Analysis with an introduction to proof 5th by steven lay ch01a

Specific Cases General Principle If n is a positive integer and pn represents the statement “n2 + n + 17 is a prime number,” have we shown that  n, pn is true?. Since our reasoning at e

Chapter 1

Logic and Proof

Section 1.3 Techniques of Proof l

In fact, we know that p(n) is true for many n

Example 1.3.1

Consider the function f (n) = n2 + n + 17.

If we evaluate this function for various positive integers, we observe that we always

seem to obtain a prime number

All of these numbers (and the ones skipped over) are prime

On the basis of this experience we might conjecture that the function f (n) = n2 + n +17

will always produce a prime number when n is a positive integer.

Drawing a conclusion of this sort is an example of inductive reasoning

Specific Cases General Principle

If n is a positive integer and p(n) represents the statement “n2 + n + 17 is a prime number,”

have we shown that  n, p(n) is true?

No, we have not We have shown that  n p(n) is true.

But we have not proved it is true for all n.

If p(n): “n2 + n + 17 is a prime number,” how can we prove “ n, p(n)” is true?

In this case we can’t because, in fact, “n, p(n)” is false.

To prove it is false, we need to find one example where n2 + n + 17 is not prime.

Such an example is called a counterexample

One such example is n = 17:

172 + 17 + 17 = 17(17 + 1 + 1) = 17 · 19

There are others as well For example when n = 16:

162 + 16 + 17 = (16)(16 + 1) + 17

= (16)(17) + 17 = (16 + 1)(17) = 172

In this case there is more than one counterexample, but it only takes one counterexample

to prove that “n, p(n)” is false.

Example 1.3.3

Consider the function g(n, m) = n2 + n + m.

In Example 1.3.1 we saw that g(16,17) = 162 +16 +17 = 172

We also note that

g(1,2) = 12 +1 +2 = 4 = 22

g(2,3) = 22 +2 +3 = 9 = 32

.

g(5,6) =

52 +5 +6 = 36 = 62

.

g(12,13) =

122 +12 +13 = 169 =

132

On the basis of these examples (using inductive reasoning) we form the conjecture

 n, q(n) where q(n) is the statement “g(n, n+1) = (n + 1)2

This time our conjecture is true We can prove it using the familiar laws of algebra

We have g(n, m) = n2 + n + m and q(n): g(n, n + 1) = (n + 1)2.

It follows that g (n, n+1) = n2 + n + (n +1) [definition of g (n, n+1)]

= n2 +2n +1 [since n + n = 2n]

= (n + 1)(n +1) [by factoring]

= (n + 1)2 [definition of (n + 1)2]

Since our reasoning at each step does not depend on n being any specific integer,

we conclude that “"n, q (n)” is true

Now that we have proved the general statement “"n, q (n),” we can apply it to any

particular case

For example, we know that g(124, 125) = 1252 without having to do any computation

This is an example of deductive reasoning

The most common type of mathematical theorem can be symbolized as p  q, where

p and q may be compound statements

To assert that p  q is a theorem is to claim that p  q is a tautology; that is, that it is

always true

From Section 1.1 we know that p  q is true unless p is true and q is false

So to prove that p implies q, we have to show that whenever p is true it follows that

q must be true

When an implication p  q is identified as a theorem, it is customary to refer to

p as the hypothesis and q as the conclusion

The construction of a proof of the implication p  q can be thought of as building a

bridge of logical statements to connect the hypothesis p with the conclusion q

The building blocks that go into the bridge consist of four kinds of statements:

(1) definitions, (2) assumptions or axioms that are accepted as true, (3) theorems that have previously been established as true, and (4) statements that are logically implied by the earlier statements in the proof

In building a bridge from the hypothesis p to the conclusion q, it is often useful to start at

both ends and work toward the middle

We might begin by asking,

“What must I know in order to conclude that q is true?” Call this q1 Then ask,

“What must I know to conclude that q1 is true?” Call this q2 Continue this process as long as it is productive, thus obtaining a sequence of implications:

Then look at the hypothesis p and ask,

“What can I conclude from p that will lead me toward q?” Call this p1 Then ask, “What can I conclude from p1?” And continue this process as long as it is

productive

 q2  q1  q

…

p  p1  p2  …

Given any conditional statement p  q, there are three related conditional statements.

Original statement: p  q if p, then q

Converse : q  p if q, then p

Inverse : ~ p  ~ q if not p, then not q

Contrapositive : ~ q  ~ p if not q, then not p

It is easy to show using a truth table that the original statement and the contrapositive

are logically equivalent

Equivalent

Equivalent

It is also true that the converse and the inverse are equivalent to each other

But neither the converse nor the inverse are equivalent to the original statement

Since the contrapositive of a conditional statement is equivalent to the original

statement, it can be useful in proving theorems

Example 1.3.7

To prove the theorem “If 7m is an odd number, then m is an odd number”

we look at its contrapositive:

“If m is not an odd number, then 7m is not an odd number.”

OR “If m is an even number, then 7m is an even number.”

Here is a simple proof of the contrapositive

Hypothesis: m is an even number.

m = 2k for some integer k [definition]

7m = 7(2k) [known property of equality]

7m = 2(7k) [known property of multiplication]

7k is an integer [since k is an integer]

Example 1.3.12 in the text gives 17 tautologies that are useful in constructing proofs.

It is not necessary to memorize all of them, but they should be studied to see what

each one says Here is a sampling:

p  q and its converse q  p.

~ q  ~ p.

If you know p implies q, but you don’t have q, then you can’t have p

(j) [(p  q)  ~ p]  q

If you know p or q is true, but you don’t have p, then you must have q

One way to prove that p implies q or r is to prove if you have p and don’t