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Previous Lecture Summary
• Floor and Ceiling Functions
• Definition of Proof
• Methods of Proof
• Direct Proof
• Disproving by Counterexample.
• Indirect Proof: Proof by
Contradiction
Today's Lecture
• Mod Functions
• Divisibility and Floor
• Mod Congruence
• Indirect Proofs
• Proof by Contra-positive
• Relation between Contradiction and Contra-positive
methods of Proof
Compute following1. 113 mod 242. -29 mod 7
1. 113 mod 24: 113 div 24
2. -29 mod 7: -29 div 7
Mod FunctionsMod Functions
411324
17
96
5297
35
6
Let a, b be integers and n be a positive integer. We say
that a is congruent to b modulo n (i.e. a b(mod n) )
iff n | (b-a), implies that there exist some integer k such
that b-a = n·k.
Note: a mod n = b mod n
Which of the following are true?1. 3 3 (mod 17)2. 3 -3 (mod 17)3. 172 177 (mod 5)4. -13 13 (mod 26)
Mod Congruence's
Cont…
1. 3 3 (mod 17) True: any number is congruent to
itself (3-3 = 0, divisible by all)
2. 3 -3 (mod 17) False: (-3-3) = 6 isn’t divisible by
17.
3. 172 177 (mod 5) True: 177-172 = 5 is a multiple
of 5
4. -13 13 (mod 26) True: 13-(-13) = 26 divisible by
26.
Congruence's IdentitiesCongruence's Identities
Let n > 1 be fixed and a, b, c, d be arbitrary integers. Then the following properties holds:
a) (Reflexive Property ) a a (mod n). b) (Symmetric Property) If a b(mod n) then b a(mod
n). c) ( Transitive Property) If a b(mod n) and b c
(mod n) then a c(mod n).d) If a b(mod n) and c d (mod n) then a + c
(b + d ) (mod n) and a·c b·d(mod n).e) If a b(mod n) then a + c b+c(mod n) and a·c
b·c(mod n).f) If a b(mod n) then a k b k (mod n) for any
positive integer k.
Theorem
If k is any integer such that k 1 (mod 3), then k3 1 (mod 9).
Proof: k Z, k 1(mod 3) k 31(mod 9)
k 1(mod 3)
n, k-1 = 3n n, k = 3n + 1
n, k 3 = (3n + 1)3
n, k 3 = 27n 3 + 27n 2 + 9n + 1
n, k 3-1 = 27n 3 + 27n 2 + 9n
n, k 3-1 = (3n 3 + 3n 2 + n)·9
m, k 3-1 = m·9 where m = 3n 3 + 3n 2 + n
k 31(mod 9)
1. Express the statement to be proved in the form
∀x in D, if P(x) then Q(x).
2. Rewrite this statement in the contra positive form
∀x in D, if Q(x) is false then P(x) is false.
3. Prove the contra-positive by a direct proof.
a. Suppose x is a (particular but arbitrarily chosen)
element of D such that Q(x) is false (or ¬Q(x) is true).
b. Show that P(x) is false (or ¬P(x) is true).
Indirect Proofs
Method of Proof by Contra-Positive
Proof by Contra-positiveProof by Contra-positive
Proposition: For all integers n, if n2 is even then n is even.
Contra positive: For all integers n, if n is not even then n2 is not even.
Proof: Suppose n is any odd integer. [We must show that n2 is odd.] By definition of odd, n = 2k + 1 for some integer k. By substitution and algebra,
n2 = (2k+1)2 = 4 k2 + 4k + 1 = 2(2 k2 + 2k) + 1.
But 2k2 + 2k is an integer because products and sums of integers are integers. So n2 = 2·(an integer) + 1, and thus, by definition of odd, n2 is odd.
Relation ship between Contra-positive and Relation ship between Contra-positive and Contradiction Proofs Contradiction Proofs
In a proof by contraposition, the statement∀x in D, if P(x) then Q(x)
is proved by giving a direct proof of the equivalent statement
∀x in D, if Q(x) then P(x).∼ ∼To do this, you suppose you are given an arbitrary element x of D such that Q(x). You then show that ∼
P(x). This is illustrated in Figure∼
To rewrite the proof as a proof by contradiction, you suppose
there is an x in D such that P(x) and ¬Q(x). You then follow
the steps of the proof by contraposition to deduce the
statement ¬P(x). But ¬P(x) is a contradiction to the
supposition that P(x) and ¬Q(x). (Because to contradict a
conjunction of two statements, it is only necessary to
contradict one of them.) This process is illustrated in Figure
Cont…. Cont….
Proof by Contradiction
Proposition: For all integers n, if n2 is even then n is even.
Proof: Suppose n is not even integer. Then n is odd integer. By definition of odd, n = 2k + 1 for some integer k. By substitution and algebra,
n2 = (2k+1)2 = 4 k2 + 4k + 1 = 2(2 k2 + 2k) + 1.
But 2k2 + 2k is an integer because products and sums of integers are integers. So n2 = 2·(an integer) + 1, and thus, by definition of odd, n2 is odd. But n2 is even in hypothesis. Which is a contradiction because any integer cannot be both even and odd. Thus our supposition was wrong. Hence n is even.
When to use which method…???
In the absence of obvious clues suggesting indirect argument, Try first to prove a statement directly. Then, if that does not succeed, look for a counterexample. If the search for a counterexample is unsuccessful, look for a proof by contradiction or contraposition.
Cont…
m = 2k for some integer k.
m2 = (2k)2 = 4k2 = 2n2.
n2=2k2
Consequently, n2 is even, and so n is even. But we also know that m is even. Hence both m and n have a common factor of 2. But this contradicts the supposition that m and n have no common factors. [Hence the supposition is false and so the theorem is true.]