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

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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 Pr...
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Lecture 11

1.5, 3.1 Methods of Proof

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Last time in 1.5 To prove theorems we use rules of inference such as: p, pq, therefore, q NOT q, pq, 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 AND...pn)  q 2) Every rule of inference you use is correct. Some proof strategies: To proof pq 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 pF i.e. it leads to a contradiction.

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Vacuous –Trivial Proofs Lets say we want to prove pq but the premise p can be shown to be false! Then pq is always true because (FT) = T and (FF) = 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!

S∅⊆

( )x x x S∀ ∈∅→ ∈

Trivial Proof: We want to prove pq, and we can show that q is true. Then, because (TT) = T and (FT) = 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 pq is true.

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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, pq

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Example of Proof by Contradiction def: rational number is a number that can be written as a/b for integers a,b, where b should not be 0. Reals that are not rational are irrational. sqrt(2) is irrational. (note: we are not proving an implication now, although we could have written it as: if x=sqrt(2)  x = irrational.) Assume sqrt(2) is not irrational. sqrt(2) = a/b where there is no common divisor (otherwise divide by this number). 2 = a^2 / b^2 2b^2 = a^2 a^2 = even a = even a = 2c 2b^2 = 4c^2 b^2 = 2c^2 b^2 = even b=even both a and b can be divided by 2! (contradiction) p r r¬ → ∧¬

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1.5 An indirect proof is in fact also a prove by contradiction: Assume p = T and (NOT q) = T indirect prove: (NOT q)  (NOT p) Therefore: p AND (NOT p) = T, contradiction

Proof by cases: (p1 OR p2 OR p3 ... OR pn)  q (if at least one of the premises hold, q follows) This is equivalent to (p1q) AND (p2q) AND ... AND (pnq) Example: Prove x^2 >= 0 p1: x<0  x^2 >=0 p2: x=0  x^2 >=0 p3: x>0 x^2 >=0

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Equivalence Proofs These are bidirectional statements of the form pq. Equivalent to proving 2 cases: pq qp since (pq) AND (qp)  (pq) is a tautology.

Example: Prove integer n is odd if and only if (iff) n^2 is odd. Lecture 10: if n is odd  n^2 is odd Lecture 11: if n^2 is odd  n is odd. 

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Lists of Equivalences

p1 p2p3.....pn This means they are all equivalent. There are C(n,2) pairs to prove! There is however a smarter way: Design one path through all pi’s that can bring you from any pi to any other then you are done:

p1 p2

p3

p4 p5

p6

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Theorems with Quantifiers Existence proofs: Proofs of the form: There exists an element x such that..... these proofs may be constructive (construct some x) or non-constructive.

( )xP x

Uniqueness: Proofs of the form: there exists a unique element x such that...

! ( ) ( ( ) ( ( )))xP x x P x y y x P y∃ = ∃ ∧∀ ≠ →¬

Example: Prove that ! ( 0)x y x y∀ ∃ + = proof: for arbitrary x, y=-x makes the proposition true. Is it unique. Assume it is not true and show contradiction: Let say there is a r such that x+r=0 but r is not –x. Then it follows that x+r=x+y  r=y which contradicts our assumption.

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3.1

self reading

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The Halting Problem Is it possible to design an algorithm that always predicts for a given program P and a given input to that program I, if it will stop run forever? Proposition: there isn’t Proof. Assume there is such a predictor H(P,I). A program can be represented as a bit-string, and therefore as input to another program. Imagine a program P that can take itself as input. H(P,P) should predict if it stops: H = 1 if it stops, H=0 if it doesn’t stop. Define a new program K(P) as follows: that is K(P) loops forever if H(P,P) = 1 K(P) = stops if H(P,P) = 0 Can H(K,K) predict whether this program stops? K(K) loops forever if H(K,K) = 1 K(K) stops if H(K,K) = 0 So H(K,K) does not predict correctly which is in contradiction with the assumption!

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Sequences & Summations

definition:A sequence is a function from the set of integers to a set S: a1,a2,a3,.... (function from 1,2... a1,a2,...) an is a term in the sequence which is sometimes denoted with {an} (not a set!)

Example: {an} with an= 1/n over n=1,.... 1,1/2,1/3, ...

Geometric progression: Sequence of the form: a,ar,ar^2,...,ar^n,... a = initial term, r = common ratio are real numbers 3,6,12,24,... (a=3,r=2): ratios are constant Arithmetic progression Sequence of the form a,a+d,a+2d,...,a+nd,... a = initial term and d = common difference are real numbers 3,5,7,9,11,...(a=3,r=2): differences are constant A string: finite sequence a1,...,an.

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Sums Notation

2 2 2

1 1 1

1 2 1

2

2 2

, , 1

,...,

( .... )

n n n

i a k m n i m a m k m n n n

ij i m i m in i m j m i m

a a a a a

a a a a

+ = = = −

+ = = =

= = =

= + +

∑ ∑ ∑

∑ ∑ ∑

notation:

single sum

double sum

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Summations

Theorem

1

0

1 1

( 1) 1

n n

j

j

ar a if rar r n a if r

+

=

 − ≠= −

 + = ∑

0

1

0

1

1

1

0 1

1

1 1

n j

j

n j

j

n k

k n

n k

k n

n

S ar

rS ar

rS ar

rS ar a ar

rS ar a S ar aS if r

r

=

+

=

+

=

+

=

+

+

= ⇔

= ⇔

= ⇔

= − + ⇔

= − + ⇔

− = ≠

proof

trivial

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Summations Theorem:

0

( 1)( 1) 2

n

j

n na jd n a d =

+ + = + +∑

0 1 2 ... (0 ) (1 1) (2 2) ...

( 1) 2

n n n n

n n

+ + + + = + + + − + + − + +

=

sketch of proof: Split into 2 cases: even and odd n. In both cases argue as follows:

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