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Proofs
Zeph Grunschlag
Copyright © Zeph Grunschlag,
2001-2002.
Agenda
Proofs: General Techniques
Direct Proof
Indirect Proof
Proof by Contradiction
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Proof Nuts and Bolts
A proof is a logically structured argument which
demonstrates that a certain proposition is
true. When the proof is complete, the
resulting proposition becomes a theorem, or
if it is rather simple, a lemma.
For example, consider the proposition:
If k is any integer such that k  1 (mod 3),
then k 3  1 (mod 9).
Let’s prove this fact:
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Proofs
Example
PROVE: kZ k 1(mod 3)  k 31(mod 9)
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Proofs
Example
PROVE: kZ k 1(mod 3)  k 31(mod 9)
1. k 1(mod 3)
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Proofs
Example
PROVE: kZ k 1(mod 3)  k 31(mod 9)
1. k 1(mod 3)
2. n k-1 = 3n
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Proofs
Example
PROVE: kZ k 1(mod 3)  k 31(mod 9)
1. k 1(mod 3)
2. n k-1 = 3n
3. n k = 3n + 1
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Proofs
Example
PROVE: kZ k 1(mod 3)  k 31(mod 9)
1. k 1(mod 3)
2. n k-1 = 3n
3. n k = 3n + 1
4. n k 3 = (3n + 1)3
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Proofs
Example
PROVE: kZ k 1(mod 3)  k 31(mod 9)
1. k 1(mod 3)
2. n k-1 = 3n
3. n k = 3n + 1
4. n k 3 = (3n + 1)3
5. n k 3 = 27n 3 + 27n 2 + 9n + 1
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Proofs
Example
PROVE: kZ k 1(mod 3)  k 31(mod 9)
1. k 1(mod 3)
2. n k-1 = 3n
3. n k = 3n + 1
4. n k 3 = (3n + 1)3
5. n k 3 = 27n 3 + 27n 2 + 9n + 1
6. n k 3-1 = 27n 3 + 27n 2 + 9n
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Proofs
Example
PROVE: kZ k 1(mod 3)  k 31(mod 9)
1. k 1(mod 3)
2. n k-1 = 3n
3. n k = 3n + 1
4. n k 3 = (3n + 1)3
5. n k 3 = 27n 3 + 27n 2 + 9n + 1
6. n k 3-1 = 27n 3 + 27n 2 + 9n
7. n k 3-1 = (3n 3 + 3n 2 + n)·9
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Proofs
Example
PROVE: kZ k 1(mod 3)  k 31(mod 9)
1. k 1(mod 3)
2. n k-1 = 3n
3. n k = 3n + 1
4. n k 3 = (3n + 1)3
5. n k 3 = 27n 3 + 27n 2 + 9n + 1
6. n k 3-1 = 27n 3 + 27n 2 + 9n
7. n k 3-1 = (3n 3 + 3n 2 + n)·9
8. m k 3-1 = m·9
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Proofs
Example
PROVE: kZ k 1(mod 3)  k 31(mod 9)
1. k 1(mod 3)
2. n k-1 = 3n
3. n k = 3n + 1
4. n k 3 = (3n + 1)3
5. n k 3 = 27n 3 + 27n 2 + 9n + 1
6. n k 3-1 = 27n 3 + 27n 2 + 9n
7. n k 3-1 = (3n 3 + 3n 2 + n)·9
8. m k 3-1 = m·9
9. k 31(mod 9)
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Direct Proofs
Previous was an example of a direct
proof. I.e., to prove that a proposition
of the form “k P(k)  Q(k)” is true,
needed to derive that “Q(k) is true” for
any k which satisfied “P(k) is true”.
Three basic steps in a direct proof:
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Direct Proofs
1) Deconstruct Axioms
Take the hypothesis and turn it into a usable
form. Usually this amounts to just
applying the definition.
EG: k 1(mod 3) really means 3|(k-1) which
actually means n k-1 = 3n
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Direct Proofs
2) Mathematical Insights
Use your human intellect and get at “real
reason” behind theorem.
EG: looking at what we’re trying to prove,
we see that we’d really like to understand
k 3. So let’s take the cube of k ! From
here, we’ll have to use some algebra to
get the formula into a form usable by the
final step:
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Direct Proofs
3) Reconstruct Conclusion
This is the reverse of step 1. At the end of
step 2 we should have a simple form that
could be derived by applying the
definition of the conclusion.
EG. k 31(mod 9) is readily gotten from
m k 3-1 = m·9 since the latter is the
definition of the former.
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Proofs
Indirect
In addition to direct proofs, there are two
other standard methods for proving
k P(k)  Q(k)
1. Indirect Proof
For any k assume: Q(k)
and derive: P(k)
Uses the contrapositive logical
equivalence: P Q  Q  P
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Proofs
Reductio Ad Absurdum
2.
Proof by Contradiction (Reductio Ad
Absurdum)
For any k assume: P(k)  Q(k)
and derive: P(k)  Q(k)
Uses the logical equivalence:
  P  Q  P  Q  P  Q
 (P  Q )  (P  Q )  (P  Q )  (P  Q )
 (P  Q )  (P  Q )
P Q
Intuitively: Assume claim is false (so P must be true
and Q false). Show that assumption was absurd
(so P false or Q true) so claim true!
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Indirect Proof
Example
PROVE: The square of an even number is
even. (Ex. 1.16.b)
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Indirect Proof
Example
PROVE: The square of an even number is
even. (Ex. 1.16.b)
1. Suppose k 2 is not even.
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Indirect Proof
Example
PROVE: The square of an even number is
even. (Ex. 1.16.b)
1. Suppose k 2 is not even.
2. So k 2 is odd.
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Indirect Proof
Example
PROVE: The square of an even number is
even. (Ex. 1.16.b)
1. Suppose k 2 is not even.
2. So k 2 is odd.
3. n k 2 = 2n + 1
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Indirect Proof
Example
PROVE: The square of an even number is
even. (Ex. 1.16.b)
1. Suppose k 2 is not even.
2. So k 2 is odd.
3. n k 2 = 2n + 1
4. n k 2 - 1 = 2n
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Indirect Proof
Example
PROVE: The square of an even number is
even. (Ex. 1.16.b)
1. Suppose k 2 is not even.
2. So k 2 is odd.
3. n k 2 = 2n + 1
4. n k 2 - 1 = 2n
5. n (k - 1)(k + 1) = 2n
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Indirect Proof
Example
PROVE: The square of an even number is
even. (Ex. 1.16.b)
1. Suppose k 2 is not even.
2. So k 2 is odd.
3. n k 2 = 2n + 1
4. n k 2 - 1 = 2n
5. n (k - 1)(k + 1) = 2n
6. 2 | (k - 1)(k + 1)
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Indirect Proof
Example
PROVE: The square of an even number is even.
(Ex. 1.16.b)
1. Suppose k 2 is not even.
2. So k 2 is odd.
3. n k 2 = 2n + 1
4. n k 2 - 1 = 2n
5. n (k - 1)(k + 1) = 2n
6. 2 | (k - 1)(k + 1)
7. 2 | (k - 1)  2 | (k + 1) since 2 is prime
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Indirect Proof
Example
PROVE: The square of an even number is even.
(Ex. 1.16.b)
1. Suppose k 2 is not even.
2. So k 2 is odd.
3. n k 2 = 2n + 1
4. n k 2 - 1 = 2n
5. n (k - 1)(k + 1) = 2n
6. 2 | (k - 1)(k + 1)
7. 2 | (k - 1)  2 | (k + 1) since 2 is prime
8. a k - 1 = 2a  b k+1 = 2b
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Indirect Proof
Example
PROVE: The square of an even number is even.
(Ex. 1.16.b)
1. Suppose k 2 is not even.
2. So k 2 is odd.
3. n k 2 = 2n + 1
4. n k 2 - 1 = 2n
5. n (k - 1)(k + 1) = 2n
6. 2 | (k - 1)(k + 1)
7. 2 | (k - 1)  2 | (k + 1) since 2 is prime
8. a k - 1 = 2a  b k+1 = 2b
9. a k = 2a + 1  b k = 2b – 1
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Indirect Proof
Example
PROVE: The square of an even number is even.
(Ex. 1.16.b)
1. Suppose k 2 is not even.
2. So k 2 is odd.
3. n k 2 = 2n + 1
4. n k 2 - 1 = 2n
5. n (k - 1)(k + 1) = 2n
6. 2 | (k - 1)(k + 1)
7. 2 | (k - 1)  2 | (k + 1) since 2 is prime
8. a k - 1 = 2a  b k+1 = 2b
9. a k = 2a + 1  b k = 2b – 1
10. In both cases k is odd
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Indirect Proof
Example
PROVE: The square of an even number is even. (Ex.
1.16.b)
1. Suppose k 2 is not even.
2. So k 2 is odd.
3. n k 2 = 2n + 1
4. n k 2 - 1 = 2n
5. n (k - 1)(k + 1) = 2n
6. 2 | (k - 1)(k + 1)
7. 2 | (k - 1)  2 | (k + 1) since 2 is prime
8. a k - 1 = 2a  b k+1 = 2b
9. a k = 2a + 1  b k = 2b – 1
10. In both cases k is odd
11. So k is not even
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Rational Numbers
an Easier Characterization
Recall the set of rational numbers Q =
the set of numbers with decimal
expansion which is periodic past some
point (I.e. repeatinginginginginging…)
Easier characterization
Q = { p/q | p,q are integers with q  0 }
Prove that the sum of any irrational
number with a rational number is
irrational:
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Reductio Ad Absurdum
Example –English!
You don’t have to use a sequence of formulas.
Usually an English proof is preferable! EG:
Suppose that claim is false. So [x is rational and
y irrational] [=P] and [x+y is rational] [=
Q]. But y = (x+y ) - x. The difference of
rational number is rational since
a/b – c/d = (ad-bc)/bd.
Therefore [y must be rational] [implies P ].
This contradicts the hypotheses so the
assumption that the claim was false was
incorrect and the claim must be true. •
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Proofs
Disrefutation
Disproving claims is often much easier
than proving them.
Claims are usually of the form k P(k).
Thus to disprove, enough to find one k
–called a counterexample– which
makes P(k) false.
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Disrefutation by
Counterexample
Disprove: The product of irrational
numbers is irrational.
1
1. Let x  2 and y 
. Both are irrational.
2
1
2. Their product xy  2 
 1 is rational.
2
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Blackboard Exercise for 3.1
Proof that the square root of 2 is
irrational
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