Transcript Mathematical Proofs

Chapter 4 More on Directed Proof and Proof by
Contrapositive
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4.1 Proofs Involving Divisibility of Integers
4.2 Proofs Involving Congruence of Integers
4.3 Proofs Involving Real Numbers
4.4 Proofs Involving sets
4.5* Fundamental Properties of Set Operations
4.6* Proofs Involving Cartesian Products of Sets
Section 4.1 Proofs Involving Divisibility of Integers
In general, for integers a and b with a≠0, we say that a divides b if there
is an integer c such that b=ac. In this case, we write a | b.
If a | b, then we also say that b is a multiple of a and that a is a divisor
of b. If a does not divide b, then we write a | b.
Result: Let a, b, and c be integers with a≠0 and b ≠0. If a|b and b|c,
then a|c.
Proof. Assume that a|b and b|c. Then b=ax and c=by, where x, yZ.
Therefore, c=by=(ax)y=a(xy). Since xy Z, a|c.
Examples
Result: Let a, b, c, x, y Z, where a≠0. If a|b and a|c, then a|(bx+cy).
Exercise.
Result: Let x, y Z, If 3 | xy, then 3 | x and 3 | y.
Proof: Assume that 3 | x or 3 | y. WLOG, assume that 3 divides x.
Then x=3z for some integer z. Hence xy=(3z)y=3(zy). Since zy is
an integer, 3 | xy.
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Examples
Let x Z. If 3 | (x2-1), then 3 | x.
Proof. Assume that 3 | x. Then either x=3q+1 for some integer q, or
x=3q +2 for some integer q. We consider these two cases.
Case 1. x =3q+1 for some integer q. Then
x2-1=(3q+1)2-1=3(3q2+2q).
Since 3q2+2q is an integer, 3 | x2-1.
Case 2. x=3q+2 for some integer q. Then
x2-1=(3q+2)2-1=3(3q2+4q+1).
Since 3q2+4q+1 is an integer, 3 | x2-1.
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Section 4.2 Proofs Involving Congruence of Integers
For integers a, b, and n≥2, we say that a is congruent to b modulo n,
written a b (mod n), if n | (a-b).
For example: 157 (mod 4) since 4 | (15-7),
but 14  4 (mod 6) since 6 | (14-4).
Note that: since every integer can be expressed as x=2q or as x=2q+1
for some integer q, it follows that either 2|(x-0) or 2|(x-1); that is,
x  0 (mod 2) or x  1 (mod 2).
Similarly, we have x  0(mod 3), x  1(mod 3), or x  2(mod 3).
Etc.
Examples
Result: Let a, b , k, and b be integers, where n≥2. If a  b (mod n), then
ka  kb (mod n).
Proof: Assume that a  b(mod n). Then n | (a-b). Hence a-b =nx for
some integer x. Therefore,
ka-kb=k(a-b)=k(nx)=n(kx).
Since kx is an integer, n | (ka-kb) and so ka  kb (mod n).
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Examples
Result: Let a, b, c, d, n Z, where n ≥2. If a  b (mod n) and c  d (mod
n), the ac  bd (mod n).
Proof: Exercise.
Examples
Let n  Z. If n2 n (mod 3), then n  0(mod 3) and n 1(mod 3).
Proof. Let n be an integer such that n  0 (mod 3) or n  1(mod 3). We
consider these two cases.
Case 1. n  0(mod 3). Then n=3k for some integer k. Hence
n2-n=(3k)2-(3k)=3(3k2-k).
Since 3k2-k is an integer, 3 | n2-n. Thus n2  n (mod 3).
Case 2. n  1(mod 3). Then n=3x+1 for some integer x. Hence
n2-n=(3x+1)2-(3x+1)=3(3x2+x).
Since 3x2-x is an integer, 3 | n2-n and so n2  n (mod 3).
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Section 4.3 Proofs Involving Real Numbers
Some facts about real numbers that can be used without justification.
• a2≥0 for every real number a.
• an≥0 for every real number a if n is a positive even integer.
• If a<0 and n is a positive odd integer, then an<0.
• If the product of two real numbers is positive if and only if both
numbers are positive or both are negative.
• If the product of two real numbers is 0, then at least one of these
numbers is 0.
• Let a, b, c R. If a ≥b and c ≥0, then ac ≥ bc.
Indeed, if c>0, then a/c ≥b/c.
• If a>b and c>0, then ac>bc and a/c>b/c.
• If a>b and c<0, then ac<bc and a/c<b/c.
Theorem
Theorem: If x and y are real numbers such that xy=0, then x=0 or y=0.
Proof. Assume that xy=0. We consider two cases, x=0 or x≠0.
Case 1. x=0. Then we have the desired result.
Case 2. x ≠0. Multiplying xy=0 by the number 1/x, we obtain
1/x(xy)=1/x(0)=0. Since 1/x(xy)= ((1/x)x)y=y, it follows that y=0.
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Result: Let x R. if x5-3x4+2x3-x2+4x-1 ≥0, then x ≥0.
Proof. Assume that x<0. Then x5<0, 2x3<0, and 4x<0. In addition,
-3x4<0, -x2<0. Thus x5-3x4+2x3-x2+4x-1<0-1<0, as desired.
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Examples
Result. If x, y R, then 1/3x2+3/4y2 ≥xy.
Proof. Since (2x-3y)2 ≥0, it follows that 4x2-12xy+9y2 ≥0 and so
4x2+9y2 ≥12xy.
Dividing this inequality by 12, we obtain 1/3x2+3/4y2 ≥xy.
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Section 4.4 Proofs Involving sets
Recall, for set A and B contained in some universal set U,
A B={x: x  A or x  B}.
A  B={x: x  A and x  B}.
A - B={x: x  A and x  B}.
A  U  A  {x U and x  A}.
To show the equality of two sets C and D, we can verify the two sets
inclusions CD and D C.
To establish the inclusion C D, we show that every element of C is
also an element of D; that is, if x  C then x D.
Examples
Result. For every two sets A and B, A-B=A  B
Proof. First we show that A-B A  B . Let x A and xB. Since x
B, it follows that x  B . Therefore, x A and x  B ; so x  A  B .
Hence A-B A  B .
Next we show that A  B A- B. Let y  A  B . Then y  A and y  B
. Since y  B , we see that y B. Now because y  A and y  B, we
conclude that y  A- B. Thus, A  B A-B.
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Examples
Result. Let A and B be sets. Then AB= A if and only if B A.
Proof. First we prove that if AB= A, then B A. We use a proof by
contrapositive. Assume that B is not a subset of A. Then there must
be some element x B such that x A. Since x B, it follows that x
A B. However, since x A, we have A B≠A.
Next we prove the converse, namely, if B  A, then A B=A. We use a
direct proof here. Assume that B  A. To verify that AB= A, we
show that A  AB and AB A. The set inclusion A  AB is
immediate. It remains only to show then that AB A. Let y A B.
Thus y A or y B. If y A, then we already have the desired result.
If y B, then since B  A, it follows that y A. Thus AB  A.
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