Mathematics for Computer Science
MIT 6.042J/18.062J
Proof by Contradiction
Proof by Cases
Copyright © Albert R. Meyer, 2009. All rights reserved.
February 5, 2009
lec 1R.1
Proof by Contradiction
Is
3
1332  11?
(i)
If so, 1332 /
 1331
11
3
That’s not true,
true so (i)
better not be true either
Copyright © Albert R. Meyer, 2009. All rights reserved.
February 5, 2009
lec 1R.2
Proof by Contradiction
Is
3
1333  11?
If so, 1333  1332
11
3
That’s not true, so
3
1332  11
Copyright © Albert R. Meyer, 2009. All rights reserved.
February 5, 2009
lec 1R.3
Proof by Contradiction
If an assertion implies
something false, then
the assertion itself
must be false!
Copyright © Albert R. Meyer, 2009. All rights reserved.
February 5, 2009
lec 1R.4
Proof by Contradiction
Theorem:
2 is irrational.
• Suppose 2 was rational
• So have n, d integers without common
prime factors such that
n
2=
d
• We will show that n & d are both even.
This is contradicts no common factor.
Copyright © Albert R. Meyer, 2009. All rights reserved.
February 5, 2009
lec 1R.5
Proof by Contradiction
Theorem:
2 is irrational.
so can assume
n
2=
d
2d = n
2d2 = 4k 2
2d2 = n2
d = 2k
n = 2k
n2 = 4k2
2
So n is even
Copyright © Albert R. Meyer, 2009. All rights reserved.
2
So d is even
February 5, 2009
lec 1R.6
Quickie
Proof assumes that
2
if n is even, then n is even.
Why is this true?
Copyright © Albert R. Meyer, 2009. All rights reserved.
February 5, 2009
lec 1R.7
Mathematics for Computer Science
MIT 6.042J/18.062J
Proof by Cases
Copyright © Albert R. Meyer, 2009. All rights reserved.
February 5, 2009
lec 1R.9
Java Logical Expression
AND
OR
if ((x>0) || (x <= 0 && y>100))
(more code)
better: if ((x>0) || y>100)
(more code)
Copyright © Albert R. Meyer, 2009. All rights reserved.
February 5, 2009
lec 1R.10
Case 1: x>0
true OR
if ((x>0) || (x <= 0 && y>100))
true OR
if ((x>0) || y>100)
so both are true
Copyright © Albert R. Meyer, 2009. All rights reserved.
February 5, 2009
lec 1R.11
Case 2: x
falseOR
0
if ((x>0) || (x <= 0 && y>100))
falseOR
if ((x>0) || y>100)
Copyright © Albert R. Meyer, 2009. All rights reserved.
February 5, 2009
lec 1R.12
Case 2: x
if
0
true AND
(x <= 0 && y>100)
(
if ((x>0)
|| y>100)
Copyright © Albert R. Meyer, 2009. All rights reserved.
February 5, 2009
lec 1R.13
Case 2: x
if
0
(
y>100)
(
if ((x>0)
|| y>100)
so both still the same
Copyright © Albert R. Meyer, 2009. All rights reserved.
February 5, 2009
lec 1R.14
Proof by Cases
Reasoning by cases can break a
complicated problem into
easier subproblems.
Some philosophers* think
reasoning this way is worrisome.
*intuitionists
Copyright © Albert R. Meyer, 2009. All rights reserved.
February 5, 2009
lec 1R.26
$1,000,000 Question
Is P = NP ?
Copyright © Albert R. Meyer, 2009. All rights reserved.
February 5, 2009
lec 2M.29
$1,000,000 Question
The answer is on my
desk!
(Proof by Cases)
Copyright © Albert R. Meyer, 2009. All rights reserved.
February 5, 2009
lec 1R.31
Team Problems
Problems
1―4
Copyright © Albert R. Meyer, 2009. All rights reserved.
February 5, 2009
lec 1R.33