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Fundamental Mathematics - 347 G1
Homework 4 – Solutions
1. 4.6. We compute
f (Monday) = 6
f (Tuesday) = 7
f (Wednesday) = 9
f (Thursday) = 8
f (Friday) = 6
f (Saturday) = 8
f (Sunday) = 6
This is not an injection: there are several days which map to 6.
2. 4.8. We compute
(f ◦ g)(x) = f (g(x)) = f (x2 − 1) = (x2 − 1) − 1 = x2 − 2,
and
(g ◦ f )(x) = g(f (x)) = g(x − 1) = (x − 1)2 − 1 = x2 − 2x.
3. 4.10. As shown in class, if a is not zero, then f (x) = ax + b is both injective and surjective. The
reason for this is that if we write
y = ax + b,
we can solve for x to obtain
y−b
.
a
For any y, there is thus exactly one x which maps to that y. The same argument holds for g if c 6= 0.
x=
Now, we compute:
(f ◦ g)(x) = f (g(x)) = f (cx + d) = a(cx + d) + b = acx + (ad + b),
(g ◦ f )(x) = g(f (x)) = g(ax + b) = c(ax + b) + d = acx + (bc + d),
and then
(f ◦ g − g ◦ f )(x) = ad − bc + b − d,
which is a constant function. This is clearly neither injective nor surjective, since every x maps to
the same number; there are infinitely many numbers which map to ad − bc + b − d, and so f is not
injective. Moreover, if we pick any y 6= ad − bc + b − d, there are no numbers which map to y, so f is
not surjective.
4. 4.11. First we show f (x) = 2x is a bijection from R to R. If we write
y = 2x,
then we have x = y/2, and we see that for any real number y, there is a unique real number x which
maps to y under the function f .
However, now consider f : Z → Z where f (x) = 2x. Notice now that for some y ∈ Z (in fact, the odd
ones), y/2 6∈ Z, so there is no x ∈ Z which maps to y. (As a specific example, notice that there is no
integer x such that 2x = 3.)
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5. 4.24. No. For example, let f (x) = g(x) = x. Clearly f and g are surjections. But (f g)(x) = x2 , and
as a function from Z to Z, this is not a surjection (for example, x2 = 3 has no integer solution, and
x2 = 1 has two).
6. 4.25.
(a) Yes. For example, set a = 0, and we have f (0, b) = b for any b ∈ N.
(b) Yes. Similar idea to above, set a = 1, then f (1, b) = b for all b ∈ N.
(c) Yes. Plug in b = 1, and we get
f (a, 1) =
a∗1∗2
= a,
2
and we can then get any a.
(d) No. There is no way to get the number one from this function. The way to see this is pick a as
small as possible, thus a = 1, then we have
f (1, b) = b(b + 1).
If we consider the sequence of b(b + 1) for b ∈ N, we have
0, 2, 6, 12, 20, . . .
So it is never true that b(b + 1) = 1 for b ∈ N and therefore f (1, b) is never 1.
Now, fix a as any natural number bigger than one. Then, letting b run through the integers in N,
we have
f (a, b) = 0, 2a, 6a, 12a, 20a, . . .
and clearly there is no choice of integer a for which one of those numbers is 1.
(e) No. To see why, notice that the function is monotone in each argument, meaning more simply that
if I increase either a or b, then f (a, b) increases. Now, let us compute for some natural numbers:
a = 1, b = 1,
ab(a + b)/2 = 1,
a = 1, b = 2,
a = 2, b = 1,
ab(a + b)/2 = 3,
ab(a + b)/2 = 3,
a = 2, b = 2,
ab(a + b)/2 = 8, . . .
So, we might ask, how could we get a 2? (Or a 4, 5, 6, 7?) This is not possible, since we’ve already
missed 2, and since f (2, 2) = 8, if we plug in any of the inputs that we’ve not yet checked, we
must get something larger than 8, so not 2.
7. 4.27. If f : R → R is a quadratic polynomial, then it must be of the form
f (x) = ax2 + bx + c,
with a 6= 0. Let us first assume that a > 0.
A quick proof, if we’re allowed to use calculus, is like this: Notice that f has exactly one critical point:
f ′ (x) = 2ax + b,
so the critical point is at
x=−
b
.
2a
By checking the second derivative, we have
f ′′ (x) = 2a > 0,
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so that this critical point is a local minimum. However, we know more: since the second derivative is
positive everywhere, we know f ′ is positive to the right of −b/2a and negative to the left of −b/2a,
so this must also be a global minimum. If a function has a global minimum then it is clearly not
surjective.
Now, if a < 0, then we know that g(x) := −f (x) has a global minimum from the above argument, and
thus f has a global maximum. Again, it cannot be surjective.
Of course, in this class, we haven’t proven anything using calculus (we’ve not yet even defined the
derivative!) so let’s see if we can find an algebraic proof. Again assume a > 0. We write
c
b
2
2
ax + bx + c = a x + +
a a
b
c
b2
b2
2
=a x + x+ 2 + − 2
a
4a
a 4a
!
2
b
4ac − b2
=a
x+
.
+
2a
4a2
We have “completed the square”. Now, noting that the first term in the last line is guaranteed to be
nonnegative, we have
4ac − b2
.
f (x) = ax2 + bx + c ≥
4a
So f has a global minimum, and thus it cannot be surjective.
If, on the other hand, a < 0, then we argue as before. Define g(x) := −f (x), and g has a global
minimum, and therefore f has a global maximum. Again, f cannot be surjective.
8. 4.37. If f : A → A, then f ◦ f : A → A. We will prove the statement by using the contrapositive,
namely that we will show that if f is not injective, then f ◦ f is not injective.
So, let us assume that f is not injective, namely that there exist x 6= y such that f (x) = f (y). If
f (x) = f (y), then f (f (x)) = f (f (y)). So, then we have x 6= y with f (f (x)) = f (f (y)), and therefore
f ◦ f is not injective.
9. 4.47. It is easy to see that the natural numbers are countable, since if we choose the identity map
from N to N, this is a bijection, thus N is countable.
Now, let us consider the even numbers. We write the following function:
f (n) = 2n.
Clearly, f : N → E + where E + is the even natural numbers:
E + = {2, 4, 6, 8, 10, . . . }.
We need to show that f is a bijection. First of all, it is certainly injective, since if f (n) = f (m) then
2n = 2m and thus n = m. To show it is surjective, let n be any natural even number, and notice then
that n/2 ∈ N, and we have f (n/2) = n. Since f is injective and surjective, then it is bijective. Since
there is a bijection between N and E + , they have the same cardinality.
Now consider the odd numbers. We now define f as
f (n) = 2n − 1.
Again, we see that f : N → O+ where O+ is the set of odd natural numbers:
O+ = {1, 3, 5, 7, 9, . . . }.
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Similarly to above, we see that f is injective because if 2n − 1 = 2m − 1, then n = m. Furthermore, f
is surjective because if n is odd and positive, then (n + 1) is even and positive, so (n + 1)/2 ∈ N. But
then
n+1
n+1
=2
− 1 = n + 1 − 1 = n.
f
2
2
So f is surjective, and then f is a bijection. Since we have defined a bijection between N and O+ , this
means that N and O+ have the same cardinality.
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