Math 3220-1
Exam 0
1. Let a and b be distinct primes. Prove that x =
Solutions
√
ab is not rational.
Proof. Suppose x = p/q with p, q ∈ Z in lowest terms. Then square both sides and multiply by q 2 .
Thus we obtain
q 2 ab = p2 .
Note that a is a factor in the left hand side, thus a divides p2 . Since a is prime, it must therefore divide
p. Say p = ac, and substitute. We get
q 2 ab = a2 c2 .
We may cancel an a from both sides to get
q 2 b = ac2 .
Now a is a factor in the right hand side, so a must divide the left hand side. But since b is prime, and
not equal to a, we may conclude that a divides q 2 . Hence a divides q, and p and q have a common
factor of a. This contradicts our assumption that p and q are in lowest terms.
2. Let S =
∞ [
1
.
n2
n=1
(a) Prove that each x ∈ S is a boundary point.
(b) Prove that although 0 ∈
/ S, it too is a boundary point.
Proof. (a) Let x ∈ S. Then x = 1/n2 for some n ∈√N. Let ε > 0, and consider N (x, ε). Since
/ Q, let y = x + 2ε . Then, since x
x ∈ N (x, ε), S ∩ N (x, ε) 6= ?. If ε ∈ Q, let y = x + 22 ε; and if ε ∈
is rational, y is not rational. Therefore y ∈
/ S. But we notice that y is within ε of x. We conclude
that N (x, ε) ∩ (R r S) 6= ?. Thus, x is a boundary point. (Note that S has no interior points).
(b) Let ε > 0. Then there is a natural number n such that n1 < ε, by the Archimedean Property.
Then certainly n12 is also less than ε. Thus there is a point in S within ε of 0. Similarly, let
y = −ε/2. Then y ∈
/ S (since it is negative) but is within ε of 0. Therefore, 0 is also a boundary
point for S.
3. Let s1 = 0, and let sn+1 =
√
6 + sn .
(a) Prove that (sn ) converges.
(b) Calculate its limit.
Proof. (a) We will show that (sn ) is increasing and bounded. This will imply that it converges. We
first
√ want to show that sn+1 ≥ sn . We will do so by induction. For n = 1, this is clear, since
6 > 0. Now suppose that sk+1 ≥ sk . Consider the following
p
√
sk+2 = 6 + sk+1 ≥ 6 + sk = sk+1 .
Thus, by induction, we conclude that (sn ) is an increasing sequence.
Now we claim that (sn ) is bounded by 3. That is, sn ≤ 3 for all n. Certainly s1 ≤ 3. Now suppose
that sk ≤ 3. Then
√
√
sk+1 = 6 + sk ≤ 6 + 3 = 3.
Therefore, by induction, we have shown that (sn ) is bounded by 3. Hence it converges.
(b) We claim that (sn ) → 3. To see this, we note that (sn ) converges to the supremum s of its range.
Also we note that
lim sn+1 = lim sn .
n→∞
Thus, s must satisfy the equation
s=
n→∞
√
6 + s.
Square both sides, and write as a quadratic to get
s2 − s − 6 = 0.
This has solutions s = −2 and s = 3. Obviously s = −2 is not the limit of the sequence. Therefore,
we conclude that (sn ) → 3.
You should notice that (sn ) is a sequence of irrational numbers which converges to the rational
number 3.
4. Let f (x) =
x cos(x),
0,
if x ∈ Q
if x ∈
/Q
(a) Show that f is continuous at 0.
(b) Show that f is not differentiable at 0.
Proof. (a) Note that −|x| ≤ f (x) ≤ |x| for all x. Thus, by the Squeeze Theorem, lim f (x) = 0 = f (0).
x→0
Therefore, f is continuous at 0.
(b) To see that it is not differentiable at 0, note that on the rationals, lim f (x)/x is 1, while on the
irrationals, it is zero. Thus, the limit does not exist.
5. Define g(x) =
sin(x) sin(1/x),
0,
x→0
if x 6= 0
if x = 0
(a) Is g continuous at 0?
(b) Prove your assertion.
Yes, g is continuous at 0.
Proof. Note that for all x 6= 0, −| sin x| ≤ g(x) ≤ | sin x|. Thus, since ±| sin x| both have limit 0 as x
goes to 0, we may conclude, by the Squeeze Theorem, that lim g(x) = 0 = g(0). Thus g is continuous
x→0
at 0.
6. Let p(x) = 3x3 − 5x2 + 1. Prove, using the Intermediate Value Theorem, that p has three real roots.
Note that we know (by the Fundamental Theorem of Algebra) that p has exactly three complex roots.
I am asking you to show that in fact, all three roots are real. You should be able to find intervals in
which each of these roots lies.
Proof. We first note that p is a continuous function because it is polynomial. Second, we note that if
[a, b] is an interval where p(a) and p(b) have different signs, then the Intermediate Value Theorem tells
us that p has at least one zero on [a, b]. Therefore, if we can find three such intervals, then we know
that p has three real roots. Now for some easy computations:
p(−1) =
p(0) =
−7,
1,
p(1) =
p(2) =
−1,
5.
So we see that p changes sign on the intervals [−1, 0]; [0, 1]; and [1, 2].
7. Let f and g be differentiable functions on (a, b), and continuous on [a, b]. Further suppose f (a) =
f (b) = 0. Define h(x) = f (x)eg(x) .
(a) Is h continuous on [a, b]? Prove your answer.
(b) Is h differentiable on (a, b)? Prove your answer.
(c) Prove that there is some point x ∈ (a, b) such that f ′ (x) + f (x)g ′ (x) = 0. Hint: We defined h for
a reason; this is that reason.
Proof. (a) Note that f , g and exp are all continuous functions. Products and compositions are
continuous, therefore h is continuous.
(b) Likewise products and compositions of differentiable functions are differentiable. Therefore h is
differentiable as well.
(c) Now h satisfies the hypotheses of the Mean Value Theorem on [a, b]. Notice that since f (a) =
f (b) = 0, we also have h(a) = h(b) = 0. Thus, applying the Mean Value Theorem tells us that
there is at least one x0 ∈ (a, b) such that h′ (x0 ) = 0. If we calculate h′ (x) we get
h′ (x) = f ′ (x)eg(x) + f (x)eg(x) g ′ (x).
We note that eg(x) is always positive, therefore we may conclude that at x = x0 , f ′ (x0 ) +
f (x0 )g ′ (x0 ) = 0.
8. Let f be continuous on [0, ∞). Suppose that f (x) 6= 0 if x > 0, and that
Z x
2
[f (x)] = 2
f (t) dt,
0
for all x ≥ 0. Prove that f (x) = x for all x ≥ 0. Hint: What is f ′ (x)?
Proof. Let’s begin by differentiating the given equation (using the Second Fundamental Theorem of
Calculus):
2f (x) · f ′ (x) = 2f (x).
Therefore, we may conclude that either f (x) = 0, or f ′ (x) = 1. The first case is ruled out by the given
assumption that f (x) 6= 0 if x > 0. Then f (x) is a continuous function whose derivative is 1. Therefore
f (x) = x + C for some constant C. But we may turn to the equation
Z x
2
[f (x)] = 2
f (t) dt,
0
once more to see that f (0) = 0. Thus, C = 0 and f (x) = x for all x ≥ 0.
9. Let G(x) =
Z
x3
2
et sin t dt. Calculate G′ (x). You may use any theorems you wish.
0
Solution. Let F (y) =
Z
0
y
2
et sin t dt, and let h(x) = x3 . Then G(x) = (F ◦ h)(x). We must use both
the Chain Rule, and the Second Fundamental Theorem of Calculus. Together, they yield
6
G′ (x) = 3x2 ex sin(x3 ).