Download MATH 311-02 Midterm Examination #2 solutions 1. (20 points

MATH 311-02
Midterm Examination #2 solutions
1. (20 points) Prove that the equation x3 + 3x = 14 has exactly one positive real solution.
Proof. We can prove that at least one exists simply by furnishing an example: 23 + 3 · 2 = 14.
Now, to prove that this example is unique, let us suppose by way of contradiction that there
are distinct positive values x and y such that x3 + 3x = 14 and y 3 + 3y = 14. WLOG we may
suppose x < y. Then x3 < y 3 and 3x < 3y as well (since all numbers involved are positive, the
multiplication of inequalities will preserve their direction), and adding these inequalities yields
x3 + 3x < y 3 + 3y. Since both the left and right sides of this inequality are equal to 14, we
arrive at the patently absurd arithmetic contradiction that 14 < 14.
Alternative algebra-maven’s proof. We can factor a rational root out of x3 + 3x − 14 to find that
it is equal to (x − 2)(x2 + 2x + 7). Thus (since a product is zero only when one of its factors is
zero, as has been proven in class) x3 + 3x − 14 = 0 if and only if x − 2 = 0 or x2 + 2x + 7 = 0.
The former is true when x = 2 (which is verifiably a single solution of x3 + 3x = 14); the latter
is true when x2 + 2x + 1 = −6, or in other words when (x + 1)2 = −6, which can never occur
since the square of any real number is non-negative. Thus x3 + 3x − 14 = 0 only when x = 2,
which is the sole positive real root.
2. (20 points) Prove that for any positive integer n, it is the case that
1 · 21 + 2 · 22 + 3 · 23 + 4 · 24 + · · · + n · 2n = (n − 1)2n+1 + 2
Proof. We proceed by induction on n; note that the base case n = 1 can be demonstrated by
arithmetic: 1 · 21 = 2 = (0)22 + 2. Now, we assume, for a specific n, that it is true that
1 · 21 + 2 · 22 + 3 · 23 + 4 · 24 + · · · + n · 2n = (n − 1)2n+1 + 2
and seek to prove that
1 · 21 + 2 · 22 + 3 · 23 + 4 · 24 + · · · + n · 2n + (n + 1) · 2n+1 = n2n+2 + 2
To do so, we may add (n + 1)2n+1 to both sides of our inductive hypothesis, and proceed via
arithmetic:
1 · 21 + 2 · 22 + 3 · 23 + 4 · 24 + · · · + n · 2n + (n + 1)2n+1 = (n − 1)2n+1 + 2 + (n + 1)2n+1
= (2n)2n+1 + 2
= n2n+2 + 2
which is the desired result.
3. (5 points) Disprove the following statement: If x and y are real numbers such that x < y, then
it is also the case that x2 < y 2 .
Disproof. Consider the counterexample x = −2 and y = 1; −2 is indeed less than 1, but
(−2)2 > 12 .
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MATH 311-02
Midterm Examination #2 solutions
4. (15 points) Prove that for any integer n, if n is odd, then 8 | n2 − 1.
Proof. Since n is odd, we know n = 2k +1 for some integer k; thus n2 −1 = 4k 2 −4k = 4(k 2 −k).
Then we may investigate based on the parity of k:
Case I: k is odd. Then k = 2ℓ + 1 for some integer ℓ, so that
n2 − 1 = 4[(2ℓ + 1)2 − (2ℓ + 1)] = 4(4ℓ2 + 2ℓ) = 8(2ℓ2 + ℓ)
and so 8 | n2 − 1.
Case I: k is even. Then k = 2ℓ for some integer ℓ, so that
n2 − 1 = 4[(2ℓ)2 − (2ℓ)] = 4(4ℓ2 − 2ℓ) = 8(2ℓ2 − ℓ)
and so 8 | n2 − 1.
Alternative induction-addict’s proof. We shall prove that this is true for natural odd values
of n by induction, and then since n2 = (−n)2 it will then be a simple corollary that since
n2 − 1 = (−n)2 − 1, the truth of the proposition for every positive value will guarantee truth
for negative values as well.
For our base case n = 1, note that 12 −1 = 0 and that 8 | 0. For our inductive step, let us assume
for some specific k that 8 | k 2 − 1, and we intend to show that the next odd number, k + 2, also
satisfies the condition in question, i.e., our desired conclusion will be that 8 | (k + 2)2 − 1.
Using our inductive hypothesis, we know that k 2 − 1 = 8ℓ for some integer ℓ. We now attempt
to write (k + 2)2 − 1 in terms of this expression:
(k + 2)2 − 1 = k 2 + 4k + k − 1 = (k 2 − 1) + 4(k + 1) = 8ℓ + 4(k + 1)
noting that k is odd, we then can determine that k + 1 is even, so k + 1 = 2m for some integer
m. Then (k + 2)2 − 1 = 8ℓ + 8m, which is indeed a multiple of 8.
5. (20 points) Prove that there is no integer n such that n ≡ 4 (mod 6) and n ≡ 2 (mod 9).
Proof. Let us assume counterfactually that there is such an n. Then 6 | n − 4 and 9 | n − 2,
so there are integers k and ℓ such that n − 4 = 6k and n − 2 = 9ℓ. Subtracting these two
equations from each other, we get −2 = 6k − 9ℓ = 3(2k − 3ℓ). Since 2k − 3ℓ is an integer, we
then conclude that 3 | −2, which contradicts the known arithmetic fact that 3 ∤ −2.
6. (a) (15 points) Prove that for any sets A, B, and C, (A ∪ B) − C ⊆ A ∪ (B − C).
Proof. Consider x ∈ (A ∪ B) − C. By the definition of a set difference, x ∈ A ∪ B and
x ∈
/ C. Since x ∈ A ∪ B, it is either the case that x ∈ A or x ∈ B; we address these
separately:
Case I: x ∈ A. Membership in the set A guarantees membership in any union involving
set A, so x ∈ A ∪ (B − C).
Case I: x ∈ B. Then since x ∈ B and x ∈
/ C, x ∈ B − C. Membership in a set guarantees
membership in a union involving that set, so x ∈ A ∪ (B − C).
In either case, x ∈ A ∪ (B − C), so (A ∪ B) − C ⊆ A ∪ (B − C)
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MATH 311-02
Midterm Examination #2 solutions
(b) (5 points) Demonstrate that for sets A, B, and C, it is not necessarily true that (A ∪
B) − C = A ∪ (B − C).
A simple counterexample is A = {1}, B = ∅, and C = {1}. Then (A ∪ B) − C = ∅ whereas
A ∪ (B − C) = {1}.
In general, any case where some value x is in both A and C but not in B will be a
counterexample, since x ∈ A ∪ (B − C) and x ∈
/ (A ∪ B) − C.
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