Math 220
Exam 2 Partial Solutions
November 8, 2013
S. Witherspoon
1. (a) f g(u) = f (g(u)) = f (w) = u, f g(v) = f (g(v)) = f (v) = w, f g(w) =
f (g(w)) = f (w) = u
(b) f is invertible, and f −1 (u) = w, f −1 (v) = u, f −1 (w) = v
2. (a) Not invertible. (For example, 3 is not in the image of f , since if it were, then
3 = 3x + 2 for some integer x, which implies x = 31 , a contradiction since 31 is not an
integer.)
(b) Invertible. (f −1 (x) = 13 (x − 2) for all x ∈ Q)
(c) Not invertible. (For example, 2 is not in the image
of f , since if it were, then
√
√
3
3
2 = x for some rational number x, which implies x = 2, a contradiction since 3 2
is not a rational number.) √
(d) Invertible. (f −1 (x) = 3 x for all x ∈ R)
3. (a) Yes: For all a, b ∈ Q, a ∗ b = 2a + 2b = 2b + 2a = b ∗ a (by commutativity of
addition of rational numbers).
(b) No: For example, 1 ∗ (2 ∗ 3) = a ∗ (2 · 2 + 2 · 3) = 1 ∗ 10 = 2 + 20 = 22 while
(1 ∗ 2) ∗ 3 = (2 + 4) ∗ 3 = 6 ∗ 3 = 12 + 6 = 18.
(c) Yes: Let a, b be even integers, so that a = 2x and b = 2y for some integers x, y.
Then
a ∗ b = 2(2x) + 2(2y) = 2(2x + 2y),
which is even since 2x + 2y is an integer.
(d) No: For example, 1, 3 ∈ O but 1 ∗ 3 = 2 + 6 = 8, which is not in O.
4. We first check that the statement is true when n = 1: Since 2+(4·1+2) = 2+6 = 8
and 2(1 + 1)2 = 8, it is true.
Now assume that the statement is true when n = k for some positive integer k,
that is, assume that
2 + 6 + 10 + · · · + (4k + 2) = 2(k + 1)2 .
Then
2 + 6 + 10 + · · · + (4k + 2) + (4(k + 1) + 2) =
=
=
=
2(k + 1)2 + (4(k + 1) + 2)
2(k 2 + 2k + 1) + 4k + 6
2k 2 + 8k + 8
2(k + 2)2 = 2((k + 1) + 1)2 ,
which shows that the statement is true when n = k + 1. Therefore it is true for all
positive integers n.
1
2
5. (a)
48
28
20
8
=
=
=
=
28 · 1 + 20
20 · 1 + 8
8·2+4
4·2
Therefore (48, 28) = 4.
(b) Using the results of part (a):
4 =
=
=
=
=
20 − 8 · 2
20 − (28 − 20) · 2
20 · 3 − 28 · 2
(48 − 28) · 3 + 28 · (−2)
48 · 3 + 28 · (−5)
Therefore we may let x = 3 and y = −5.
6. (a) There are several possible answers. One is a = 4, b = 9.
(b) There are several possible answers. One is a = 2, b = 18.
(c) Recall from a homework problem (5.4 #9) that a positive integer n is a perfect
square if and only if, when n is written in standard form, all of the exponents are
even. Setting n = x2 in this problem, then, we may write x2 in standard form:
2nr
2
x2 = p12n1 p2n
2 · · · pr
for distinct primes p1 , p2 , . . . , pr and positive integers n1 , n2 , . . . , nr .
Since x2 = ab, for each prime factor pi , either pi divides a or pi divides b. Since
i
divides a. Similarly, if
(a, b) = 1, if pi divides a, then it does not divide b, and so p2n
i
2ni
pi divides b, then pi divides b. Therefore, by relabeling if necessary, a = p12n1 · · · ps2ns
2n
r
and b = ps+1s+1 · · · p2n
r , for some s (1 ≤ s ≤ r). Again, by the homework problem,
since in the standard form of a, all exponents are even, a is a perfect square (in
fact, a = y 2 where y = pn1 1 · · · pns s ). Similarly, b is a perfect square (b = z 2 where
ns+1
z = ps+1
· · · pnr r ).