Download Only Prime Numbers?

Mathematics for Computer Science
MIT 6.042J/18.062J
Truth and Proof
Math vs. Reality
Propositions & Predicates
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.1
Only Prime Numbers?
Let p(n) :: n  n  41.
2
Hypothesis:
n 
Copyright © Albert R. Meyer, 2002. All rights reserved.
p (n ) is a prime number
February 7, 2002
L1-2.2
Only Prime Numbers?
Evidence:
p(0)  41
p (1)  43
prime
p (2)  47
prime
p(3)  53
prime
p(20)  461
prime
looking good!
p (39)  1601
prime
enough already!
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
prime
L1-2.3
Only Prime Numbers?
n 
.
p(n) :: n  n  41
2
is a prime number
This can’t be a coincidence.
The hypothesis must be true.
BUT IT’S NOT:
p(40)  1681 is NOT PRIME.
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.4
Only Prime Numbers?
Quickie:
Prove that 1601 is prime,
and 1681 is not prime.
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.5
Further Extreme Example
EULER'S CONJECTURE (1769)
a b c  d
has no solution for a, b, c ,d positive integers:
4
a 

4
4

b 
4
c 

d 
a b c  d
4
Copyright © Albert R. Meyer, 2002. All rights reserved.
4
February 7, 2002
4

4
L1-2.6
Further Extreme Example
Counterexample: 218 years later by Noam
Elkies at Liberal Arts school up Mass Ave:
95800  217519  414560  422481
4
4
4
4
(= (+ (expt 95800
4)
(expt 217519 4)
(expt 414560 4))
(expt 422481 4))
;Value: #t
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.7
Further Extreme Example
Hypothesis:
313  ( x  y )  z
3
3
3
has no positive integer solution.
False. But smallest counterexample has
MORE THAN 1000 digits!
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.8
Evidence vs. Proof
Claim: All odd numbers greater than 1 are prime.
MATHEMATICIAN: 3 is prime, 5 is prime,
7 is prime, but 9  3  3 is not prime, so the
proposition is false!
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.9
Evidence vs. Proof
Claim: All odd numbers greater than 1 are prime.
PHYSICIST: 3 is prime, 5 is prime, 7 is
prime, 9 is not prime, but 11 is prime, 13 is
prime. So 9 must be experimental error; the
proposition is true!
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.10
Evidence vs. Proof
Claim: All odd numbers greater than 1 are prime.
LAWYER: Ladies and Gentleman of the
jury, it is beyond all reasonable doubt that
odd numbers are prime. The evidence is
clear: 3 is prime, 5 is prime, 7 is prime, 9 is
prime, 11 is prime, and so on.
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.11
Math
Sets
Numbers
T, F
Booleans
Strings
F ( x)  x 2  2
Functions


F  m a
Relations
Copyright © Albert R. Meyer, 2002. All rights reserved.
7 , , i  1
a b
ab
Vectors
February 7, 2002
L1-2.12
Not Math
Solar System
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.13
Not Math
Physical Motion
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.14
Not Math
Family
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.15
Not Math
Cats
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.16
Not Math: Cogito ergo sum
René Descartes'
MEDITATIONS
(Picture source: http://www.btinternet.com/~glynhughes/squashed/descartes.htm)
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.17
Not Math: Cogito ergo sum
René Descartes'
MEDITATIONS
on First Philosophy in which the Existence of God and
the Distinction Between Mind and Body are Demonstrated.
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.18
Propositional (Boolean) Logic
Proposition is either True or False
Examples:
Nonexamples:
Copyright © Albert R. Meyer, 2002. All rights reserved.
22  4
1 1  4
True
False
Wake up!
Where am I?
February 7, 2002
L1-2.19
Operators
 :: AND
 :: OR
 :: NOT
 :: IMPLIES
 :: IFF (if and only if)
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.20
Deductions
A student is trying to prove that propositions P, Q,
and R are all true. She proceeds as follows.
First, she proves three facts:
• P implies Q
• Q implies R
• R implies P.
Then she concludes,
``Thus P, Q, and R are obviously all true.''
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.21
Deductions
From:
P implies Q, Q implies R, R implies P
Conclude: P, Q, and R are true.
(( P  Q)  (Q  R)  ( R  P))
( P  Q  R)
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.22
Truth Table
Could use a truth table.
Conclusion (below the line) must be true
whenever Hypothesis (above the line) is true.
(( P  Q)  (Q  R)  ( R  P))
( P  Q  R)
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.23
Truth Table
Conclusion (below the line) must be true
whenever Hypothesis (above the line) is true.
P Q R ( P  Q)  (Q  R)  ( R  P)
T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
T
F
F
F
F
F
F
T
T
F
T
F
T
F
T
F
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
OK
OK
OK
OK
OK
OK
OK
NOT OK!
L1-2.24
Goldbach Conjecture
Every even integer greater than 2 is the
sum of two primes.
Evidence:
4  22
5  3 2
6  3 3
8  53
20  ? 13  7
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.25
Goldbach Conjecture
True for all even numbers with
up to 13 digits!
(Rosen, p.182)
It remains an OPEN problem:
no counterexample, no proof.
UNTIL NOW!…
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.26
Goldbach Conjecture
The answer is on my desk!
(Proof by Cases)
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.27
Quicker by Cases
(( P  Q)  (Q  R)  ( R  P))
( P  Q  R)
Case 1: P is true. Now, if Hypothesis is true,
then Q must be true (because P implies Q).
Then R must be true (because Q implies R).
So the conclusion P  Q  R is true.
This case is OK.
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.28
Quicker by Cases
(( P  Q)  (Q  R)  ( R  P))
( P  Q  R)
Case 2: P is false. To make Hypothesis true,
R must be false (because R implies P), so
Q must be false (because Q implies R).
This assignment does make the Hypothesis true,
but the conclusion P  Q  R is (very) False.
This case is not OK.
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.29
Problems
Tutorial Problems 1 & 2
Copyright © Albert R. Meyer, 2002. All rights reserved.
February 7, 2002
L1-2.30