Download PROOFS Math 174 May 2017 I. Introduction. In the natural sciences

PROOFS
Math 174 May 2017
I. Introduction. In the natural sciences, we cannot prove anything to be true. We can only
support hypotheses with data and make decisions based on the most strongly-supported
hypotheses, but the fact remains that fringe philosophical problems still apply (can we trust
our senses? Are we living in a simulation?)
The great thing about mathematics is that we can prove statements, called theorems
(or propositions or lemmas depending on their place in the hierarchy), to be true.
II. Direct proofs. Most theorems have the form p → q, where p is some assumption we make
(even one as seemingly innocent as “numbers exist”). We apply other theorems—the laws
of algebra, or facts about whole numbers, or other statements we’ve proven true—and arrive
at the fact that q is true.
Here’s an example.
Theorem. If n is an even integer, then so is n2 .
This is a p → q statement where p is the statement that n is some even integer and q is the
statement that n2 is an even integer. Before we dive into the proof, though, we must be
comfortable with the definitions. (In fact, writing proofs is a better way to learn definitions
and theorems than memorizing them!) An integer is a whole number, e.g. 21, −4, 0. An
even integer is one that can be divided by 2 to yield another integer. In other words,
Definition. Let n be an integer. If n is even, then there exists an integer m such that
n = 2m. If n is odd, then there exists an integer m such that n = 2m + 1.
(WARNING: n and m are different integers. Don’t let me catch you writing n = 2n + 1,
because that equation is never true!)
Definitions don’t need to be proven—we take them for granted. However, we need to be
sure our audience agrees with our definitions. (Most definitions in math, especially at the
undergraduate level, are standard, and everyone uses the same ones.)
Now that we know what integers and even integers are, we can try to prove the theorem.
In blue, I will explain each step.
Proof.
Let n be an even integer.
By definition, there exists another integer m where n = 2m.
state assumption
unpack definition
Now what? We are trying to prove something about n2 , so let’s square n. Be careful that
you apply algebra correctly.
Therefore, n2 = (2m)2 = 4m2 = . . .
To prove n2 is even, we need to be able to factor out a 2. Can we?
square n
. . . = 2(2m2 ). Since m is a whole number, so is 2m2 .
Hence, n2 is 2 times an integer.
conclusion
Thus if n is an even integer, then so is n2 .
restate theorem
III. Proofs by contradiction. Sometimes, no matter how hard you try, you cannot get from p
to q with the normal methods. Maybe, worse, the theorem relies on a “negative” definition
that only says what something is not. In this case, we’ll have to rely on some logical trickery.
Recall from the homework that the statement [q ∧ (¬p → ¬q)] → p is a tautology. (If
you haven’t done this problem yet, go do it!) Here, p is a statement we’d like to prove and
q is something we know to be true. It turns out that if we assume ¬p—the opposite of what
we want to prove!—and derive ¬q, we come to a contradiction. We can’t have both q and
¬q true simultaneously. What caused the problem? It was assuming ¬p, and hence p must
have been true all along.
Theorem. If n is an integer and n2 is even, then n is even.
This theorem is the converse of the first theorem we proved. To prove it, we will assume
that n is odd and derive a contradiction: namely, that n2 is odd, when it was supposed to
be even.
Proof.
Let n be an integer such that n2 is even.
Suppose n is odd.
There exists an integer m where n = 2m + 1.
Thus, n2 = (2m + 1)2 = 4m2 + 4m + 1.
Since 4m2 + 4m is even, n2 is odd.
This contradicts the assumption that n2 is even.
Thus, if n is an integer and n2 is even, n must be even too.
state assumption
set up contradiction
definition
apply algebra correctly!
get contradiction
conclusion
IV. Biconditional proofs. A theorem including the phrase “if and only if,” “is necessary and
sufficient for,” “are equivalent,” etc. is a biconditional p ↔ q. These are proven by proving
p → q and q → p together.
Theorem. Let n be an integer. Then n is even if and only if n2 is even.
We have shown that if n is even then n2 is even, and we have shown that if n2 is even then
n is even. Together, they prove that n is even if and only if n2 is even.
V. Irrational numbers exist. We are now prepared to prove an important theorem in mathematics: namely, that not all real numbers are rational.
Theorem. The positive square root of 2 is irrational.
√
Again, let’s work out the definitions.√The positive square root of 2, denoted 2, is the
positive number whose square is 2: ( 2)2 = 2.
A rational number is the quotient p/q of two integer p and q where q 6= 0, and if we like
we may take p and q to have no factors in common so that the quotient is in lowest terms.
Finally, an irrational number is one that isn’t rational.
Notice that the core word in the theorem,
√ “irrational,” is a negative one: “not rational.”
The only course of action is to suppose that 2 is rational and come up with a contradiction.
Let’s get started!
Proof. There is no assumption in this proof, so we just start with the contradiction right out
the gate.
√
Suppose that 2 is rational.
√
By definition, there exist integers p and q where q 6= 0 and 2 = p/q.
Additionally, we assume p/q is in lowest terms.
Again, we’ve hit a wall. The only thing we know about
use that and see what happens.
√
assume ¬p
definition
it doesn’t hurt!
2 is that its square is 2, so let’s
If we square both sides, we see 2 = p2 /q 2 .
We can rearrange this equation into p2 = 2q 2 , which tells us that p2 is even.
Thus, p is even, too.
earlier theorem
Let’s stop and taken inventory of what we know: p is even, p/q is in lowest terms, and
somewhere lurks a contradiction. It’s going to turn out that q is even, too; and since p and
q are both even, p/q isn’t in lowest terms as assumed. This is our contradiction. Let’s see
how it comes up.
Because p is even, there exists an integer m with p = 2m.
Substituting p = 2m into p2 = 2q 2 , we get
(2m)2 = 2q 2 ⇒ 4m2 = 2q 2 ⇒ 2m2 = q 2 .
Hence, q 2 and thus q are even.
However, we assumed that p/q was in lowest terms, so p and q had no common factors.
Therefore, p and q cannot both be even.
√
By assuming 2 = p/q in lowest terms, we arrived at the conclusion
that both p and q are
√
even. This is contradictory. It must thus be impossible that 2 = p/q.
Therefore by contradiction, no such p and q exist, and so
√
2 is irrational.