Math 3200: Introduction to Higher Mathematics
Spring 2014, Section 26-761
In-Depth Examples for Week 3
Multiplying a Bunch of Odd Numbers
In this exercise, we’ll solve Problem 7.24 from the textbook: for every three odd integers
a, b, and c, their product abc is odd.
Brainstorming:
The immediate thing to try is to write out all of a, b, and c in the form given by the
definition of odd numbers. Thus, maybe we write
a = 2k + 1
b = 2l + 1
c = 2m + 1
where k, l, m ∈ Z. However, this means we have to expand out the product (2k + 1)(2l +
1)(2m + 1). That looks like a giant nuisance!
Are there any alternatives? Well, on a more intuitive level, here’s how we would probably
reason this out: “Any time you multiply odd integers, you get something odd. Thus, ab is
odd, and we multiply it by c to still get something odd.” That suggests that we really should
focus most of our attention on this key statement:
The product of any two odd integers is odd.
Once we have that, we can reuse that result! (We say that this result is a generalization of
several facts we’ve already established, such as “If x is odd, then 5x is odd” or “If x is odd,
then x2 is odd”.)
We’ll prove this “helper result” (also called a lemma) first. We’ll prove it using the
approach we started with, but there will be fewer variables and hence less expansion to do.
Once we’ve finished this lemma, we’ll apply it to our three numbers.
Solution:
First, we prove a helpful lemma.
Lemma. The product of any two odd numbers is odd. In symbols,
∀x, y ∈ Z, (x odd) ∧ (y odd) ⇒ (xy odd)
Proof. Let x, y ∈ Z, and assume x and y are odd. We may choose k, l ∈ Z so that x = 2k + 1
and y = 2l + 1. Therefore,
xy = (2k + 1)(2l + 1) = 4kl + 2k + 2l + 1 = 2(2kl + k + l) + 1
Because 2kl + k + l is an integer, this shows xy is odd.
Now we prove the main claim. Let a, b, c ∈ Z, and assume all three are odd. We will
prove abc is odd.
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Because a and b are odd, our lemma says ab is odd. (In other words, we plug x = a and
y = b into the lemma.) Now we know ab and c are odd, so our lemma says (ab)c is odd. (i.e.
We plug x = ab and y = c.) This proves abc is odd. Remarks upon the finished solution:
This problem shows how useful it can be to reuse a very general statement! Our conclusion
is also very general: it says any three odd numbers produce an odd product. It applies no
matter which odd numbers I plug in; for instance, −3xy is odd when x and y are odd.
Can you see how we can similarly show the product of any four odd numbers is odd?
Any five? What’s the pattern here?
A Difficult Four-Variable Solving Problem
In this exercise, we’ll solve Problem 7.26 from the textbook: prove that for every positive
real number a and positive rational number b, there exist a real number c and an irrational
number d such that ac + bd = 1. To write this in symbols, I’ll use a subscript of + on a set
to refer to the positive elements of the set. With that in mind, our symbolic sentence is:
∀a ∈ R+ , ∀b ∈ Q+ , ∃c ∈ R, ∃d ∈ R − Q, ac + bd = 1
(Recall the irrationals are the real numbers without the rational numbers: R − Q.)
Brainstorming:
We’re going to start the proof by letting a and b be given, of course. However, we then
need to produce our own witnesses for c and d, which may or may not use a and b.
(Warning: I’m deliberately going to go down a wrong path at first, to be honest about
what I tried. I’ll mark the point when I finally hit the key insight, though!)
In all likelihood, we can’t just guess valid numbers for this right off the bat. However, we
can try some kind of naive guess (using simple numbers), see what’s wrong with it, and fix
it. For instance, what if I do c = d = 0? Well, that’s too simplistic, because a(0) + b(0) = 0,
but we want ac + bd to be 1. Well, what if only one of the variables is 0?
We can’t make d equal to 0 because d must be irrational, so maybe we can make c = 0.
This would give us a(0) + bd = 1, so bd = 1. (In effect, I’m trying to say “does ac even
matter?”) From bd = 1, we must have d = 1/b. Does this work?
Nope! The problem is: if b is rational, then so is its reciprocal. In fact, it turns out
that no matter what irrational number I pick for d, bd will be irrational (we’ll have a full
proof of this later in the course). This suggests that perhaps it doesn’t really matter which
irrational number we pick for d; no matter what, bd won’t even be rational, so it definitely
won’t be 1. This tells us that c can’t be 0, and ac most definitely matters. In fact, in order
for ac + bd = 1, we are forced to make ac = 1 − bd.
Based on this, though, we finally hit the key insight: once d is picked, we can uniquely
solve for c! The previous paragraph
suggests that we can really use any irrational number
√
we like for d, so we pick d = 2 and solve for c. The good news is that c can be any real
number, so as long as we don’t divide by 0 or do anything similarly silly, our choice of c will
be fine. (Also, remember that c is allowed to depend on a or b, so it’s fine to leave them in
the answer.)
Solution:
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Let a ∈ R+ and b ∈ Q+ . We wish to pick c ∈ R and d ∈ R − Q satisfying ac + bd = 1.
Our choices are
√
√
1−b 2
d = 2 and c =
a
We know d is irrational, and c is a well-defined real number because a is positive and hence
not zero. Algebra shows that
√
√
√
√
1−b 2
+ b 2 = (1 − b 2) + b 2 = 1
ac + bd = a ·
a
as desired. Remarks upon the finished solution:
It shouldn’t be particularly surprising, in retrospect, that the choices had to use both a
and b in them somehow. It may be more surprising that c ended up using both of them,
whereas d didn’t have to use any of them necessarily. However, this makes sense due to the
fact that d was chosen first, since it is the variable that has the stricter condition on it (it
must be irrational, as opposed to just being real). We also saw a couple surprising instances
where it pays for a number to be nonzero.
The main moral might just be: with two unknowns, you can usually write one in terms
of the other, so you really only have to pick one. This allowed us to first make sure d was
irrational, then we could make sure ac + bd = 1 was satisfied. This is an instance of a general
lesson paraphrased from the great educator Polyá:
When you have multiple conditions to satisfy, solve one condition first, then solve the
second condition in the space of the first.
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