest example, and the possible example are
exactly the multiples of 10 – not part of this problem, but interesting).
Problem 8.6 Let A, B, C be sets. Prove that if A ⊆ B then A−C ⊆ B −C.
Solution. Let A ⊆ B. We want to prove A − C ⊆ B − C. So let x ∈ A − C.
This means x ∈ A but x 6∈ C. By hypothesis, x is then an element of B and
taken together, x is in B − C.
Problem 8.28 Prove that {12a + 25b : a, b ∈ Z} = Z.
Solution. Let X = {12a + 25b : a, b ∈ Z}. Equality between sets: we show
X ⊆ Z and Z ⊆ X. The first of these is obvious, since 12a + 25b is always an
integer when a, b ∈ Z. Now for the second. Let x ∈ Z. To show that x ∈ X,
note that for a = −2 and b = 1
12(−2) + 25(1) = 1
(you can say this proves that 1 ∈ X). Multiply both sides of this equation
by x, and you get
12(−2x) + 25(x) = x.
So the choices a = −2x and b = x demonstrate that x ∈ X (there are other
possible choices, too).
Problem 9.8 Let A, B, C be sets. Is A − (B ∪ C) = (A − B) ∪ (A − C)?
(Disproof).
Solution. Example: A = {1, 2}, B = {1}, C = {2}. This gives B ∪ C = A
and
A − (B ∪ C) = ∅
(A − B) ∪ (A − C) = {1, 2}
and these are obviously not equal.
A Venn diagram for the three sets A, B, C with the two given regions highlighted will show how to construct lots more examples. But any one example
is enough.
Problem 9.28 Let a, b ∈ Z. Is it true that if a|b and b|a then a = b?
(Disproof).
Solution. No, eg a = 4 and b = −4 (not part of this problem: generally a|b
and b|a imply only a = ±b).
Problem 10.10 For any integer n ≥ 0, 3 divides 52n − 1.
Solution. Proof by induction. Base Step: n = 0 gives 3|0 which is true.
Induction Step: Suppose 3|(52n − 1) for some value of n. Therefore there
exists k ∈ Z with
52n − 1 = 3k.
So 52n = 1 + 3k and
52(n+1) = 25(1 + 3k) = 25 + 75k = 1 + 3(8 + 25k)
which shows 3|(52(n+1) − 1.
Problem 10.19 Prove that for all n ≥ 1
1 1 1
1
1
+ + + ··· + 2 ≤ 2 − .
1 4 9
n
n
Solution. Proof by Induction. Base Step n = 1: check
Step: suppose that for some value of n ≥ 1
1
1
≤ 2− 11 . Induction
1 1 1
1
1
+ + + ··· + 2 ≤ 2 − .
1 4 9
n
n
Add 1/(n + 1)2 to both sides:
1 1 1
1
1
+ + + ··· + 2 +
≤2−
1 4 9
n
(n + 1)2
1
1
−
n (n + 1)2
.
(1)
Next,
1
1
(n + 1)2 − n
n2 + n
1
−
=
≥
=
.
2
2
2
n (n + 1)
n(n + 1)
n(n + 1)
n+1
So the amount that we subtract on the right-hand side of Equation (1) is
greater than 1/(n + 1). Hence the right-hand side is less than 2 − 1/(n + 1),
completing the Induction Step.
Problem 10.25 Concerning the Fibonacci sequence, prove that
F1 + F2 + · · · + Fn = Fn+2 − 1.
Solution. Proof by Induction. Base Step: Check that this is true for n = 1.
Induction Step: suppose the statement is true for some value of n. Then
consider
F1 + F2 + · · · + Fn + Fn+1 = Fn+2 − 1 + Fn+1
= Fn+3 − 1.
(2)
(3)
Equation (2) used the induction hypothesis. The next step used the recurrence relation defining the Fibonacci numbers.