123
(f) 43 mod 9 = 7
43 29 (mod 9).
29 mod 9 = 2
18. The statement is false. Let maxcfw_a, b, c = c. Then 0 < |a b| < c and c
a b (mod c).
(a b). So
Exercises for Section 7.5. Introduction to Cryptography
1. DTZ
FWJ
HTWWJHY
XNW.
2. SOS RADAR!
3. The likel
120
21. Proof. Let p 5 be a prime. Dividing p by 6, we obtain p = 6k + r for some integers
k and r, where 0 r 5. If r = 0, then 6 | p, which is impossible. If r = 2, then
p = 6k + 2 = 2(3k + 1). Since 3k + 1 is an integer, 2 | p. Since p 5, we have a cont
115
11. Proof. If n = 1, then n3 + 1 = 2 is a prime. Assume, to the contrary, that there exists a prime
p = n3 + 1 for some integer n 2. Then p = (n + 1)(n2 n + 1). Therefore, either n + 1 = 1
or n2 n + 1 = 1. If n + 1 = 1, then n = 0, which is impossible
76
(b) The distinct equivalence classes resulting from R are S1 , S2 , , Sk .
16. Observe that
(1) tells us that there are three equivalence classes.
(2) tells us that no equivalence class has exactly three elements.
(3) tells us that 3 and 4 are in diere
73
(d) The relation R4 is symmetric but is neither reexive nor transitive. For example, (1)
|cfw_1, 2cfw_1, 2| = 2 and (2) |cfw_1, 2cfw_1| = 1 and |cfw_1cfw_1, 2| = 1 but |cfw_1, 2cfw_1, 2| = 2.
Exercises for Section 5.2. Equivalence Relations
1. (a) Proo
58
By the Principle of Mathematical Induction,
Fn1 Fn+2 = Fn Fn+1 + (1)n
for every integer n 2.
Section 4.4. The Strong Principle of Mathematical Induction
1. (a) a2 = 2, a3 = 4, a4 = 8 and a5 = 16.
(b) Claim: an = 2n1 for every positive integer n.
(c) Pr
127
Hence gcd(ak+1 , ak+2 ) = 1.
By the Strong Principle of Mathematical Induction, gcd(an , an+1 ) = 1 for every positive integer
n.
26. Proof. Assume, to the contrary, that log2 3 is rational. Then log2 3 = a , where a, b N. We
b
a
a
b
may assume that g