more number theory proofs rosen 1.5, 3.1. prove or disprove if m and n are even integers, then mn is...
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![Page 1: More Number Theory Proofs Rosen 1.5, 3.1. Prove or Disprove If m and n are even integers, then mn is divisible by 4. The sum of two odd integers is odd](https://reader033.vdocument.in/reader033/viewer/2022042717/56649d255503460f949fc682/html5/thumbnails/1.jpg)
More Number Theory Proofs
Rosen 1.5, 3.1
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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 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.
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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.
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The sum of two odd integers is odd.
This is false. A counter example is 1+3 = 4
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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.
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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.
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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.
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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.
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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.
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Prove |x| + |y| |x + y| when x,y R.
Note: |z| is equal to z if z0 and equal to -z if z < 0
There are four cases
x y
0 0
<0 0
0 <0
<0 <0
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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.
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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|
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Prove |x| + |y| |x + y| when x,y R.
Case 3 x0 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 x0, then x-x , so that
|x| + |y| = x + -y -x + -y = |x+y|
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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.
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Prove that 33 is irrational.
Proof (by contradiction): Assume that 33 is rational, i.e. that 33 = a/b for a,bZ and b0. 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.
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Prove that 33 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 kZ a = 3k.
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Prove that 33 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.
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Prove that 33 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 33 a/b for some a,bZ, b0.
Therefore 33 is irrational.
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Some Other Proof Strategies
Rosen 3.1
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Backward Reasoning
We have used mostly forward reasoning strategies up to now.
However, sometimes it is unclear how to proceed from the initial assumptions.
Backward reasoning may help--
Motto: If you can’t prove the original proposition, equate it to one you can prove.
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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 Original Assumption
(a+b)2/4> ab Why?
(a+b)2> 4ab Why?
a2+2ab+b2 > 4ab Why?
a2-2ab+b2 > 0 Why?
(a-b)2> 0 Why?
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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.
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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?)
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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).
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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!
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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 hundreds of pages of advanced mathematics.