Proof By ContradictionProof By Contradiction

Chapter 3Indirect Argument

Contradiction

Theorems 3.6.1 and 3.6.2

pg. 171

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Robert Di Battista Introduction, slides

Da’niel Rowan Theorem 3.6.1

Annette Stiller Theorem 3.6.2

Proof By Contradiction

Indirect Argument Prove the negation is false. reductio ad absurdum Assume its negation is true, until it is

reduced to an impossibility or absurdity. This leaves only one possibility.

Method of Proof By Contradiction

Suppose the statement to be proved is false, i.e. suppose the negation of the statement is true.

Show that this supposition leads logically to a contradiction.

Conclude that the statement to be proved is true.

Analogous to…

Shows truth by discounting the opposite. Sarcasm Reverse psychology

Theorem 3.6.1

There is no greatest integer.There is no greatest integer.

Theorem 3.6.1

[We take the negation of the theorem

and suppose it to be true]

Suppose not. That is, suppose there is a greatest integer N.

[We must deduce

a contradiction]

Then N > n for every integer n.

Theorem 3.6.1

Let M = N + 1.

M is an integer since it is the sum of integers.

Also M > N since M = N + 1.

Theorem 3.6.1

Thus M is an integer that is greater than N. So N is the greatest integer and N is not the greatest integer, which is a contradiction.

[This contradiction shows that the

supposition is false and, hence,

that the theorem is true.]

Theorem 3.6.1

Bill Gates is disgustingly rich, but someone can always have $1 more than him.

Theorem 3.6.1

Exam question:Prove by contradiction

“There is no greatest odd integer.”

WAIT!

THERES MORE

Theorem 3.6.2

There is no integer that There is no integer that

is both even and odd.is both even and odd.

Theorem 3.6.2

[We take the negation of the theorem

and suppose it to be true]

Suppose not. That is, suppose there is an integer N that is both even and odd.

[We must deduce

a contradiction.]

Theorem 3.6.2

By definition of even, n = 2a for some integer a, and by definition of odd, n = 2b + 1 for some integer b. Consequently,

2a = 2b +1 [By equating the two expressions for n.]

And so2a – 2b = 12(a – b) = 1 (a – b) = 1/2 [by algebra]

Theorem 3.6.2

Now since a and b are integers, the difference a – b must also be an integer. But a – b = 1/2, and 1/2 is not an integer. Thus a – b is an integer and a – b is not an integer, which is a contradiction.

[This contradiction shows that the

supposition is false and, hence,

that the theorem is true.]

Analogous to…

Mutually exclusive groups: Male or female Positive or negative True or false

Theorem 3.6.2

Exam question:

Prove by contradiction

“There is no real number that is both positive and negative.”

Related Homework

Section 3.6 3 -15 21 – 27 32