Download Redemption Answer Key

Redemption
Answer Key
March 1, 2005
4.1) Show that there exists a natural number M such that for all natural
numbers n, n2 + 3n + 1 > M .
Attack: This is a there exists proof, so we need to provide a value for M .
The inequality is supposed to be true for all values of n, so our value of M
should be smaller than any n2 + 3n + 1. A bit of experimenting should show
that the minimum value of the quadratic is 5, so any natural number less
than 5 will be a good choice of M . Let’s pick M = 1, and go with that.
Proof: Let M = 1. Let n be a natural number. Since n ≥ 1, n2 ≥ 1 and
3n ≥ 1. Hence n2 + 3n + 1 ≥ 1 + 1 + 1 = 3 > 1, and so n2 + 3n + 1 > M for
all any natural number n.
4.2) Show that for every real number > 0, there is a natural number M
such that 1/M 2 < .
Attack: We begin with an arbitrary real number > 0, and then we find
a value of M that satisfies 1/M 2 < . Rearranging the inequality, we need
M 2 > 1/epsilon. Since we can find natural numbers that are arbitrarily big,
we can find an M to satisfy this equation.
Proof: Let be an arbitrary positive real number. Pick M to be the first
natural number larger than 1/. Since M > 1, we know M 2 > M , and so
M 2 > 1/. Rearranging, we get 1/M 2 < . Thus every positive real number
has a corresponding natural number M with 1/M 2 < .
4.3) Show that for every natural number n and every natural number m with
m < n, there is a rational number q with m < q < n.
Attack: As in the last problem, we start with arbitrary natural numbers m
and n satisfying m < n, and then we have to find our rational number q.
All we need is fraction between m and n. An obvious choice would be the
average (m + n)/2, so that will be our choice of q.
Proof: Let m and n be natural numbers with m < n. Let q = (m + n)/2.
Since q is the ratio of two natural numbers, q is a rational number. Since q
is the average of m and n, we have m < q < n. Thus q is a rational number
that satisfies the conditions in the theorem.
5.1) Show that if x is irrational and a is a rational number with a 6= 0, then:
• x + a is irrational.
• ax is irrational.
• x/a is irrational.
• xa could be rational.
Attack: I’m just going to the first and last problems. If you know what to
do on the first, then you know what to do on the second and third.
Irrational proofs are one of the best places to break out the proof by
contradiction. So we should assume x is irrational, a is rational, but x + a is
rational. Then we’ll use the definition of rational (it means it’s a fraction),
and then lead to a contradiction.
Proof: Let x be irrational and a be rational, and assume that x + a is
rational. Then x + a = rs for some integers r and s. Since a is rational,
for some integers m and n. Hence we have x + m
= rs , and so
then a = m
n
n
rn−sm
x = sn . Both the top and bottom of this fraction are integers, so x is a
rational number. But x is irrational, hence a contradiction. Therefore x + a
is irrational.
Last part of problem: To show it could
√ be, you only need to give an
example. The quickest one is to let x = 2 (which we know is irrational)
and a = 2 (which is rational). Hence xa = 2, which is rational.
√
3+1 is irrational. You may not assume that 3 is irrational.
√
√
Attack: If we show 3 is irrational, then we can use 5.1 to claim that 3+1
is irrational. We will do that. Feeling lazy? Copy the example on page 39 in
the book, replacing each 2 with a 3.
√
Proof: Let’s first show that 3 is irrational. Assume√that it is rational.
Then there exists a reduced fraction m/n with m/n = 3. So m2 /n2 = 3,
or m2 = 3n2 . The number of times 3 is a factor of m2 is twice the number
of times that 3 is a factor of m, so m2 has 3 as a factor an even number of
times. But n2 also has 3 as a factor an even number of times, so 3n2 = m2
has 3 as a factor an odd
so our
√ number of times. This is a contradiction,
√
initial assumption that 3 is rational is false. √
Therefore, 3 is irrational.
By the previous problem, we can say that 3 + 1 is also irrational.
5.2) Show that
√
6.1) Prove that if A ⊆ B ∩ C, then A ∪ B ⊆ B ∪ C.
Attack: We’re trying to prove a subset relation, and that means we should
start with an x ∈ A ∪ B (WE DO NOT START WITH x ∈ A), and then
work with the definitions and the assumption to show x ∈ B ∪ C.
Proof: Assume that A ⊆ B ∩ C. Let x be an element in A ∪ B. Then
x ∈ A or x ∈ B. If x ∈ B, then x ∈ B ∪ C. If x ∈ A, then by assumption,
x ∈ B ∩ C, and so x ∈ B, and so x ∈ B ∪ C. In either case, any element of
A ∪ B is also in B ∪ C, and so A ∪ B ⊆ B ∪ C.
6.2) Prove by example that A ∪ (B ∩ C) is not always equal to (A ∪ B) ∩ C.
Attack: Most choices of three distinct sets will lead to this. So just choose
an example, and then work out the two sides.
Proof: Let A = {1}, B = {2}, and C = {3}. Then (A ∪ B) ∩ C = ∅, while
A ∪ (B ∩ C) = {1}. Clearly the two sets are not equal.