Mathematics for Computer Science
MIT 6.042J/18.062J
Proof by Contradiction
Proof by Contradiction
Proof by Cases
Is
3
1332 ≤ 11?
3
1331
If so, 1332 ≤ 11
That’s not true, so
Albert R Meyer
lec 1F.1
February 5, 2010
Proof by Contradiction
February 5, 2010
Albert R Meyer
lec 1F.3
Proof by Contradiction
Theorem:
If an assertion implies
something false, then
the assertion itself
must be false!
•  Suppose
is irrational.
was rational
•  So have n, d integers without common
prime factors such that
•  We will show that n & d are both even.
This contradicts no common factor.
Albert R Meyer
lec 1F.4
February 5, 2010
is irrational.
Albert R Meyer
lec 1F.5
Proof assumes that
if n2 is even, then n is even.
Why is this true?
so can assume
So n is even
February 5, 2010
Quickie
Proof by Contradiction
Theorem:
Albert R Meyer
So d is even
February 5, 2010
lec 1F.6
Albert R Meyer
February 5, 2010
lec 1F.7
1
Mathematics for Computer Science
MIT 6.042J/18.062J
Java Logical Expression
if
((x>0)
||
(x
<=
0
&&
y>100))


 AND
OR
Proof by Cases
(more code)
better: if
((x>0)
||
y>100)

(more code)
February 5, 2010
Albert R Meyer
lec 1F.8
Case 1: x > 0
false
if
((x>0)
||
(x
<=
0
&&
y>100))


if
((x>0)
||
(x
<=
0
&&
y>100))


AND
AND
OR
true
false
if
((x>0)
||
y>100)

if
((x>0)
||
y>100)

OR
OR
so both are true
Albert R Meyer
lec 1F.9
Case 2: x ≤ 0
true
OR
February 5, 2010
Albert R Meyer
February 5, 2010
lec 1F.10
Case 2: x ≤ 0
Albert R Meyer
February 5, 2010
lec 1F.11
Case 2: x ≤ 0
true
if
(x
<=
0
&&
y>100)

if
(
y>100)
AND
if
(
y>100)
if
((x>0)
||
y>100)
if
(
y>100)
if
((x>0)
||
y>100)
so both still the same
Albert R Meyer
February 5, 2010
lec 1F.12
Albert R Meyer
February 5, 2010
lec 1F.13
2
$1,000,000 Question
Proof by Cases
Reasoning by cases can break a
complicated problem into
easier subproblems.
Some philosophers* think
reasoning this way is worrisome.
Is P = NP ?
*intuitionists
Albert R Meyer
February 5, 2010
lec 1F.25
$1,000,000 Question
February 5, 2010
February 5, 2010
lec 2M.28
Team Problems
Problems
1―4
The answer is on my
desk!
(Proof by Cases)
Albert R Meyer
Albert R Meyer
lec 1F.30
Albert R Meyer
February 5, 2010
lec 1F.31
3
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6.042J / 18.062J Mathematics for Computer Science
Spring 2010
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