Mathematical Proofs
Chapter 4 More on Directed Proof and Proof by
Contrapositive
• 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
|
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
| |
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 (x 2 -1), then 3 | x.
x=3q +2 for some integer q. We consider these two cases.
Case 1. x =3q+1 for some integer q. Then x 2 -1=(3q+1) 2 -1=3(3q 2 +2q).
Since 3q 2 +2q is an integer, 3 | x 2 -1.
Case 2. x=3q+2 for some integer q. Then x 2 -1=(3q+2) 2 -1=3(3q 2 +4q+1).
Since 3q 2 +4q+1 is an integer, 3 | x 2 -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),
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 n 2
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 n 2 -n=(3k) 2 -(3k)=3(3k 2 -k).
Since 3k 2 -k is an integer, 3 | n 2 -n. Thus n 2
n (mod 3).
Case 2. n
1(mod 3). Then n=3x+1 for some integer x. Hence n 2 -n=(3x+1) 2 -(3x+1)=3(3x 2 +x).
Since 3x 2 -x is an integer, 3 | n 2 -n and so n 2
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.
• a 2 ≥0 for every real number a.
• a n ≥0 for every real number a if n is a positive even integer.
• If a<0 and n is a positive odd integer, then a n <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.
#.
Result: Let x
R. if x 5 -3x 4 +2x 3 -x 2 +4x-1 ≥0, then x ≥0.
Proof. Assume that x<0. Then x 5 <0, 2x 3 <0, and 4x<0. In addition,
-3x 4 <0, -x 2 <0. Thus x 5 -3x 4 +2x 3 -x 2 +4x-1<0-1<0, as desired.
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Examples
Result. If x, y
R, then 1/3x 2 +3/4y 2 ≥xy.
Proof. Since (2x-3y) 2 ≥0, it follows that 4x 2 -12xy+9y 2 ≥0 and so
4x 2 +9y 2 ≥12xy.
Dividing this inequality by 12, we obtain 1/3x 2 +3/4y 2 ≥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 }.
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
Proof. First we show that A-B
A
. Let x
A and x
B. Since x
B, it follows that x
. Therefore, x
A and x
; so x
A
.
Hence A-B
A
.
Next we show that A
A- B. Let y
A
. Then y
A and y
. Since y
, 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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