More Number Theory Proofs

1
More Number Theory Proofs

• Rosen 3.1

2
Prove or Disprove

• If m and n are even integers, then mn is
  divisible by 4.
• The sum of two odd integers is odd.
• The sum of two odd integers is even.
• If n is a positive integer, then n is even iff
  3n28 is even.
• n2 n 1 is a prime number whenever n is a
  positive integer.
• n2 n 1 is a prime number whenever n is a
  prime number.
• x y ? x y when x,y ? R.
• ?3 is irrational.

3
If m and n are even integers, then mn is
divisible by 4.

• Proof
• m and n are even means that there exists integers
  a and b such that m 2a and n 2b
• Therefore mn 4ab. Since ab is an integer, mn
  is 4 times an integer so it is divisible by 4.

4
The sum of two odd integers is odd.

• This is false. A counter example is 13 4

5
The sum of two odd integers is even.

• Proof
• If m and n are odd integers then there exists
  integers a,b such that m 2a1 and n 2b1.
• m n 2a12b1 2(ab1). Since (ab1) is
  an integer, mn must be even.

6
If n is a positive integer, then n is even iff
3n28 is even.

• Proof We must show that n is even ? 3n28 is
  even, and that 3n28 is even ? n is even.
• First we will show if n is even, then 3n28 is
  even.
• n even means there exists integer a such that n
  2a. Then 3n28 3(2a)2 8 12a2 8 2(6a2
  4) which is even since (6a2 4) is an integer.

7
If n is a positive integer, then n is even iff
3n28 is even (cont.).

• Now we will show if 3n28 is even, then n is even
  using the contrapositive (indirect proof).
• Assume that n is odd, then we will show that
  3n28 is odd. n odd means that there exists
  integer a such that n 2a1.
• 3n28 3(2a1)2 8 3(4a2 4a 1) 8
  2(6a2 2a 4) 1, which is odd.
• Therefore, by the contrapositive if 3n28 is
  even, then n is even.

8
n2 n 1 is a prime number whenever n is a
positive integer.

• Try some examples
• n 1, 111 3 is prime
• n 2, 421 7 is prime
• n 3, 931 13 is prime
• n 4, 1641 21 is not prime and is a counter
  example.
• Not true.

9
n2 n 1 is a prime number whenever n is a
prime number.

• Try some examples
• n 1, 111 3 is prime
• n 2, 421 7 is prime
• n 3, 931 13 is prime
• n 5, 2551 31 is prime
• n 7, 4971 57 is not prime (193).
• Not true.

10
Prove x y ? x y when x,y ? R.

• Note z is equal to z if z?0 and equal to -z if
  z lt 0
• There are four cases
• x y
• ?0 ?0
• lt0 ?0
• ?0 lt0
• lt0 lt0

11
Prove x y ? x y when x,y ? R.

• Case 1
• x,y are both ?0
• Then
• x y x y xy since both x and y
  are positive.

12
Prove x y ? x y when x,y ? R.

• Case 2 x lt 0 and y ?0 so x y -x y
• If y ? -x, then xy is nonnegative and xy
  xy
• Since x is negative, -x gt x, so that
• x y -x y gt xy xy
• If y lt -x, then xy -(xy) -x -y.
• Since y ?0, then y ? -y, so that
• x y -x y ? -x -y xy

13
Prove x y ? x y when x,y ? R.

• Case 3 x?0 and y lt0 so x y x -y
• If x ? -y, then xy is nonnegative and xy
  xy
• Since y is negative, -y gt y, so that
• x y x -y gt xy xy
• If x lt -y, then xy -(xy) -x -y.
• Since x?0, then x?-x , so that
• x y x -y ? -x -y xy

14
Prove x y ? x y when x,y ? R.

• Case 4 x,y are both lt 0
• Then x y -x - y -(xy) xy
• Therefore the theorem is true. This is know in
  mathematics as the Lipschitz condition.

15
Prove that 3?3 is irrational.

• Proof (by contradiction) Assume that 3?3 is
  rational, i.e. that 3?3 a/b for a,b?Z and b?0.
  Since any fraction can be reduced until there are
  no common factors in the numerator and
  denominator, we can further assume that a and b
  have no common factors.

16
Prove that 3?3 is irrational. (cont.)

• Then 3 a3/b3 which means that 3b3 a3 so a3 is
  divisible by 3.
• Lemma When m is a positive integer, then if m3
  is divisible by 3, then m is divisible by 3.
  (Left as an exercise).
• By the lemma since a3 is divisible by 3, then a
  is divisible by 3. Thus ?k?Z ? a 3k.

17
Prove that 3?3 is irrational. (cont.)

• Now, we will show that b is divisible by 3.
• From before, a3/b3 3 ? 3b3 a3 (3k)3.
• Dividing by 3 gives b3 9k3 3(3k3). Therefore
  b3 is divisible by 3 and from the Lemma , b is
  divisible by 3.

18
Prove that 3?3 is irrational. (cont.)

• But, if a is divisible by 3 and b is divisible by
  3, then they have a common factor of 3. This
  contradicts our assumption that our a/b has been
  reduced to have no common factors.
• Therefore 3?3 ? a/b for some a,b?Z, b?0.
• Therefore 3?3 is irrational.