More Number Theory Proofs
Rosen 1.5, 3.1
Prove or Disprove
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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 3n2+8 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.
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.
The sum of two odd integers is odd.
This is false. A counter example is 1+3 = 4
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 = 2a+1 and n =
2b+1.
m + n = 2a+1+2b+1 = 2(a+b+1). Since
(a+b+1) is an integer, m+n must be even.
If n is a positive integer, then n is
even iff 3n2+8 is even.
Proof: We must show that n is even  3n2+8
is even, and that 3n2+8 is even  n is even.
First we will show if n is even, then 3n2+8 is
even.
n even means there exists integer a such that n
= 2a. Then 3n2+8 = 3(2a)2 + 8 = 12a2 + 8 =
2(6a2 + 4) which is even since (6a2 + 4) is
an integer.
If n is a positive integer, then n is
even iff 3n2+8 is even (cont.).
Now we will show if 3n2+8 is even, then n is even using
the contrapositive (indirect proof).
Assume that n is odd, then we will show that 3n2+8 is
odd. n odd means that there exists integer a such that n
= 2a+1.
3n2+8 = 3(2a+1)2 + 8 = 3(4a2 + 4a + 1) + 8 = 2(6a2 + 6a +
5) + 1, which is odd.
Therefore, by the contrapositive if 3n2+8 is even, then n is
even.
n2 + n + 1 is a prime number
whenever n is a positive integer.
Try some examples:
n = 1, 1+1+1 = 3 is prime
n = 2, 4+2+1 = 7 is prime
n = 3, 9+3+1 = 13 is prime
n = 4, 16+4+1 = 21 is not prime and is a
counter example.
Not true.
n2 + n + 1 is a prime number
whenever n is a prime number.
Try some examples:
n = 1, 1+1+1 = 3 is prime
n = 2, 4+2+1 = 7 is prime
n = 3, 9+3+1 = 13 is prime
n = 5, 25+5+1 = 31 is prime
n = 7, 49+7+1 = 57 is not prime (19*3).
Not true.
Prove |x| + |y|  |x + y| when x,y  R.
Note: |z| is equal to z if z0 and equal to -z if
z<0
There are four cases
x y
0 0
<0 0
0 <0
<0 <0
Prove |x| + |y|  |x + y| when x,y  R.
Case 1
x,y are both 0
Then
|x| + |y| = x + y = |x+y| since both x and y are
positive.
Prove |x| + |y|  |x + y| when x,y  R.
Case 2 x < 0 and y 0 so |x| + |y| = -x + y
If y  -x, then x+y is nonnegative and |x+y| = x+y
Since x is negative, -x > x, so that
|x| + |y| = -x + y > x+y = |x+y|
If y < -x, then |x+y| = -(x+y) = -x + -y.
Since y 0, then y  -y, so that
|x| + |y| = -x + y  -x + -y = |x+y|
Prove |x| + |y|  |x + y| when x,y  R.
Case 3 x0 and y <0 so |x| + |y| = x + -y
If x  -y, then x+y is nonnegative and |x+y| = x+y
Since y is negative, -y > y, so that
|x| + |y| = x + -y > x+y = |x+y|
If x < -y, then |x+y| = -(x+y) = -x + -y.
Since x0, then x-x , so that
|x| + |y| = x + -y  -x + -y = |x+y|
Prove |x| + |y|  |x + y| when x,y  R.
Case 4 x,y are both < 0
Then |x| + |y| = -x + - y = -(x+y) = |x+y|
Therefore the theorem is true. This is know in
mathematics as the Lipschitz condition.
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.
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.
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.
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.
Some Other Proof Strategies
Rosen 3.1
Backward Reasoning
We have used mostly forward reasoning
strategies up to now.
However, sometimes it is unclear how to
proceed from the initial assumptions.
Bacward reasoning may help-Motto: If you can’t prove the original
proposition, equate it to one you can prove.
Prove (a,b  Z+, a≠b)[(a+b)/2> ab]
(i.e., the arithmetic mean is always greater than the
geometric mean for this universe of discourse.)
Backward reasoning proof
(a+b)/2> ab
(a+b)2/4> ab
(a+b)2> 4ab
a2+2ab+b2 > 4ab
a2-2ab+b2 > 0
(a-b)2> 0
Original Assumption
Why?
Why?
Why?
Why?
Why?
Prove (a,b  Z+, a≠b)[(a+b)/2> ab]
Backward reasoning proof (cont.)
But, a,b  Z+, (a-b)2> 0 implies a≠b.
We can now easily start from a≠b and go
backwards to reconstruct the path to
prove the original proposition.
Conjecture and Proof
A conjecture is a plausible statement that has
not been proved. A conjecture may result
from recognizing that there are multiple
examples for which it is true.
For some conjectures, counterexamples are
eventually found.
However, even if a conjecture is valid for very
many examples, this does not usually
constitute a valid proof. (Why?)
Conjecture and Proof
Sometimes a complex proof is constructed as a
series of conjectures that are then proved.
Sometimes a proof is found by referring to
the proof of a similar problem or class of
problem.
There are many famous conjectures that are
still not proved (or only recently proved).
Fermat’s Last Theorem
xn + yn = zn has no solution for x,y,z,n  Z,
x,y,z≠0, n>2
This is a conjecture made by Pierre de Fermat
(1601-1665), the French mathematician. He
wrote in the margin of his copy of the works of
Diophantus, an ancient Greek mathematician,
that he had a “wondrous proof”, but that it
wouldn’t fit in the margin. He then died, leaving
no record of the proof!
Fermat’s Last Theorem
Attempts at proof over the years led to new
fields, such as algebraic number theory.
Finally, in 1994, Andrew Wiles provided a
correct proof that required hundred of pages
of advanced mathematics.