HW 2 Solutions, MATH 55, Spring 2016
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1.7, 1 Let 2k + 1 and 2j + 1 be two odd integers. Then 2k + 1 + 2j + 1 = 2(j + k) + 2 = 2(j + k + 1),
which two divides, and so is even.
1.7, 8 Let n = a2 and suppose n+2 = b2 . We can assume that a, b are posiive. Then 2 = (n+2)−n =
b2 − a2 = (b − a)(b + a). One of these factors is 1 and one of them is 2. It must be that,
a + b = 2 and b − a = 1. There are no integers that satisfy these equations.
1.7, 12 The product is irrational. Let a = p/q be rational (p, q are integers) and b irrational. Then
ab = pb/q. If this is rational, then pb/q = x/y for integers x, y. But then, b = xq/yp,
contradicting that b is irrational.
1.7, 16 We prove the contrapositive. That is, suppose m and n are odd. Then let m = 2k + 1 and
n = 2j + 1. Then mn = 4jk + 2k + 2j + 1 = 2(jk + j + k) + 1. This is odd.
1.7, 18 (a) Suppose that n is odd, i.e. n = 2k +1. Then 3n+2 = 3(2k +1)+2 = 6k +5 = 2(3k +2)+1,
which is odd. (b) Suppose that 3n + 2 is even and n is odd. Then 3n + 2 = 2k and n = 2j + 1.
Then substituting the second equation into the first, we find that 3(2j + 1) + 2 = 6j + 5 = 2k.
Rearranging, we find that 2(k − 3k) = 5. The left hand side is even and the right hand side
is odd, the desired contradiction.
1.7, 24 Suppose the opposite is true, and that every month only has two or fewer days selected. Since
there are 12 months, this means that the number of days selected is less than or equal to
24 = 2 · 12. This contradicts that we have selected 25 days.
1.8, 10 More generally, we can show that either n or n + 1 is not a perfect square, where n > 0 is an
integer. Suppose that n = a2 and n + 1 = b2 with a and b positive. Then 1 = n + 1 − n =
b2 − a2 = (b − a)(b + a). Then b − a = b + a = 1, which cannot happen for positive integers.
This is a nonconstructive proof.
√
1.8, 14 This is false. The square root of 2 is not rational, and 2 = 21/2 .
1.8, 18 Let r be an irrational number. In particular, r is not an integer, so there is an integer n
such that n < r < n + 1. Further, since n + 21 = 2n+1
is rational, r 6= n + 12 . We will adopt
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the notation d(x, y) = |x − y| for easy reading (d stands for distance). We claim that either
d(r, n) < 21 or d(r, n + 1) < 12 . Suppose otherwise; since d(r, n) 6= 12 and d(r, n + 1) 6= 12 (since
r 6= n + 21 ), we have that d(r, n) > 21 and d(r, n + 1) > 12 . Summing these two inequalities,
we find that 1 = d(r, n) + d(r, n + 1) > 1, which is a contradiction, proving our claim. This
proves existence of an integer which is less than 12 distance from r.
For uniqueness, let us note that, since d(r, n) + d(r, n + 1) = 1, if one of d(r, n) or d(r, n + 1)
is less than 21 then the other is greater than 12 . I claim that for integers m 6= n, n + 1, one has
that d(r, m) > 1. To see this, if m < n, then d(r, m) = d(r, n) + d(n, m) ≥ d(r, n) + 1 > 1. If
m > n + 1 then d(r, m) = d(r, n + 1) + d(n + 1, m) ≥ d(r, n + 1) + 1 > 1.
1.8, 24 The quadratic mean is always greater than or equal to the arithmetic mean. We can write
the arithmetic mean as the square root of its own square (the reason is so we can more easily
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HW 2 Solutions, MATH 55, Spring 2016
compare with the quadratic mean):
x+y
2
s
=
x+y
2
r
2
=
x2 + 2xy + y 2
4
Then, we would like to show that, with the square of the arithmetic mean on the left and the
square of the quadratic mean on the right,
x2 + 2xy + y 2
x2 + y 2
≤
4
2
Multiplying by 4,
2x2 + 4xy + 2y 2 ≤ 4x2 + 4y 2
Rearranging terms,
2xy ≤ x2 + y 2
Finally, we find that
0 ≤ x2 − 2xy + y 2 = (x − y)2
The last statement is true, since squares of real numbers are always nonnegative. Since
every step we have performed is invertible (i.e. we can undo them by rearranging terms, or
multiplying by 4), this implies that the square of the arithmetic mean is less than or equal
to the square of the quadratic mean. Since all numbers involved are nonnegative this implies
that the arithmetic mean is less than or equal to the quadratic mean.
√
1.8, 34 Suppose that 3 2 = p0 /q0 . Then 2 = p30 /q03 , i.e. 2q03 = p30 . Thus, p30 is even. We claim
that p0 is even. Suppose otherwise, that it is odd, and write p0 = 2k + 1. Then p30 =
8k 3 + 6k 2 + 6k + 1 = 2(4k 3 + 3k 2 + 3k) + 1, which is odd, contradiction that p30 is even. Thus
we must have that p0 is even.
Since p0 is even, we can write p0 = 2p1 . Going back to our equation 2q03 = p30 and substituting,
we find that 2q03 = 8p31 , i.e. q03 = 4p31 . Thus, by a very similar argument as before, q0 is also
even, and we can write q = 2q1 . Doing a similar substitution, we find that 2q13 = p31 . Now
note that q1 < q0 and p1 < p0 , and all are positive integers. We can continue this process
indefinitely, and obtain sequences
p0 > p1 > p2 > p3 > · · ·
q0 > q1 > q2 > q3 > · · ·
where each qi , pi is a positive integer. However there are not an infinite number of positive
integers pi strictly less than p0 , and so we cannot have an infinite number, and we have a
contradiction.
2.1, 10 (a) T (b) T (c) F (d) T (e) T (f) T (g) F (it is not a strict subset)
2.1, 18 A = {1} and B = {1, {1}}.
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HW 2 Solutions, MATH 55, Spring 2016
2.1, 22 Yes. For A a set, the union of all sets in P(A) is equal to A. Thus, if P(A) = P(B), then the
union of all the elements of the power sets (which are sets themselves) are equal, so A = B.
(Alternate proof: A ∈ P(A) = P(B), so A ⊂ B. Similarly, B ∈ P(B) = P(A), so B ⊂ A. So,
A = B).
2.1, 26 The elements of A × B are pairs (a, b) with a ∈ A and b ∈ B. Since A ⊂ C and B ⊂ D, we
have that a ∈ C and b ∈ D. The (a, b) ∈ C × D.
2.1, 42 (a) There is a real number x such that x3 = −1. It is true; x = −1 works.
(b) There is an integer x such that x + 1 > x. This is true; any integer works.
(c) For all integers x, x − 1 is an integer. This is true.
(d) For all integers x, x2 is an integer. This is true.
2.1, 46 (a) Suppose that S ∈ S. Then, S 6∈ S by definition of S. (b) Suppose that S 6∈ S. Then
S ∈ S by definition of S.
2.2, 12 (a) A ∩ b (b) A − B (c) A ∪ B (d) A ∪ B
2.2, 16de (d) Automatically, ∅ ⊂ A ∩ (B − A). Suppose that a ∈ A and a ∈ B − A. If a ∈ B − A then
a 6∈ A. Thus, A ∩ (B − A) = ∅.
(e) First we show that A ∪ (B − A) ⊂ A ∪ B. If x ∈ A then clearly x ∈ A ∪ B. If x ∈ B − A
then x ∈ B so x ∈ A ∪ B. Now we show that A ∪ B ⊂ A ∪ (B − A). If x ∈ A, then clearly
x ∈ A ∪ (B − A). If x ∈ B, then either x ∈ A or x 6∈ A. If x 6∈ A then x ∈ B − A and so
x ∈ A ∪ (B − A). If x ∈ A then x ∈ A ∪ (B − A¿
2.2, 48 (a) An (b) A1
2.2, 26bc (b)
(c)
2.2, 18cd (c) If x ∈ (A − B) − C, then x ∈ A but x 6∈ B and x 6∈ C. Then, x ∈ A − C. (d) If x ∈ A − C
then x 6∈ C. If x ∈ C − B, then x ∈ C. Thus (A − C) ∩ (C − B) = ∅.
2.3, 1 (a) It is not defined for x = 0 (b) It is not define for negative x (c) f (x) has two possible
values.
2.3, 4bc (b) The domain is N and the range is {x ∈ Z | x ≥ 2}. (c) The domain is the set of bit
strings. The range is {x ∈ Z | x ≥ 0}.
2.3, 14 (a) This is onto. Let m = 0 and let n ranger over all integers.
(b) This is not onto; 2 is not in the range. Suppose f (m, n) = (m − n)(m + n) = 2. Then we
have the cases: (i) m − n = 2 and m + n = 1, (ii) m − n = 1 and m + n = 2, (iii) m − n = −2
and m + n = −1, and (iv) m − n = −1 and m + n = −2. We see that none of these are
solvable in integers.
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HW 2 Solutions, MATH 55, Spring 2016
(c) This is onto. Let m = 0 and let n range over all integers.
(d) This is onto. For positive integers, let n = 0 and let m be the desired integer. For negative
integers, let m = 0 and n be the desired integers.
(e) This is not onto. For example, there is no integer m such that m2 − 4 = 1.
2.3, 20 (a) f (n) = n + 1 (b) f (n) = dn/2e (the ceiling function on n/2) (c) f (1) = 2, f (2) = 1, and
f (n) = n for n 6= 1, 2.
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