Indirect proofs

Consider an implication: p→q

– It’s contrapositive is ¬q→¬p

• Is logically equivalent to the original implication!

– Thus, show that if ¬q is true, then ¬p is true

• To perform an indirect proof, do a direct proof on the contrapositive

Proof by contradiction

• Given a statement p, assume it is false

– Assume ¬p

• Prove that ¬p leads to false

– A contradiction exists

• Given a statement of the form p→q

– To assume it’s false, you only have to consider the case where p is true and q is false

Vacuous proofs

• Consider an implication: p→q

• If it can be shown that p is false, then the implication is always true

– By definition of an implication

• Note that you are showing that the antecedent is false

Trivial proofs

• Consider an implication: p→q

• If it can be shown that q is true, then the implication is always true

– By definition of an implication

• Note that you are showing that the conclusion is true

Let's suppose √2 were a rational number. Then we can write it √2 = a/b where a,b are whole numbers, b not zero.

From the equality √2 = a/b it follows that 2 = a^2/b^2, or a^2 = 2 * b^2.

if a itself is an even number, then a is 2 times some other whole number, or a = 2k where k is this other number. 2 = a^2/b^2, this is what we get:

B^2 =2k^2. b^2 is even. It is contradiction. So √2 cannot be rational.

33.Mathematical Induction. Well-ordering property. Basic and inductive steps. Why mathematical induction is valid. Second principle of mathematical induction.

Show that 1² + 2² + … + n² = n(n+1)(2n+1)/6.

1)Mathematical induction is a a specialized form of deductive reasoning used to prove a fact about all the elements in an infinite set by performing a finite number of steps.

2)The well-ordering property states, that every nonempty subset of the Natural Numbers has a least member.

∀S ≠ ∅ ⊆ N ∃x ∈ S ∀z ∈ S (x ≤ z)

Mathematical induction for proving ``the proposition P(n) is true for all positive integers n'':

Basis step: show that P(1) is true

Inductive step: show that the proposition P(k) -> P(k+1) is true, where k is an arbitrary positive integer.

After we complete the basis step and inductive step, we proved that P(n) must be true for all positive integers n.

Let us show by using ``proof-by-contradiction''.

Assume that there is at least one positive integer for which P(n) is false. As a result, the set S of positive integers for which P(n) is false is nonempty.

Thus, by the well-ordering property, S has a least element, which will be denoted by m. Correspondingly, P(m) is false.

We know that m cannot be 1, because P(1) is true.

Because m is positive and greater than 1, m-1 is a positive integer.

Since m-1 is less than m, it is no in S. Therefore P(m-1) must be true.

Because the conditional statement P(m-1) -> P(m) is true, it must be the case that P(m) is true.

This is a contradiction. Hence, P(n) must be true for every positive integers n.

Recall: well-ordering property: Every nonempty subset of the set of positive integers has a least element.

Formally the second principle of induction states that

if n [ k [ k < n P(k) ] P(n) ] , then n P(n) can be concluded.

Here k [ k < n P(k) ] is the induction hypothesis.

The reason that this principle holds is going to be explained later after a few examples of proof.