a, b c c != 0 gcd(a, c) = 1 c|abc|(ab+ bx) x " Zgcd(a, c) = 1 c|ab c|b b = ck

k ab+ bx = ack + ckx = c(ak + kx) c|(ab+ bx)

a, b, x, y ax + by = 7 gcd(a, b) = 1gcd(a, b) = 7

d = gcd(a, b) d|a d|b da b d|(ax + by) d|7 71 7 d = 1 d = 7

R1 R2 A R1 # R2

A

(x, y) " R1 # R2 (x, y) " R1 (x, y) " R2 R1

(y, x) " R1 R2 (y, x) " R2 (y, x) "R1 #R2 (x, y) R1 #R2

A = {1, 2, 3, 4, 5} A

R1 = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (1, 2)}• R1

• (1, 2) " R1 (2, 1) /" R1

••

R2 = {(1, 2), (2, 1), (2, 3), (3, 2), (3, 1), (1, 3), (4, 4), (5, 5)}• R2 (1, 1) /" R2

•• (1, 2) " R2 (2, 1) " R2 1 != 2

• (1, 2) " R2 (2, 1) " R2 (1, 1) /" R2

R N $ N (a, b)R(c, d) ad = bcR N$ N

R R(a, b)R(a, b) (a, b) " N$N ab = ba (a, b)R(a, b)

(a, b) " N$ NR (a, b)R(c, d) ad = bc cb = da

(c, d)R(a, b) R

R (a, b)R(c, d) (c, d)R(e, f) ad = bccf = de (a, b)R(e, f) af = be ad = bc

adf = bcf cf = de adf = bcf = bde daf = be (a, b)R(e, f) R

R

A = {a, b, c, d, e} P= {{a, b, c}, {d, e}} P AR A P

R

R = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c), (d, d), (d, e), (e, d), (e, e)}

A = {1, 2, 3, 4} B = {2, 4} P(A) AR P(A) XRY X, Y " P(A) X # B = Y # B R

P(A)R

R RX " P(A) X # B = X # B XRX

R XRY X # B = Y # BY # B = X #B Y RX R

R XRY Y RZ X#B = Y #B =Z # B XRZ

R X % A X # B& {2} {4} {2, 4}

[&] = {&, {1}, {3}, {1, 3}},

[{2}] = {{2}, {1, 2}, {2, 3}, {1, 2, 3}},[{4}] = {{4}, {1, 4}, {3, 4}, {1, 3, 4}},

[{2, 4}] = {{2, 4}, {1, 2, 4}, {2, 3, 4}, {1, 2, 3, 4}}.

f : N ' Q f(n) = nn+1

f

f(x) = f(y) x/(x+ 1) = y/(y + 1) x(y + 1) =y(x+ 1) xy + x = yx+ y xy x = yf

f

n/(n+1) nf(n) = (1 f

g : Z6 ' Z10 g([x]6) = [3x]10

[0]6 = [6]6 f([0]6) = [3 · 0]10 = [0]10 f([6]6) = [3 · 6]10 = [18]10 =[8]10 [0]10 != [8]10 f

f : Z12 ' Z12 f([x]) = [5x + 1]f!1 f

f f([x]) = f([y]) [5x+1] = [5y+1][5x+1] = [5][x]+[1] [5y+1] = [5][y]+[1] [5][x] = [5][y]

[5] [5][5][x] = [5][5][y] [25][x] = [25][y][1][x] = [1][y] 25 ) 1 (mod 12) [1x] = [1y] [x] = [y]

ff([x]) = f([y]) [5x + 1] = [5y + 1]

12|[(5x+1)( (5y+1)] 12|5(x( y) gcd(12, 5) = 112|(x( y) x ) y (mod 12) [x] = [y]

ff !

f!1 [y] = f([x]) = [5x + 1] = [5][x] + [1][1] Z12 [11] [y + 11] = [5][x] [5]

[5][y + 11] = [25][x] = [1][x] = [x] [x] = [5y + 55] = [5y + 7]f!1([y]) = [5y + 7]

gcd(12, 5) = 1 Z12

[5] Z12

s t 5s+ 12t = 1 5s( 1 = 12((t) 12|(5s( 1)5s ) 1 (mod 12) [5][s] = [1] Z12

s12 = 5(2) + 25 = 2(2) + 1

1 = 5 + 2((2)= 5 + [12 + 5((2)]((2)= 5(5) + 12((2).

s = 5