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Basic Definitions
With examples from Number Theory
Rosen (6th Ed.) 1.6, 1.7, 3.4, 3.5
Theorem - A statement that can be shown to be
true.
Proof - A series of statements that form a valid
argument.
•  Each statement in the series must be:
–  Basic fact or definition
–  Hypothesis of the theorem you are proving
–  Previously proved theorem (lemma or corollary)
•  Must end with what you are trying to prove.
∍ = “such that”
Basic Number Theory Definitions
• 
• 
• 
• 
• 
2/3, 100, √2,
Z = Set of all Integers 1.4,
-5.6, 1,000,000,
Z+ = Set of all Positiveetc.Integers
N = Set of Natural Numbers (Z+ and Zero)
R = Set of Real Numbers
Addition and multiplication on integers
produce integers. (a,b ε Z) → [(a+b) ε Z] ∧
[(ab) ε Z]
Methods of Proof
p→ q (Example: if n is even, then n2 is even)
•  Direct proof: Assume p is true and use a series of
previously proven statements to show that q is
true.
•  Indirect proof: Show ¬q →¬p is true, using any
proof technique (usually direct proof). Your book
calls this proof by contraposition.
•  Proof by contradiction: Assume negation of what
you are trying to prove (p∧¬q). Show that this
leads to a contradiction to your original
assumptions.
¬(p→q) ⇔ ¬(¬p∨q) ⇔ ¬¬p∧¬q ⇔ p∧¬q
Number Theory Defs (cont.)
•  n is even is defined as ∃k ε Z ∍ n = 2k
•  n is odd is defined as ∃k ε Z ∍ n = 2k+1
•  x is rational is defined as ∃a,b ε Z ∍ x = a/b,
b≠0
•  x is irrational is defined as ¬∃a,b ε Z ∍ x =
a/b, b≠0 or ∀a,b ε Z, x ≠ a/b, b≠0
•  p ε Z+ is prime means that the only positive
factors of p are p and 1. If p is not prime we
say it is composite.
Direct Proof
Prove: ∀n∈Z, if n is even, then n2 is even.
Tabular-style proof:
n is even
hypothesis
n=2k for some k∈Z definition of even
n2 = 4k2
algebra
n2 = 2(2k2) which is algebra and mult of
2*(an integer)
integers gives integers
n2 is even
definition of even
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Same Direct Proof
Prove: ∀n∈Z, if n is even, then n2 is even.
Sentence-style proof:
Assume that n is even. Thus, we know that n
= 2k for some integer k. It follows that n2
= 4k2 = 2(2k2). Therefore n2 is even since it
is 2 times 2k2 which is an integer.
Structure of a Direct Proof
Prove: ∀n∈Z, if n is even, then n2 is even.
Proof:
Assume that n is even. Thus, we know that n
= 2k for some integer k. It follows that n2 =
4k2 = 2(2k2). Therefore n2 is even since it is
2 times 2k2 which is an integer.
Another Direct Proof
Prove: The sum of two rational numbers is
a rational number.
Proof: Let s and t be rational numbers. Then
s = a/b and t = c/d where a,b,c,d ∈Z, b,d ≠0.
Then s+t = a/b + c/d = (ad+cb)/bd . But
since (ad+cb) ∈Z and bd ∈Z ≠0 (why?), then
(ad+cb)/bd is rational.
Example of an Indirect Proof
Prove: If n3 is even, then n is even.
In an indirect proof we want to prove that the
contraposition is true. What is the contrapositive of “If
n3 is even, then n is even” ?
“If n is not even, the n3 is not even”
What is “not even”?
Odd!
“If n is odd, then n3 is odd.”
Structure of this Direct Proof
Prove: The sum of two rational numbers is a
rational number.
Proof: Let s and t be rational numbers.
Assumed
Then s = a/b and t = c/d where a,b,c,d ∈Z , b,d ≠0. Def
Then s+t = a/b + c/d = (ad+cb)/bd .
Basic facts of
arithmetic
But since (ad+cb) ∈Z and bd ∈Z ≠0, then (ad+cb)/
bd is rational.
Conclusion from Def
Example of an Indirect Proof
Prove: If n3 is even, then n is even.
Proof: The contrapositive of “If n3 is even, then n is
even” is “If n is odd, then n3 is odd.” If the
contrapositive is true then the original statement must
be true.
Assume n is odd. Then ∃k∈Z ∍ n = 2k+1. It follows that
n3 = (2k+1)3 = 8k3+8k2+4k+1 = 2(4k3+4k2+2k)+1.
(4k3+4k2+2k) is an integer. Therefore n3 is 1 plus an
even integer. Therefore n3 is odd.
Assumption, Definition, Arithmetic, Definitions, Conclusion
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Discussion of Indirect Proof
Proof by Contradiction
Could we do a direct proof of If n3 is even,
then n is even?
Assume n3 is even . . . then what?
We don’t have a rule about how to take n3 apart!
Proof by Contradiction
1.  Want to prove that p→q.
2.  Assume ¬(p→q) ⇔ p∧¬q is true
3.  Show that if p is true, p∧¬q cannot be true
(the contradiction).
4.  The Punch line! Since ¬(p→q) is false,
then (p→q) has to be true.
Example: Proof by Contradiction
Prove: The sum of an irrational number and a
rational number is irrational.
Proof: Let t be an irrational number and r be a
rational number. Assume that their sum is
rational, i.e., t+r=s where s is a rational number.
Then t = s-r. But by our previous proof the sum of
two rational numbers must be rational, so we have
an irrational number on the left equal to a rational
number on the right. This is a contradiction.
Therefore t+r can’t be rational and must be
irrational.
Example: Proof by Contradiction
Prove:
The sum of an irrational number and a rational number is
irrational.
(p→q)
Proof: Let t be an irrational number and r be a rational number.
Assume that their sum is rational, i.e., t+r=s where s is a rational
number. (Assume p∧¬q, find a contradiction that this can’t
HW 3 is due by 9:00am on
Friday, September 6
be true)
Then t = s-r. (Did some arithmetic)
But by our previous proof the sum of two rational numbers (i.e., s-t)
must be rational,
so we have an irrational number on the left equal to a rational number
on the right.
This is a contradiction. (p∧¬q is not true, i.e ¬(p→q) is false)
Therefore t+r can’t be rational and must be irrational.
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Structure of Proof by Contradiction
•  Basic idea is to assume that the opposite of
what you are trying to prove is true and
show that it results in a violation of one of
your initial assumptions.
•  In the previous proof we showed that
assuming that the sum of a rational number
and an irrational number is rational resulted
in the impossible conclusion that a number
could be rational and irrational at the same
time.
Proof by contradiction
Want to prove (p → q) is true when p is true.
Try to prove ¬(p→q) is true when p is true
¬(p→q) ⇔ (p∧¬q)
Show that this leads to a contradiction in
something you are assuming is true.
2nd Proof by Contradiction
Prove: If 3n+2 is odd, then n is odd.
Proof: Assume 3n+2 is odd and n is even.
Since n is even, then n=2k for some integer k.
It follows that 3n+2 = 6k+2 = 2(3k+1).
Thus, 3n+2 is even. This contradicts the
assumption that 3n+2 is odd. Therefore n must
be odd.
p∧¬q, definition, arithmetic, definition, contradiction, conclusion
Cases (cont.)
Case 2: n is odd. Then ∃k∈Z ∍ n = 2k+1.
n3 + n = (8k3 +12k2 + 6k + 1) + (2k + 1) =
2(4k3 + 6k2 + 4k + 1) which is even since
4k3 + 6k2 + 4k + 1 must be an integer.
Therefore ∀n ∈Z, n3 + n is even
Using Cases
Prove: ∀n ∈Z, n3 + n is even.
Separate into cases based on whether n is even or
odd. Prove each separately using direct proof.
Proof: We can divide this problem into two
cases. n can be even or n can be odd.
Case 1: n is even. Then ∃k∈Z ∍ n = 2k.
n3+n = 8k3 + 2k = 2(4k3+k) which is even since
4k3+k must be an integer.
Even/Odd is a Special Case of Divisibility
We say that x is divisible by y if ∃ kεZ ∍ x=yk
•  n is divisible by 2 if ∃ kεZ ∍ n = 2k (even)
•  The other case is n = 2k+1(odd,remainder of
1)
•  n is divisible by 3 if ∃ kεZ ∍ n = 3k
•  n = 3k + 1
•  n = 3k + 2
•  n is divisible by 4 if ∃ kεZ ∍ n = 4k
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Lemmas and Corollaries
•  A lemma is a simple
theorem used in the
proof of other
theorems.
•  A corollary is a
proposition that can be
established directly
from a theorem that
has already been
proved.
Divisibility Example
Prove: n2 - 2 is never divisible by 3 if n is an
integer.
Discussion: What does it mean for a number to be
divisible by 3? If a is divisible by 3 then ∃ bεZ ∍ a =
3b. Remainder when n is divided by 3 is 0. Other
options are a remainder of 1 and 2.
So we need to show that the remainder when n2 - 2 is
divided by 3 is always 1 or 2 but never 0.
n2-2 is never divisible by 3 if nεZ
Proof:
Case 1: n = 3k for k∈Z then
n2-2 = 9k2 - 2 = 3(3k2) - 2 =
3(3k2 - 1) + 1
The remainder when dividing by 3 is 1.
Remainder Lemma
Lemma: Let a=3k+1 where k is an integer.
Then the remainder when a2 is divided by 3
is 1.
Proof: Assume a =3k+1. Then
a2 = 9k2 + 6k + 1 = 3(3k2+2k) + 1.
Since 3(3k2+2k) is divisible by 3, the
remainder must be 1.
Divisibility Example (cont.)
Prove: n2 - 2 is never divisible by 3 if n is an
integer.
Let’s use cases!
There are three possible cases:
•  Case 1: n = 3k
•  Case 2: n = (3k+1)
•  Case 3: n= (3k+2); k∈Z
n2-2 is never divisible by 3 if nεZ
Case 2: n = 3k+1 for k∈Z
n2-2 = (3k+1)2 - 2 = 9k2 + 6k +1-2 =
3(3k2 + 2k) - 1 = 3(3k2 + 2k -1) + 2
Thus the remainder when dividing by 3 is 2.
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n2-2 is never divisible by 3 if nεZ
Case 3: n = 3k+2 for k∈Z
n2-2 = (3k+2)2 - 2 = 9k2 + 12k +4 -2 =
3(3k2 + 4k) + 2
Thus the remainder when dividing by 3 is 2.
In each case the remainder when dividing n2-2
by 3 is nonzero. This proves the theorem.
More Complex Proof
Prove: √2 is irrational.
Direct proof is difficult.
Must show that there are no a,b ∈Z such that
a/b = √2.
Try proof by contradiction.
More Complex Proof (cont.)
More Complex Proof (cont.)
Proof by Contradiction of √2 is irrational:
Assume √2 is rational, i.e., √2 = a/b for some
a,b ∈Z, b≠0.
Since any fraction can be reduced until there
are no common factors in the numerator and
denomination, we can further assume that:
√2 = a/b for some a,b ∈Z, b≠0 and a and b
have no common factors.
(√2)2 = (a/b)2 = a2/b2 = 2.
Now what do we want to do? Let’s show that
a2/b2 = 2 implies that both a and b are even!
Since a and b have no common factors, this is
a contradiction since both a and b even
implies that 2 is a common factor.
Clearly a2 is even (why?). Does that mean a
is even?
More Complex Proof (cont.)
More Complex Proof (cont.)
Lemma 1: If a2 is even, then a is even.
Proof (indirect): If a is odd, then a2 is odd.
Assume a is odd. Then ∃k∈Z ∍ n = 2k+1.
a2 = (2k+1)2 = 4k2 + 4k + 1= 2(2k2+2k) + 1.
Therefore a is odd. So the Lemma must be
true.
Back to the example!
So far we have shown that a2 is even. Then
by Lemma 1, a is even. Thus ∃k∈Z ∍ a =
2k.
Now, we will show that b is even.
From before, a2/b2 = 2 ⇔ 2b2 = a2 = (2k)2.
Dividing by 2 gives b2 = 2k2. Therefore b2 is
even and from Lemma 1, b is even.
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More Complex Proof (cont.)
But, if a is even and b is even then they have a
common factor of 2. This contradicts our
assumption that our a/b has been reduced to
have no common factors.
Therefore √2 ≠ a/b for some a,b ∈Z, b≠0.
Therefore √2 is irrational.
Fallacies
•  Fallacy of affirming the conclusion
•  [(p → q) ∧ q] → p
•  Fallacy of denying the hypothesis
•  [(p → q) ∧ ¬p] → ¬q
•  Fallacy of circular reasoning
•  One or more steps in the proof are based on the
truth of the statement being proved.
Proof?
n3
Prove if is even then n is even.
Proof: Assume n3 is even.
Then ∃k∈Z ∍ n3 = 8k3 for some integer k. It
follows that n = 3√8k3 = 2k. Therefore n is
even.
Statement is true but argument is false.
Argument assumes that n is even in making
the claim n3=8k3, rather than n3 = 2k.
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