logic and proof - numeracy workshop - uwa · 2014. 4. 7. · these slides give a brief introduction...

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Logic and ProofNumeracy Workshop

Adrian Dudek, Geoff Coates

Adrian Dudek, Geoff Coates Logic and Proof 2 / 33

Introduction

These slides give a brief introduction to mathematical logic and methods of proof

Workshop resources: These slides are available online:

www.studysmarter.uwa.edu.au → Numeracy and Maths → Online Resources

Next Workshop: See your Workshop Calendar →

www.studysmarter.uwa.edu.au

Drop-in Study Sessions: Monday, Wednesday, Friday, 10am-12pm, Room 2202, SecondFloor, Social Sciences South Building, every week.

Email: geoff.coates@uwa.edu.au

Introduction

Mathematical Statements

Here are some examples of mathematical statements which are written in English:

Adding one to a number always makes the number greater. X

The square root of two is irrational. X

Squaring a number always makes it larger. ×

Which of the above are true?

Adding one to a number always makes the number greater.

The square root of two is irrational.

Squaring a number always makes it larger.

The writing and reading maths workshop covered some shorthand notation to replacecommon mathematical words and phrases. Here is another one:

“∀” stands for “for all” or “for every possible value of”.

We can use this to shorten written statements:

Adding one to a number always makes the number greater: ∀x ∈ R, x + 1 > x .

The square root of two is irrational:√

2 ∈ Q (or√

2 /∈ Q).

Squaring a number always makes it larger: ∀x ∈ R, x2 > x .

2 ∈ Q (or√

2 /∈ Q).

2 ∈ Q (or√

2 /∈ Q).

Adding one to a number always makes the number greater:

∀x ∈ R, x + 1 > x .

2 ∈ Q (or√

2 /∈ Q).

Adding one to a number always makes the number greater: ∀x ∈ R

, x + 1 > x .

2 ∈ Q (or√

2 /∈ Q).

2 ∈ Q (or√

2 /∈ Q).

The square root of two is irrational:

√2 ∈ Q (or

√2 /∈ Q).

2 ∈ Q

(or√

2 /∈ Q).

2 ∈ Q (or√

2 /∈ Q).

2 ∈ Q (or√

2 /∈ Q).

Squaring a number always makes it larger: ∀x ∈ R

, x2 > x .

2 ∈ Q (or√

2 /∈ Q).

Compound Statements and Connectives

Statements can be combined, using logical connectives, to form compound statements.

Here we let A and B represent statements.

“and”

The compound statement “A and B” (denoted A ∧ B) is true if A is true and B is true.

The statement “It is raining and my socks are wet” is only true if both statements “it israining” and “my socks are wet” are true.

“and”

“or”

The compound statement “A or B” (denoted A ∨ B) is true if at least one of A or B istrue.

The statement “I want to go to the movies or I want to go to the party” is true if one ofthe following holds:

I want to go to the movies

I want to go to the party

I want to go the movies and I want to go to the party

“or”

“not”

The statement “not A” (denoted ∼ A), called the negation of A, is true if A is false.

Let A be the statement “π equals 3”. Then ∼ A is the statement π 6= 3. This is true ifthe statement “π equals 3” is false.

“not”

“implication”

The statement “If A is true, then B is true” (denoted A⇒ B) is a true statement exceptwhen A is true and B is false.

Note: If A is false, B could still be true.

Consider the statement “If I am rich, then I have at least $10”. This is of the formA⇒ B, where A is the statement “I am rich” and B is the statement “I have at least

$10”.

Note: If A⇒ B is true, the reverse implication B ⇒ A may not be.

I know lots of people that have at least $10 but are not rich!

“implication”

$10”.

“implication”

$10”.

“implication”

$10”.

“implication”

$10”.

“implication”

$10”.

“double implication”

The statement “A is true if and only if B is true” (denoted A⇐⇒ B) means A⇒ B andB ⇒ A. That is, the statements imply each other!

Sometimes we say that A and B are equivalent.

Let A be the statement “x equals zero”, and let B be the statement “x2 equals zero”.

Then we have that A⇐⇒ B as we have both A⇒ B and B ⇒ A.

Exercise

Consider the following two statements:

A : x2 = 9

B : x = 3

Which of the following are true?

A⇒ B

B ⇒ A

A⇐⇒ B

Exercise

A : x2 = 9

B : x = 3

A⇒ B ×B ⇒ A X

A⇐⇒ B ×

Exercise

A : x is an even number

B : x + 2 is an even number

A⇒ B

B ⇒ A

A⇐⇒ B

Exercise

A⇒ B X

B ⇒ A X

A⇐⇒ B

Exercise

A⇒ B X

B ⇒ A X

A⇐⇒ BX

The Converse

“converse”

B ⇒ A is the converse of A⇒ B. It’s as easy as reversing the direction of the arrow!

The truth or falsity of a converse can not be inferred from the truth or falsity of theoriginal statement.

For example,

x = 2⇒ x2 = 4

is true, but . . . its converse

x2 = 4⇒ x = 2

is false, because x could be equal to −2.

The Converse

“converse”

For example,

x = 2⇒ x2 = 4

x2 = 4⇒ x = 2

The Converse

“converse”

For example,

x = 2⇒ x2 = 4

x2 = 4⇒ x = 2

The Converse

“converse”

For example,

x = 2⇒ x2 = 4

is true, but . . .

its converse

x2 = 4⇒ x = 2

The Converse

“converse”

For example,

x = 2⇒ x2 = 4

x2 = 4⇒ x = 2

is false, because

x could be equal to −2.

The Converse

“converse”

For example,

x = 2⇒ x2 = 4

x2 = 4⇒ x = 2

The Contrapositive

“contrapositive”

∼ B ⇒∼ A is the contrapositive of A⇒ B.

A statement and its contrapositive are logically equivalent. This means that if one istrue, then the other is true!

For example,

x = 2⇒ x2 = 4

is true, and its contrapositive

x2 6= 4⇒ x 6= 2

is true.

The Contrapositive

For example,

x = 2⇒ x2 = 4

x2 6= 4⇒ x 6= 2

is true.

The Contrapositive

For example,

x = 2⇒ x2 = 4

x2 6= 4⇒ x 6= 2

is true.

The Contrapositive

For example,

x = 2⇒ x2 = 4

x2 6= 4⇒ x 6= 2

is true.

The Contrapositive

For example,

x = 2⇒ x2 = 4

x2 6= 4⇒

x 6= 2

is true.

The Contrapositive

For example,

x = 2⇒ x2 = 4

x2 6= 4⇒ x 6= 2

is true.

The Contrapositive: Example

Consider the statement:

x ≥ 2⇒ x2 > 1

The contrapositive of the above is the statement:

x2 ≤ 1⇒ x < 2

Both are true.

x ≥ 2⇒ x2 > 1

x2 ≤ 1

⇒ x < 2

Both are true.

x ≥ 2⇒ x2 > 1

x2 ≤ 1⇒ x < 2

Both are true.

Necessary and Sufficient Conditions

If the statement A⇒ B is true then we say “A is a sufficient condition for B”, i.e. thetruth of A is sufficient to guarantee that B is true.

Example: “I am rich” ⇒ “I have at least $10” so if “I am rich” (A) is true, it guaranteesthat “I have at least $10” (B).

We also say that “B is a necessary condition for A”, i.e., if B is false then A is false.

Example: If “I have at least $10” (B) is false, then I can’t be rich (A).

If the statement A⇐⇒ B is true, then we say that “A is necessary and sufficient for B”.

Some might say that mathematics is a “toolkit of truths”.

Research in mathematics consists of increasing the number of tools in our toolkit.

The way that we prove new tools, is to show that they are implications of tools which arealready in our toolkit.

There are two main kinds of tool: axioms and theorems.

Axioms

Axioms are statements that are simply accepted as being true without the need for proof.This is because they are so fundamental that we feel everyone must accept them.

Here are some examples:

Euclid’s First Axiom: There is only one straight line that can be drawn to join twospecific points in space.

Well Ordering Principle: Every non-empty set of positive integers contains a smallestelement.

Axioms

Theorems

Theorems are statements that can be proved to be true using accepted definitions,axioms, and other already proven theorems.

The truth of the theorem is arrived at by reasoning from other accepted truths.

Theorems

Theorems are statements that can be proved to be true using accepted definitions,axioms, and other already proven theorems.

The truth of the theorem is arrived at by reasoning from other accepted truths.

Deductive Proofs using Algebra

Look what happens when we multiply two even numbers together:

4× 6 = 24

8× 2 = 16

4× 4 = 16

40× 8 = 320

We always seem to get an even number.

However, we can not conclude that the statement “the product of any two even numbersis even” is a true statement, simply because we have seen a few examples.

We need to prove this theorem completely. This means that we have to present aconvincing argument.

4× 6 = 24

8× 2 = 16

4× 4 = 16

40× 8 = 320

4× 6 = 24

8× 2 = 16

4× 4 = 16

40× 8 = 320

4× 6 = 24

8× 2 = 16

4× 4 = 16

40× 8 = 320

4× 6 = 24

8× 2 = 16

4× 4 = 16

40× 8 = 320

4× 6 = 24

8× 2 = 16

4× 4 = 16

40× 8 = 320

4× 6 = 24

8× 2 = 16

4× 4 = 16

40× 8 = 320

We want to show that for any two even numbers we choose, their product is also even.

However, we don’t want to write out every single choice of two even numbers as this isimpossible!

Instead, we use expressions which represent arbitrary even numbers.

By definition, even numbers are multiples of two, that is, any even number is the productof 2 and some other integer.

Hence, we can represent an even number by the expression 2n, where n is an integer.

Proof: even x even = even

Here is our proof. Take two even numbers:

2n, 2m

Multiply them together:

(2n)(2m)

We can simplify and express the above product in a clever way!

(2n)(2m) = 4nm = 2(2nm)

But 2(2nm) is really just the product of 2 and an integer (ie. 2nm)! So this product isclearly even!

Note: We usually denote the end of a proof with a box.

2n, 2m

(2n)(2m)

(2n)(2m) = 4nm = 2(2nm)

2n, 2m

(2n)(2m)

(2n)(2m) = 4nm

= 2(2nm)

2n, 2m

(2n)(2m)

(2n)(2m) = 4nm = 2(2nm)

2n, 2m

(2n)(2m)

(2n)(2m) = 4nm = 2(2nm)

2n, 2m

(2n)(2m)

(2n)(2m) = 4nm = 2(2nm)

2n, 2m

(2n)(2m)

(2n)(2m) = 4nm = 2(2nm)

But 2(2nm) is really just the product of 2 and an integer (ie. 2nm)! So this product isclearly even! �

Proof: odd + odd = even

Let’s prove that the sum of two odd numbers is even. Here is our proof.

Take two odd numbers:

2n + 1, 2m + 1

Add them together:

2n + 1 + 2m + 1

We can simplify and express the above in the same clever way:

2n + 1 + 2m + 1 = 2n + 2m + 2 = 2(n + m + 1)

We have managed to write the sum of two odd numbers as a number which is a multipleof 2! So the sum of two odd numbers must be an even number. �

2n + 1, 2m + 1

Add them together:

2n + 1 + 2m + 1

2n + 1 + 2m + 1 = 2n + 2m + 2 = 2(n + m + 1)

2n + 1, 2m + 1

Add them together:

2n + 1 + 2m + 1

2n + 1 + 2m + 1 = 2n + 2m + 2 = 2(n + m + 1)

2n + 1, 2m + 1

Add them together:

2n + 1 + 2m + 1

2n + 1 + 2m + 1 = 2n + 2m + 2

= 2(n + m + 1)

2n + 1, 2m + 1

Add them together:

2n + 1 + 2m + 1

2n + 1 + 2m + 1 = 2n + 2m + 2 = 2(n + m + 1)

2n + 1, 2m + 1

Add them together:

2n + 1 + 2m + 1

2n + 1 + 2m + 1 = 2n + 2m + 2 = 2(n + m + 1)

We have managed to write the sum of two odd numbers as a number which is a multipleof 2! So the sum of two odd numbers must be an even number.

2n + 1, 2m + 1

Add them together:

2n + 1 + 2m + 1

2n + 1 + 2m + 1 = 2n + 2m + 2 = 2(n + m + 1)

Structure of algebraic deductive proof

Theorem: If statement(s) is/are true, show that statement is true.

Proof: statement(s) given

statement to be proved

clever, legal steps

Structure of algebraic deductive proof: example

(This example involves matrices and is similar to proofs in MATH1001.)

Theorem: If A is a symmetric, invertible matrix then A−1 is also symmetric.

Proof: We know A is symmetric and A is invertible

A−1 is symmetric

AT = A and A−1 exists

⇒ AA−1= I (from defn of inverse)

⇒(AA−1

)T= (I )T = I (transpose both sides)

⇒(A−1

)TAT = I (a property of transpose)

⇒(A−1

)TA = I (known property of A)

⇒(A−1

)Tis the inverse of A by defn of inverse

∴ �

(A−1

)TExpress the red and blue phrases algebraically.

Proof:

We know A is symmetric and A is invertible

A−1 is symmetric

⇒(AA−1

⇒(A−1

∴ �

(A−1

Proof: We know

A is symmetric and A is invertible

A−1 is symmetric

⇒(AA−1

⇒(A−1

∴ �

(A−1

Proof: We know A is symmetric

and A is invertible

A−1 is symmetric

⇒(AA−1

⇒(A−1

∴ �

(A−1

A−1 is symmetric

⇒(AA−1

⇒(A−1

∴ �

(A−1

A−1 is symmetric

⇒(AA−1

⇒(A−1

∴ �

(A−1

clever, legal steps

A−1 is symmetric

⇒(AA−1

⇒(A−1

∴ �

Proofs are hard because “clever” steps may be hard to spot. Here are some tips.(A−1

)TAdrian Dudek, Geoff Coates Logic and Proof 25 / 33

A−1 is symmetric

⇒(AA−1

⇒(A−1

∴ �

(A−1

)TTip: Express the red and blue phrases algebraically.

(A−1

A−1 is symmetric

AT = A

and A−1 exists

⇒(AA−1

⇒(A−1

∴ �

(A−1

A−1 is symmetric

⇒(AA−1

⇒(A−1

∴ �

(A−1

A−1 is symmetric

(A−1

)T=A−1

⇒(AA−1

⇒(A−1

∴ �

(A−1

A−1 is symmetric

(A−1

)T=A−1

⇒(AA−1

⇒(A−1

∴ �

(A−1

)TWe need to start with an equation involving A−1 . . .

(A−1

A−1 is symmetric

(A−1

)T=A−1

⇒(AA−1

⇒(A−1

∴ �

(A−1

)TWe need to start with an equation involving A−1 . . .

(A−1

A−1 is symmetric

(A−1

)T=A−1

⇒(AA−1

⇒(A−1

∴ �

(A−1

)TNow to get transpose involved . . .

(A−1

A−1 is symmetric

(A−1

)T=A−1

⇒(AA−1

⇒(A−1

∴ �

(A−1

)TNow to get transpose involved . . .

(A−1

A−1 is symmetric

(A−1

)T=A−1

⇒(AA−1

⇒(A−1

∴ �

We need to get an(A−1

)Tterm.

A−1 is symmetric

(A−1

)T=A−1

⇒(AA−1

⇒(A−1

∴ �

We need to get an(A−1

)Tterm.

A−1 is symmetric

(A−1

)T=A−1

⇒(AA−1

⇒(A−1

∴ �

(A−1

A−1 is symmetric

(A−1

)T=A−1

⇒(AA−1

⇒(A−1

∴ �

(A−1

A−1 is symmetric

⇒ (A−1

)T=A−1

⇒(AA−1

⇒(A−1

∴ �

(A−1

A−1 is symmetric

⇒ (A−1

)T=A−1

⇒(AA−1

⇒(A−1

(A−1

A−1 is symmetric

⇒ (A−1

)T=A−1

⇒(AA−1

⇒(A−1

∴ �(A−1

Proof or Counterexample

To prove that a statement is true, we need to construct a proof.

To show that a statement is false, we simply need to find a single counterexample whereit fails.

Proof or Counterexample: Example

Consider the following statement:

Squaring a number makes it larger.

This statement is false. To disprove it, we simply need to demonstrate one instancewhere it fails to be true.

We can see that if we square the number 1, we get 1, which is not larger.

So the above statement can not be true.

Types of Proof: Contrapositives

There are many common types of proof. We have already seen deductive proofs usingalgebra.

We know that a statement is true, if and only if its contrapositive is true. Sometimes itturns out that it’s easier to prove the contrapositive of a statement rather than the

original statement itself. This is useful!

Types of Proof: Contrapositives

There are many common types of proof. We have already seen deductive proofs usingalgebra.

We know that a statement is true, if and only if its contrapositive is true. Sometimes itturns out that it’s easier to prove the contrapositive of a statement rather than the

original statement itself. This is useful!

Types of Proof: Proof by Contradiction

Proof by Contradiction works as follows.

1 We are asked to prove that a statement is true.

2 We instead assume that the statement is false.

3 We then show that this assumption leads to a contradiction.

4 Reality forces the original statement to be true!

Proof by Contradiction: Example

Use Proof by Contradiction to prove that√

2 is irrational. (Euclid 500BC)

Proof: Assume that√

2 is rational, that is,√

bwhere a and b are integers with no

common factors. (If they did have common factors we could cancel them out.)

We want to show that this assumption yields a contradiction.

Squaring both sides of√

bgives us:

Rearranging this we get:

2b2 = a2

bgives us:

2b2 = a2

bgives us:

2b2 = a2

bgives us:

2b2 = a2

bgives us:

2b2 = a2

We see that a2 must be even, because it is the product of 2 and some integer b2.

Now, we proved earlier (ie. added to our maths toolkit) that the product of two evennumbers is even. It’s not hard to show that the product of two odd numbers is odd

(exercise) so, if a2 is even, then a must also be even.

As a is even, we can write a = 2n for some integer n. Substituting this in we get:

2b2 = (2n)2

Expanding the bracket we get:

2b2 = 4n2

Dividing by 2 we get:

b2 = 2n2

2b2 = a2

2b2 = (2n)2

2b2 = 4n2

b2 = 2n2

2b2 = a2

2b2 = (2n)2

2b2 = 4n2

b2 = 2n2

2b2 = a2

2b2 = (2n)2

2b2 = 4n2

b2 = 2n2

2b2 = a2

2b2 = (2n)2

2b2 = 4n2

b2 = 2n2

2b2 = a2

2b2 = (2n)2

2b2 = 4n2

b2 = 2n2

Here we can see that b2 must also be even, as it is the product of 2 and some otherinteger n2. So b must also be even.

We have established that both a and b are even. This means they both are divisible by 2.

However, we assumed that they had no factors in common. Thus, we have reached acontradiction.

∴√

2 is irrational. �

b2 = 2n2

∴√

b2 = 2n2

∴√

b2 = 2n2

∴√

b2 = 2n2

∴√

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