Math 3200 Exam #2 Practice Problem Solutions
1. Suppose x ∈ R is positive. Prove that if x is irrational, then x1/6 is also irrational. Show that this is
not an if and only if statement by giving a counterexample to the converse.
Proof. By contradiction. Suppose there exists an irrational number x so that x1/6 is rational, meaning
that x1/6 = p/q for some p, q ∈ Z with q 6= 0. Then, raising both sides to the sixth power, we see that
6 p 6
p6
1/6
= 6.
x= x
=
q
q
6
But pq6 is clearly a rational number since p6 and q 6 are integers and q 6 6= 0. But this contradicts the
fact that x is irrational.
From this contradiction, then, we conclude that if x is irrational, then so is x1/6 .
Counterexample: The converse of the given statement is “If x1/6 is irrational, then x is irrational”.
This false, as we can see by letting x = 8, since
1/2 q √
√
3
1/6
1/3
8 = 2,
8
= 8
=
which we know is irrational, even though 8 is obviously rational.
2. Show that for any sets A and B,
(A − B) ∪ (B − A) = (A ∪ B) − (A ∩ B).
Proof. To prove the two sets are equal, it suffices to show containment both ways.
To that end, suppose x ∈ (A − B) ∪ (B − A). Then either x ∈ A − B or x ∈ B − A. In the first
case x ∈
/ B, whereas in the second case x ∈
/ A; either way, x ∈
/ A ∩ B. On the other hand, in both
cases x ∈ A ∪ B, so it follows that x ∈ (A ∪ B) − (A ∩ B). Since x was arbitrary, this implies that
(A − B) ∪ (B − A) ⊆ (A ∪ B) − (B ∪ A).
On the other hand, suppose x ∈ (A ∪ B) − (A ∩ B). Then x ∈ A ∪ B and x ∈
/ A ∩ B. The former implies
that either x ∈ A or x ∈ B. If x ∈ A, then x ∈
/ A ∩ B implies x ∈
/ B, so x ∈ A − B. Alternatively, if
x ∈ B, then x ∈
/ A ∩ B implies x ∈
/ A, so x ∈ B − A. Either way, it follows that x ∈ (A − B) ∪ (B − A).
Since x was arbitrary, I can conclude that (A ∪ B) − (A ∩ B) ⊆ (A − B) ∪ (B − A).
Having proved containment both ways, I’ve shown that (A − B) ∪ (B − A) = (A ∪ B) − (A ∩ B).
3. Prove that, for any n ∈ N,
20 + 21 + 22 + . . . + 2n = 2n+1 − 1.
Proof. By induction. For each n ∈ N, let P (n) be the statement that 20 + 21 + 22 + . . . + 2n = 2n+1 − 1.
Base Case: The statement P (1) says that 20 + 21 = 21+1 − 1, which is clearly true since both sides
are equal to 3.
Inductive Step: Suppose k ∈ N and assume that P (k) is true, meaning that
20 + 21 + 22 + . . . + 2k = 2k+1 − 1.
Re-writing slightly, this implies that
2k+1 = 1 + 20 + 21 + 22 + . . . + 2k .
1
Now, I want to prove P (k + 1). To do so, consider the number 2(k+1)+1 − 1. Clearly, since
2(k+1)+1 = 2 · 2k+1 , the inductive hypothesis (that P (k) is true) implies that
2(k+1)+1 − 1 = 2 · 2k+1 − 1 = 2 1 + 20 + 21 + 22 + . . . + 2k − 1
= 2 + 2 · 20 + 2 · 21 + 2 · 22 + . . . + 2 · 2k − 1
= 2 + 21 + 22 + 23 + . . . + 2k+1 − 1
= 1 + 21 + 22 + . . . + 2k+1
= 20 + 21 + 22 + . . . + 2k+1 ,
which is exactly what P (k + 1) claims.
Therefore, the Principal of Mathematical Induction implies that P (n) is true for all n ∈ N, as desired.
4. Suppose n ≥ 2 is an integer. Prove that there exists a ∈ N with 1 < a < n so that a2 ≡ 1 (mod n).
(Hint: What is a when n = 4? n = 5? What’s the pattern? )
Proof. Consider a = n − 1. Then
a2 = (n − 1)2 = n2 − 2n + 1 = n(n − 2) + 1 ≡ 1
(mod n)
since n(n − 2) is obviously a multiple of n.
5. Prove that 3 | (52n − 1) for all integers n ≥ 0.
Proof. By induction. For each non-negative integer n, let P (n) be the statement that 3 | (52n − 1).
Base Case: Since the statement is about non-negative integers, the base case is the statement P (0),
which says that 3 divides 50 − 1 = 1 − 1 = 0, which is certainly true.
Inductive Step: Let k be a non-negative integer and suppose P (k) is true, meaning that 3 divides
52k − 1. In other words,
52k − 1 = 3`
for some `inZ. In other words, 52k = 3` + 1
Now, I want to prove P (k + 1), which says that 52(k+1) − 1 is divisible by 3. Certainly
52(k+1) − 1 = 52k+2 − 1 = 52 · 52k − 1 = 52 (3` + 1) − 1,
where I can substitute 3` + 1 for 52k by the inductive hypothesis. Continuing the calculation, I
see that
52(k+1) − 1 = 52 (3` + 1) − 1 = 25(3` + 1) − 1 = 75` + 25 − 1 = 75` + 24 = 3(25` + 8).
Since 25` + 8 is an integer, this means that 52(k+1) − 1 is a multiple of 3, which is exactly the
statement P (k + 1).
Having proved the base case and the inductive step, the Principal of Mathematical Induction allows
me to conclude that P (n) is true for all integers n ≥ 0, as desired.
6. Prove that the equation
2x3 + 6x + 1 = 0
has no integer solutions.
2
Proof. By contradiction. Suppose that there were some x ∈ Z so that
2x3 + 6x + 1 = 0.
Re-arranging, this implies that
1 = −2x3 − 6x = 2(−x3 − 3x).
Since −x3 − 3x is an integer, this implies that 1 is even, which is obviously not true.
From this contradiction, then, we can only conclude that there is no integer solution to the equation
2x3 + 6x + 1 = 0.
7. Prove that the equation from the previous problem does have a real solution.
Proof. Let f (x) = 2x3 + 6x + 1. Then f (x) is continuous, so we’ll be able to use the Intermediate
Value Theorem.
Now, notice that f (0) = 2 · 03 + 6 · 0 + 1 = 1 and that f (−1) = 2(−1)3 + 6(−1) + 1 = −7. Therefore,
since 0 is between f (−1) = −7 and f (0) = 1, the Intermediate Value Theorem guarantees that there is
some c between −1 and 0 so that f (c) = 0. This c, then, is the desired real solution of the equation.
8. Prove that there is no integer a so that a ≡ 2 (mod 6) and a ≡ 7 (mod 9).
Proof. For the sake of contradiction, suppose that there were an a ∈ Z so that a ≡ 2 (mod 6) and
a ≡ 7 (mod 9). The first implies that
a = 6k + 2
for some k ∈ Z and the second implies that
a = 9` + 7
for some ` ∈ Z. Setting these two ways of writing a equal to each other, I see that
6k + 2 = 9` + 7
or, equivalently, that
6k − 9` = 5.
Now, 6 and 9 are both divisible by 3, so I can factor the left hand side:
3(2k − 3`) = 5,
which would imply that 5 is divisible by 3. Since that’s obviously not true, this is a contradiction, so
I can see that there is no such a, which is what I wanted to prove.
9. Suppose x, y ∈ Z. Prove that if x2 (y 2 − 2y) is odd, then x and y are both odd.
Proof. I will prove the contrapositive. To that end, assume x and y are not both odd, meaning that
either x is even or y is even. If x is even, then x = 2k for some k ∈ Z, so
x2 (y 2 − 2y) = (2k)2 (y 2 − 2y) = 4k 2 (y 2 − 2y) = 2(2k 2 (y 2 − 2y)),
which is even since 2k 2 (y 2 − 2y) is an integer.
On the other hand, if y is even, then y = 2` for some ` ∈ Z and
x2 (y 2 − 2y) = x2 ((2`)2 − 2(2`)) = x2 (4`2 − 4`) = 2(x2 (2`2 − 2`))
is even since x2 (2`2 − 2`) is an integer.
Hence, I’ve shown that if either x is even or y is even, then x2 (y 2 −2y) is even, which is the contrapositive
of the given statement.
3
10. Prove that for any positive x ∈ R,
x+
1
≥ 2.
x
Proof. By contradiction. Suppose there is some x > 0 so that x + x1 < 2. Since x > 0 we can multiply
both sides by x without changing the direction of the inequality, so
x2 + 1 < 2x.
In other words,
0 > x2 − 2x + 1 = (x − 1)2 ,
which is clearly impossible since the square of any real number must be non-negative.
From this contradiction, then, I can conclude that there is no such x and hence that x +
positive x ∈ R.
4
1
x
≥ 2 for all