Download CS 173: Discrete Structures, Summer 2013 Lecture 23 selected proofs

CS 173: Discrete Structures, Summer 2013
Lecture 23 selected proofs
1. Proof by contradiction Claim: There are infinitely many prime numbers
Equivalent claim: There is not a finite set that includes all primes.
Suppose for purposes of contradiction there is a finite set of primes. We can list these
primes as p1 , p1 , . . . , pn . Now consider the number Q = p1 × p2 × · · · × pn + 1. If you
divide Q by any of the primes you have a remainder of 1. But by the Fundamental
Theorem of Arithmetic, Q must have a prime factor. This contradicts our assumption
that p1 , p1 , . . . , pn was a list of all the prime numbers therefore there must not be a
finite set that includes all primes.
2. Proof by contradiction Claim:
√
2 is irrational
Equivalent claim:
√ There does not exist a pair of integers a, b without common factors
such that ab = 2
Assume for purposes of
√ contradiction there is a pair of integers a, b without common
factors such that ab = 2.
√
a
2=
b
√ 2 a 2
2 =
b
2
a
2= 2
b
2b2 = a2
So by definition a2 is even and also a is even. Thus we can write a = 2k for some
k ∈ Z. So we get the following
2b2
2b2
2b2
b2
= a2
= (2k)2
= 4k 2
= 2k 2
Thus b2 is even and also b is even. This is a contradiction since we assumed that a, b
have no common factors and if they are both even they share the common factor of 2.
1
√
√
2 + 6 < 15
√
√
√
Assume for purposes of contradiction 2 + 6 ≥ 15. We can then work as follows.
3. Proof by contradiction Claim:
√
√
√
√
2 + 6 ≥ 15
√
√ 2
√
( 2 + 6)2 ≥ 15
√ √
2 + 2 2 6 + 6 ≥ 15
√ √
2 2 6≥7
√ √
(2 2 6)2 ≥ 72
4 × 2 × 6 ≥ 49
48 ≥ 49
Which is a contradiction therefor
√
√
√
2 + 6 < 15
4. Proof by contradiction Claim: There are infinitely many integers n of the form
n = 4k + 3 where k is an integer.
Assume for purposes of contradiction there are finitely many integers n of the form
n = 4k + 3 where k is an integer. Then there must be a largest. Let l be the largest
integer of the form. Then l = 4k + 3 where k is an integer but since k + 1 is also
an integer there is a m = 4(k + 1) + 3 which is also an integer of the form which is
a contradiction. So there must be infinitely many integers n of the form n = 4k + 3
where k is an integer.
5. Proof by contradiction Claim: There is no rational number r for which r3 +r+1 = 0
Assume for purposes of contradiction that there is a rational number r for which
r3 + r + 1 = 0. Then by definition of rational numbers there are a and b such that
r = ab where a, b ∈ Z and a and b share no common factors.
r3 + r + 1 = 0
a 3 a
+ +1=0
b
b
a3 a
+ +1=0
b3
b
3
a + ab2 = −b3
Now we have two cases either b is even or b is odd.
2
Case: b is even then b3 must be even and a must be odd
a common factor. Since a is odd a3 is odd since an even
ab2 must be eve since an even times any integer is even.
odd since an even plus an odd is even but b3 is even so we
a3 + ab2 6= −b3
since a and b don’t share
to any power is even also
Finally a3 + ab2 must be
have a contradiction since
Case: b is odd then b3 must be odd. So a3 + ab2 must be odd. For this to be true
either a3 must be odd or ab2 must be odd but not both. If a3 is odd a must be odd
and then ab2 is the product of two odds and must be odd. Other wise a3 must be even
and then a is even and so is ab2 since an even times anything is even. Thus since a can
be neither even or odd we have a contradiction.
Since in all cases we have a contradiction we have shown that the original claim that
there is no rational number r for which r3 + r + 1 = 0 must be true.
3