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

Chapter 1
Logic and Proof

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 1.3, Slide 1

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Section 1.3
Techniques of Proof l

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Section 1.3, Slide 2

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Example 1.3.1
2
Consider the function f (n) = n + n + 17.
If we evaluate this function for various positive integers, we observe that we always
seem to obtain a prime number.
f (1) = 19, f (2) = 23, f (3) = 29, f (4) = 37, …, f (8) = 89, …, f (12) = 173, …, f (15) = 257

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

2
If n is a positive integer and p(n) represents the statement “n + 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.

In fact, we know that p(n) is

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

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 1.3, Slide 3

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2
If p(n): “n + n + 17 is a prime number,” how can we prove “∀ n, p(n)”
In this case we can’t because, in fact, “∀ n, p(n)”

is true?

is false.

2
To prove it is false, we need to find one example where n + n + 17 is not prime.
Such an example is called a counterexample.
One such example is n = 17:
2
17 + 17 + 17 = 17(17 + 1 + 1) = 17 · 19
There are others as well. For example when n = 16:
2
16 + 16 + 17 = (16)(16 + 1) + 17
= (16)(17) + 17 = (16 + 1)(17) = 17

2

In this case there is more than one counterexample, but it only takes one counterexample
to prove that “∀ n, p(n)” is false.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 1.3, Slide 4

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Example 1.3.3
2
Consider the function g(n, m) = n + n + m.

In Example 1.3.1 we saw that g

 (16,  17) = 16

2

 + 16 + 17 

2
= 17 .

We also note that
g

2

 (1, 2)  =  1  +

 1 + 2  = 4 = 2

g (2, 3)  =  2 2 + 2 + 3  = 9 = 3

2

2
.
.
.
g (5, 6)  =  52 + 5 + 6  = 36 = 6

2

.
.
.
2
g (12, 13)  =  12 2 + 12 + 13  = 169 = 13 .

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.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

We can prove it using the familiar laws of algebra.

Section 1.3, Slide 5

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

It follows that

g

 (n, n

+ 1)

2
= n + n + (n + 1)

[definition of g (n, n

2
= n +

+ 1)]

2n + 1

[since n + n = 2n]

= (n + 1)(n + 1)

[by factoring]

2
= (n + 1)

2
[definition of (n + 1) ].

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) = 125

2

without having to do any computation.

This is an example of deductive reasoning.

General Principle

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Specific Case

Section 1.3, Slide 6

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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.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 1.3, Slide 7

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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.
…

⇒ q ⇒ q

⇒ q

2
1
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.
p ⇒ p1 ⇒ p2 ⇒
We hope that at some point the part of the bridge leaving

p will join with the part that arrives at q, forming a complete span:

p ⇒ p1 ⇒ p2 ⇒

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

…

…

⇒ q2 ⇒ q1 ⇒ q

Section 1.3, Slide 8

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

Symbols
Original statement:
Converse:

Words

p⇒q

if p, then q

q⇒p

if q, then p

Equivalent

Equivalent

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.
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.

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Section 1.3, Slide 9

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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]

Conclusion: 7m is an even number

[since 7m is 2 times the integer 7k]

This is much easier that trying to show directly that 7m being odd implies that m is odd.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 1.3, Slide 10

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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
a sampling:
eachHere
oneissays.
(a) ( p ⇔ q) ⇔ [( p ⇒ q) ∧ ( q ⇒ p)]
 

One way to prove the biconditional p ⇔ q is to prove the conditional p ⇒ q and its converse q ⇒ p.
 

(c) ( p ⇒ q) ⇔ (~ q ⇒ ~ p)
One way to prove the conditional p ⇒ q is to prove its contrapositive ~ q ⇒ ~ p.

( i)

[( p ⇒ q) ∧ ~ 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.
(p)

[ p ⇒ ( q ∨ r)] ⇔ [( p ∧ ~ q) ⇒ r]
 

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

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 1.3, Slide 11

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