 Mathematical Induction Assume that we are given an infinite supply of stamps of two different denominations, 3 cents and and 5 cents. Prove using mathematical.

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Mathematical Induction Assume that we are given an infinite supply of stamps of two different denominations, 3 cents and and 5 cents. Prove using mathematical induction that it is possible to make up stamps of any value  8 cents. Basis: We can make an 8 cents stamp by using one 3-cents and one 5-cents stamp. Inductive hypothesis: Assume that we can make stamps of values k = 8, 9, 10, … n. Inductive step: Show that we can compose a stamp of value n + 1. Composing Stamps: Example 3 Illustration 8 = 3 + 5 9 = 3 + 3 + 3 = 3 * 3 10 = 5 + 5 = 2 * 5 11 = 1 * 5 + 2 * 3 Outline of Proof

Composing Stamps – Actual Proof Inductive step: Show that we can compose a stamp of value n + 1. In composing n, several (or none) 3-cents and several (or none) 5- cents stamps have been used. To go from n to n + 1, we consider two cases. Case 1 : If there is at least one 5-cents stamp in the collection, replace it by two 3-cents stamps. This gives us stamps of value n + 1 cents. Case 2 : Suppose that the current collection uses only 3-cents stamps. Since n  8, there must three 3-cents stamps in the collection. Replace these three stamps by two 5-cents stamps. This gives us a stamp of value n + 1 cents. The proof is complete. Another Proof

n + 1 = 3 * (p + 2 ) + 5 * (q – 1) if q > 1 n + 1 = 3 * ( p – 3 ) + 5 * (q + 2 ) if p > 3 For all n > 8, n can be expressed as a linear combination of 3 and 5, that is n = 3 * p + 5 * q for n >= 8

Review Propositional Logic – Page 1 Implication In propositional logic "if p then q" is written as p  q and read as "p implies q". p q p  q same as  p  q F F T F T T T F F T T T How to remember this definition? Implication is false only when the premise is true and the consequence is false.

Propositional Logic – Page 2 Bi-conditional p  q means p iff q p q p  q same as F F T F T F T F F T T T How to remember this definition? Bi-conditional is a like a magnitude comparator, it is true when both inputs are identical.

More Terminology Contradiction A logical expression that is always false, regardless of what truth values are assigned to its statement variables, is called a contradiction. The statement p   p is a contradiction. Theorem If A and B are logical statements and if the statements A and A  B are true, then the statement B is true. A logical expression that is always true, regardless of what truth values are assigned to its statement variables, is called a tautology. The statement p   p is a tautology. Tautology

Simple Logic Proofs A B A  B A  (A  B) A  (A  B)  B F F T F T F T T F T T F F F T T T T T T Prove that A  (A  B)   B is a contradiction. A  (A  B)   B = A  (  A  B)   B = (A   A)  (A  B)   B using distributive law. = F  (A  B)   B = (A  B)   B = A  (B   B) using associative law = A  F = F Contradiction. Hence B must be true. Direct Proof using a truth table Proof by contradiction

An Important Theorem Among the four statements p  qstatement q  pconverse  p   qinverse  q   p contra-positive 1. The statement and its contra-positive are equivalent. 2. The converse and inverse are equivalent. 3. No other pairs in the statements given above are equivalent. Proof: Make a truth table to see that 1 and 2 are tautologies.1

Proof By Contradiction Prove that is not a rational number. (Example 1.7, page 13) Proof: Assume is a rational number. Let m and n be two integers, with no common factor, such that Or This shows that is even. Since is a multiple of 2, it is an even number. Therefore, n is even must be of the form 2k for some integer k. 1 1

Proof Continued … We have concluded that both and are even, therefore, n and m are also even They must have a common factor. This contradicts our assumption. Hence is not a rational number.

Proof of Proposition 1 pqp  q~p~q~q  ~p (p  q)  (~q  ~p) FFTTTTT FTTTFTT TFFFTFT TTTTTTT

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