Trivial Proofs - Discrete Mathematics and its Applications - Lecture Slides, Slides for Discrete Mathematics. Shoolini University of Biotechnology and Management Sciences

Discrete Mathematics

Description: During the study of discrete mathematics, I found this course very informative and applicable.The main points in these lecture slides are:Trivial Proofs, Methods of Proof, Rules of Inference, Proof Strategies, Vacuous Proofs, Empty Set, Postive Integers, Example Indirect Proof, Proof by Contradiction, Common Divisor, Equivalence Proofs, Theorems with Quantifiers
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Lecture 11
1.5, 3.1 Methods of Proof
Last time in 1.5
To prove theorems we use rules of inference such as:
p, pq, therefore, q
NOT q, pq, therefore NOT p.
p AND q, therefore p
FORALL x P(x), therefore for arbitrary c, P(c)
EXISTS x P(x), therefore for some c, P(c)
It is easy to make mistakes, make sure that:
1) All premises pi are true when you prove (p1 AND p2 q
2) Every rule of inference you use is correct.
Some proof strategies:
To proof pq
1) direct proof: assume p is true, use rules to prove that q is true.
2) indirect proof, assume q is NOT true, use rules to prove p is NOT true.
To prove p is true:
3) By contradiction: assume p is NOT true, use rules to show that NOT pF
i.e. it leads to a contradiction.
Vacuous –Trivial Proofs
Lets say we want to prove pq but the premise p can be shown to be false!
Then pq is always true because (FT) = T and (FF) = T.
This is a vacuous prove.
Old example: prove that for any set S:
Proof: The following must be shown to be true:
However: the empty set does not contain any elements
and the premise is always false. Therefore the implication
must always be true!
()xx x S ∈∅→ ∈
Trivial Proof: We want to prove pq, and we can show that q is true.
Then, because (TT) = T and (FT) = T we have proven the implication.
Example: P(n): a>=b a^n >= b^n for postive integers.
Is P(0) true?
P(0): a^0 >= b^0 is equivalent to 1>=1. Therefore, q is true and thus pq is true.
Example Indirect Proof
Prove that: if n is an integer and n^2 is odd, then n is odd.
Direct prove is hard in this case.
Indirect proof: Assume NOT q : n is even.
n = 2k
n^2 = 4k^2 = 2(2k^2) is even, is not odd.
Thus NOT q NOT p, pq
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