Download 1. Given that m and n are integers and that the number mn is not

1. Given that m and n are integers and that the number mn is not divisible by 4, prove that either m is odd or n is
odd.
Instead of making a new problem out of this exercise we can interpret it as asking us to prove that if both m
and n are even then mn has a factor 4. We did this problem in the 3.2.4.
2. Given that m and n are integers, that neither m nor n is divisible by 4 and that at least one of the numbers m
and n is odd, prove that the number mn is not divisible by 4.
We can interpret this exercise as asking us to prove the following three assertions:
a.
If m and n are even then mn must have a factor 4.
If m and n are integers and m has a factor 4 then mn has a factor 4.
If m and n are integers and n has a factor 4 then mn has a factor 4.
The first of these was handled in the previous exercise. The other two are easy and, of course, they are
analogues of one another so we need not do both of them.
3. Prove that if x  − cos 40 ∘ or x  − cos 80 ∘ then
8x 3 − 6x − 1  0.
We have two jobs to do:
a. We need to show that if x  − cos 40 ∘ then 8x 3 − 6x − 1  0.
We need to show that if x  − cos 80 ∘ then 8x 3 − 6x − 1  0.
To prove the first of these assertions, we assume that x  − cos 40 ∘ . In order to show that 8x 3 − 6x − 1  0 we
shall use the fact that if u is any number then
cos 3u  4 cos 3 u − 3 cos u.
We now observe that
8x 3 − 6x − 1  8− cos 40 ∘  3 − 6− cos 40 ∘  − 1
 −24 cos 3 40 ∘ − 3 cos 40 ∘  − 1
 −2 cos340 ∘  − 1
 −2 cos 120 ∘ − 1
 −2 − 1 − 1  0
2
4. A theorem in elementary calculus, known as Fermat’s theorem, says that if a function f defined on an interval
has either a maximum or a minimum value at a number c inside that interval then either f ′ c  0 or f ′ c
does not exist. Give a brief outline of a strategy for approaching the proof of this theorem.
A strategy for proving Fermat’s theorem is to assume that f is a function defined on an interval, that c is a
number inside that interval and that f ′ c exists. You then have two analogous jobs to do: You need to show
that if f has a maximum at c then f ′ c  0 and you need to show that if f has a minumum at c then f ′ c  0.
1. A well known theorem on differential calculus that is known as L’Hôpital’s rule may be stated as follows:
Suppose that f and g are given functions, that c is a given number, that
f ′ x
 L,
lim
x→c g ′ x
and that one or other of the following two conditions holds
a. Both fx and gx approach 0 as x → c.
b. gx →  as x → c.
1
Then
fx
 L.
gx
Describe how the proof of L’Hôpital’s rule can be broken down into two parts. For each part of the proof, say
what is being assumed and what is being proved.
lim
x→c
The purpose of this exercise is not to dig into the underlying ideas of L’Hôpital’s rule. Nor do we need to
concern ourselves with the question as to whether it is actually necessary to break the proof into two parts.
Instead, we assume that the two parts are needed. The student should be able to see that, approached this
way, the theorem requires two separate proofs. In one proof we assume that both fx and gx approach 0 as
x → c and in the other proof we assume that gx →  as x → c.
2