proof by contradiction cs 270 math foundations of cs jeremy johnson
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Decimals and FractionsTRANSCRIPT
Proof by Contradiction
CS 270 Math Foundations of CSJeremy Johnson
Outline
1. An example1. Rational numbers and repeating decimals2. Irrationality of
2. Negation Rules1. Bottom introduction and elimination2. Negation introduction and double negation
3. Indirect Proofs1. Double negation2. Law of excluded middle3. DeMorgan’s Laws
Decimals and Fractions
• 3/8 = .375 =
.3758 |3.0 2.4 60 56 40 40 0
Repeating Decimals• 9/11 = 0.81818181 … = 0.
.8111 |9.0 88 020 11 9• 7/23 = 0.
81
Repeating Decimals
• Theorem. A number r is rational iff it has a terminating or repeating decimal expansion
• Proof® If r = a/b perform long division to compute
the decimal expansion. At each step divide what is left by b, m = qb + r, 0 ≤ r < b. There are b possible remainders. If r = 0 the expansion is terminating. If r has occurred previously the expansion is repeating. After at most b steps one of these must happen.
Repeating Decimals
• Theorem. A number r is rational iff it has a terminating or repeating decimal expansion
• Proof Without loss of generality, we can assume 0 <
r < 1. If r = .a1…an, then r = a1…an/10n. If r = .b1…bk, then 10kr - b1…bk = a = . and (10n-1)a = a1…an and hence a is rational and consequently r is also rational.
Decimal Expansion of sqrt(2)
is the positive solution of
We can approximate the solution of this equation by repeated bisection
Decimal Expansion of sqrt(2)x0 := 1.0; x1 := 2.0; n := 20;for i from 1 to n do x := (x0+x1)/2; if x^2 > 2 then x1 := x; else x0 := x; end if;end do;
1.5000000001.250000000 1.375000000 1.437500000 1.406250000 1.421875000 1.414062500 1.417968750 1.416015625 1.415039062 1.414550781 1.414306640 1.414184570 1.414245605 1.414215088 1.414199829 1.414207458 1.414211273 1.414213180 1.414214134
Decimal Expansion of sqrt(2)
Does the expansion terminate or repeat? Maybe it doesn’t? How long should I look? Maybe it’s not rational?
Proof that sqrt(2) is not Rational
Proof by contradiction. Assume with gcd(a,b)=1. Then 2b2= a2. Since an odd number squared is odd, this
implies a=2p is even and 2b2= 4p2 and b2= 2p2 and b is also even.
Since a and b are both even gcd(a,b) 1 which is a contradiction
Thus we conclude that is not rational.
Negation Rules
• Introduce the symbol ( to encode a contradiction
• Bottom elimination can prove anything
• Bottom introduction
i
e.
Negation Rules
• Introduction and elimination rules
• Double negation e
… i
… e
Proof by Contradiction
• Negation elimination called proof by contradiction
Assume and derive a a contradiction
… PBC
Exercise
• Prove that A ® A and A ® A
Law of the Excluded Middle
• [derived rule LEM]
1 (p p) assumption
2 Assumption
3 (p p) ,4
4 3,15 p 2-4
6 p p ,4
7 6,1
8 p p e 1-7
De Morgan’s Law
(P Q) P Q
1 (P Q) premise
2 assumption
3 P Q i1 2
4 1,3
5 P 2-4
6 Q
7 P Q i2 6
8 1,7
9 -8
10 P Q i 5,9
De Morgan’s Law
(P Q) P Q
1 P Q premise
2 e1 1
3 e2 1
4 assumption
5 P assumption
6 e 2,5
7 Q i2 6
8 e 3,7
9 e 4,5-6, 7-8
10 (P Q) i 4-9
Exercise
• Prove (P Q) ® P Q
Exercise
• Prove P Q ® (P Q)