1. Copyright © Zeph Grunschlag, 2001-2002.
Proofs
Zeph Grunschlag
Copyright © Zeph Grunschlag,

2. Agenda Proofs: General Techniques Direct Proof Indirect Proof
Proof by Contradiction
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3. If k is any integer such that k  1 (mod 3),
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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4. Proofs Example
PROVE: kZ k 1(mod 3)  k 31(mod 9)
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5. Proofs Example
PROVE: kZ k 1(mod 3)  k 31(mod 9)
k 1(mod 3)
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6. Proofs Example PROVE: kZ k 1(mod 3)  k 31(mod 9) k 1(mod 3)
n k-1 = 3n
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7. Proofs Example PROVE: kZ k 1(mod 3)  k 31(mod 9) k 1(mod 3)
n k-1 = 3n
n k = 3n + 1
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8. Proofs Example PROVE: kZ k 1(mod 3)  k 31(mod 9) k 1(mod 3)
n k-1 = 3n
n k = 3n + 1
n k 3 = (3n + 1)3
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9. Proofs Example PROVE: kZ k 1(mod 3)  k 31(mod 9) k 1(mod 3)
n k-1 = 3n
n k = 3n + 1
n k 3 = (3n + 1)3
n k 3 = 27n n 2 + 9n + 1
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10. Proofs Example PROVE: kZ k 1(mod 3)  k 31(mod 9) k 1(mod 3)
n k-1 = 3n
n k = 3n + 1
n k 3 = (3n + 1)3
n k 3 = 27n n 2 + 9n + 1
n k 3-1 = 27n n 2 + 9n
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11. Proofs Example PROVE: kZ k 1(mod 3)  k 31(mod 9) k 1(mod 3)
n k-1 = 3n
n k = 3n + 1
n k 3 = (3n + 1)3
n k 3 = 27n n 2 + 9n + 1
n k 3-1 = 27n n 2 + 9n
n k 3-1 = (3n 3 + 3n 2 + n)·9
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12. Proofs Example PROVE: kZ k 1(mod 3)  k 31(mod 9) k 1(mod 3)
n k-1 = 3n
n k = 3n + 1
n k 3 = (3n + 1)3
n k 3 = 27n n 2 + 9n + 1
n k 3-1 = 27n n 2 + 9n
n k 3-1 = (3n 3 + 3n 2 + n)·9
m k 3-1 = m·9
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13. Proofs Example PROVE: kZ k 1(mod 3)  k 31(mod 9) k 1(mod 3)
n k-1 = 3n
n k = 3n + 1
n k 3 = (3n + 1)3
n k 3 = 27n n 2 + 9n + 1
n k 3-1 = 27n n 2 + 9n
n k 3-1 = (3n 3 + 3n 2 + n)·9
m k 3-1 = m·9
k 31(mod 9)
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14. 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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15. Direct Proofs 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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16. Direct Proofs 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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17. Direct Proofs 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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18. Proofs Indirect
In addition to direct proofs, there are two other standard methods for proving
k P(k)  Q(k)
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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19. Proofs Reductio Ad Absurdum
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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20. Indirect Proof Example
PROVE: The square of an even number is even. (Ex b)
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21. Indirect Proof Example
PROVE: The square of an even number is even. (Ex b)
Suppose k 2 is not even.
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22. Indirect Proof Example
PROVE: The square of an even number is even. (Ex b)
Suppose k 2 is not even.
So k 2 is odd.
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23. Indirect Proof Example
PROVE: The square of an even number is even. (Ex b)
Suppose k 2 is not even.
So k 2 is odd.
n k 2 = 2n + 1
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24. Indirect Proof Example
PROVE: The square of an even number is even. (Ex b)
Suppose k 2 is not even.
So k 2 is odd.
n k 2 = 2n + 1
n k = 2n
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25. Indirect Proof Example
PROVE: The square of an even number is even. (Ex b)
Suppose k 2 is not even.
So k 2 is odd.
n k 2 = 2n + 1
n k = 2n
n (k - 1)(k + 1) = 2n
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26. Indirect Proof Example
PROVE: The square of an even number is even. (Ex b)
Suppose k 2 is not even.
So k 2 is odd.
n k 2 = 2n + 1
n k = 2n
n (k - 1)(k + 1) = 2n
2 | (k - 1)(k + 1)
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27. Indirect Proof Example
PROVE: The square of an even number is even. (Ex b)
Suppose k 2 is not even.
So k 2 is odd.
n k 2 = 2n + 1
n k = 2n
n (k - 1)(k + 1) = 2n
2 | (k - 1)(k + 1)
2 | (k - 1)  2 | (k + 1) since 2 is prime
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28. Indirect Proof Example
PROVE: The square of an even number is even. (Ex b)
Suppose k 2 is not even.
So k 2 is odd.
n k 2 = 2n + 1
n k = 2n
n (k - 1)(k + 1) = 2n
2 | (k - 1)(k + 1)
2 | (k - 1)  2 | (k + 1) since 2 is prime
a k - 1 = 2a  b k+1 = 2b
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29. Indirect Proof Example
PROVE: The square of an even number is even. (Ex b)
Suppose k 2 is not even.
So k 2 is odd.
n k 2 = 2n + 1
n k = 2n
n (k - 1)(k + 1) = 2n
2 | (k - 1)(k + 1)
2 | (k - 1)  2 | (k + 1) since 2 is prime
a k - 1 = 2a  b k+1 = 2b
a k = 2a + 1  b k = 2b – 1
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30. Indirect Proof Example
PROVE: The square of an even number is even. (Ex b)
Suppose k 2 is not even.
So k 2 is odd.
n k 2 = 2n + 1
n k = 2n
n (k - 1)(k + 1) = 2n
2 | (k - 1)(k + 1)
2 | (k - 1)  2 | (k + 1) since 2 is prime
a k - 1 = 2a  b k+1 = 2b
a k = 2a + 1  b k = 2b – 1
In both cases k is odd
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31. Indirect Proof Example
PROVE: The square of an even number is even. (Ex b)
Suppose k 2 is not even.
So k 2 is odd.
n k 2 = 2n + 1
n k = 2n
n (k - 1)(k + 1) = 2n
2 | (k - 1)(k + 1)
2 | (k - 1)  2 | (k + 1) since 2 is prime
a k - 1 = 2a  b k+1 = 2b
a k = 2a + 1  b k = 2b – 1
In both cases k is odd
So k is not even
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32. 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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33. 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. 
You don’t need the bracketed symbols:
[=P]
[= Q]
[ P  P  Q ]
In the actual proof. These are added to point out where the symbolic version of Reductio Ad Absurdum is really being used.
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34. 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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35. Disrefutation by Counterexample
Disprove: The product of irrational numbers is irrational.
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36. Blackboard Exercise for 3.1
Proof that the square root of 2 is irrational
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