Solutions to Quiz 1 (April 20, 2007) 1. P, Q, R (P Q) R Q (P R) P Q R (P Q) R Q (P R) X T T T T T T T T T T F T F F F T T F T F T T T T T F F F T T F
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- みさき はぎにわ
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1 Quiz 1 Due at 10:00 a.m. on April 20, 2007 Division: ID#: Name: 1. P, Q, R (P Q) R Q (P R) P Q R (P Q) R Q (P R) X T T T T T T F T T F T T T F F T F T T T F T F T F F T T F F F T (1) (7) p.44 (1)-(4) P Q ( P ) Q 3. Q (P R) ( ),, 4. T X P, Q, R T Message
2 Solutions to Quiz 1 (April 20, 2007) 1. P, Q, R (P Q) R Q (P R) P Q R (P Q) R Q (P R) X T T T T T T T T T T F T F F F T T F T F T T T T T F F F T T F T F T T F T T T T F T F F T T T T F F T F T T T T F F F F T T T T (1) (7) p.44 (1)-(4) P Q ( P ) Q. NetCommons Note 1.1 (P Q) R (7) (P Q) R (3) (Q P ) R (6) ( Q P ) R (4) Q ( P R) (7) Q ( P R) (7) Q (P R) 3. Q (P R) ( ),,. Q (P R) ( Q P ) R. P Q Q P R ( ) P (Q R) (P Q) R, P (Q R) (P Q) R. 4. T X P, Q, R T. (P Q) R Q (P R) (P Q) R Q (P R) T P P (P P ) Q R
3 Quiz 2 (Due at 10:00 a.m. on Fri. April 27, 2007) Division: ID#: Name: A, B, C Venn 1. A B A (B C) 2. A B = A (B C) B B C 3. A C A B A B = A (B C) 4. A B = A (B C) A C A B 5. A B = A (B C) A C A B Message HP
4 Solutions to Quiz 2 (April 27, 2007) 1. A B A (B C). x A B A B x A x B x B B C x B C x A x A (B C) A B A (B C) X, Y X Y x X x Y x B B C A B A (B C) X X Y Y Z X Y X Z 2. A B = A (B C) B B C. A =, B =, C = { }(= P( )). A B = A (B C) B = B C ( xp (x)) x( P (x)) 3. A C A B A B = A (B C). A B = A (B C) A B A (B C) A B A (B C) 1 x A (B C) x A x B C x B x C x B x A x A B x C x A C A C A B x A B x A B A B A (B C) A B = A (B C) P Q P Q Q P P P Q P 4. A B = A (B C) A C A B. C B C A C A (B C) A B = A (B C) A C A B x A C x A B 5. A B = A (B C) A C A B. A C A B x A C x A B x C B C x A (B C) A B = A (B C) x A B x A (B C) A C A B
5 Quiz 3 (Due: 7:00 p.m. May 7, 2007) Division: ID#: Name: a, b Z b = ca c Z a b n n b a a b [a] = {x Z x a} 1. a b Z For all a, b, c Z, (i) a a, (ii) a b b a, (iii) a b b c a c. 2. a, b, c, d Z a b c d ac bd 3. a, p, r a = pn + r [a] = {mn + r m Z} 4. x, y, z x 2 + y 2 = 3z 2 n = 3 (a) x y 0 z z (b) x = y = z = 0 x, y, z 1 Message ICU HP
6 Solutions to Quiz 3 (May 7, 2007) a, b Z b = ca c Z a b n n b a a b [a] = {x Z x a} 1. a b Z For all a, b, c Z, (i) a a, (ii) a b b a, (iii) a b b c a c.. a b (I) a 0 (0 = 0a) (II) a b (b = ca c ) a b ( b = ( c)a) (III) a b (b = b a b ) a c (c = c a c ) x, y a xb + yc (xb + yc = (xb + yc )a) (i) (I) n a a a a (ii) (II) a b n b a n a b b a (iii) a b b c n b a n c b (III) n (c b) + (b a) a c 7.6 (p.153) 2. a, b, c, d Z a b c d ac bd. a b c d n b a n d c bd ac = (bd bc) + (bc ac) = b(d c) + c(b a). 1 (III) a n, x b, y c, b d c, c b a n (bd ac) ac bd 4.10 (p.84) 3. a, p, r a = pn + r [a] = {mn + r m Z}. A = {mn + r m Z} [a] A A [a] x [a] [a] n a x a = pn+r x a r 1 (iii) (ii) x r r x n x r x = mn + r m Z x A x A m Z x = mn + r x mn + r r pn + r a 1 (iii) x a x [a] 4. x, y, z x 2 + y 2 = 3z 2 n = 3 (a) x y 0 z z n = 3 z 3 r z = 3q+r r = 0, 1, 2 z 0, 1 or 2 2 z 2 0, z x 2, y 2 x 2 + y 2 = 3z 2 0 x 2 y x 2 y 2 0 x y z 0
7 (b) x = y = z = 0 x, y, z 1. x, y, z x, y, z d x = dx, y = dy, z = dz x, y, z 1 x 2 + y 2 = 3z 2 x 2 + y 2 = 3z 2 x, y, z x 2 + y 2 = 3z 2 x, y, z (a) x y 0 9 x 2 9 y 2 9 3z 2 3 z 2 (a) 3 z 3 x, y, z x = y = z = 0 x, y, z a a a a 1, a 2,..., a n Z d Z a 1, a 2,..., a n (the greatest common divisor) d = gcd{a 1, a 2,..., a n } gcd{a 1, a 2,..., a n } = 1 a 1, a 2,..., a n (relatively prime) (i) d 0. (ii) d a 1, d a 2,..., d a n. (iii) c a 1, c a 2,..., c a n c d. a 1, a 2,..., a n Z d = gcd{a 1, a 2,..., a n } d = a 1 x 1 + a 2 x a n x n x 1, x 2,..., x n Z d d gcd{a 1, a 2,..., a n } d (ii) d (iii) d d d d (i) d = d a 1, a 2,..., a n, a n+1 Z gcd{a 1, a 2,..., a n, a n+1 } = gcd{gcd{a 1, a 2,..., a n }, a n+1 }. n = 2 d = a 1 x 1 + a 2 x a n x n a = 132, b = = ( 36)( 3) + 24, 36 = 24( 2) + 12, 24 = = = 36 + (132 + ( 36)3) 2 = ( 36)7.
8 Quiz 4 (Due at 10:00 a.m. on Friday May 11, 2007) Division: ID#: Name: 1. (Z Z, Q): Z = Z {0} 0 (a, b), (c, d) Z Z 0 ad bc ad = bc (a, b)q(c, d) (a) Q Z Z (b) (1, 2) (c) (a, b)q(a, b ) (c, d)q(c, d ) (ad + bc, bd)q(a d + b c, b d ) (d) (a), (b) 2. (Z Z, Q): (a, b), (c, d) Z Z 0 ad bc ad = bc (a, b) Q(c, d) Q Z Z Message 25
9 Solutions to Quiz 4 (May 11, 2007) 1. (Z Z, Q): Z = Z {0} 0 (a, b), (c, d) Z Z 0 ad bc ad = bc (a, b)q(c, d) (a) Q Z Z. (a, b) Z Z ab = ba Q (a, b)q(a, b) (a, b), (c, d) Z Z (a, b)q(c, d) Q ad = bc cb = da Q (c, d)q(a, b) (a, b), (c, d), (e, f) Z Z (a, b)q(c, d) (c, d)q(e, f) Q ad = bc cf = de (af be)d = adf bde = bcf bcf = 0 d 0 af = be Q (a, b)q(e, f) (b) (1, 2). (a, b) Z Z, (a, b)q(1, 2) Q 2a = b b Z (a, 2a) Z Z (1, 2)Q(a, 2a) (1, 2) [(1, 2)] { [(1, 2)] = {(a, 2a) a Z } = (a, b) Z Z a b = 1 }. 2 1/2 (c) (a, b)q(a, b ) (c, d)q(c, d ) (ad + bc, bd)q(a d + b c, b d ). b, d, b, d Z (ad + bc, bd), (a d + b c, b d ) Z Z (a, b)q(a, b ) (c, d)q(c, d ) ab = ba cd = dc (ad + bc)b d bd(a d + b c ) = adb d bcb d bda d bdb c (ad + bc, bd)q(a d + b c, b d ) = ba dd bb dc ba dd bb dc = 0. (d) (a), (b). Q (a, b)q(a, b ) (c, d)q(c, d ) ab = ba cd = dc a/b = a /b c/d = c d (ad + bc, bd)q(a d + b c, b d ) (ad + bc)/bd = (a d + b c )/b d a/b, c/d a/b a /b c/d c /d a b + c ad + bc = d bd = a d + b c b d = a b + c d 2. (Z Z, Q): (a, b), (c, d) Z Z 0 ad bc ad = bc (a, b) Q(c, d) Q Z Z. (0, 1) Q(0, 0) (0, 0) Q(1, 0) (0, 1) Q(1, 0)
10 Quiz 5 Due at 10:00 a.m. on May 18, 2007 Division: ID#: Name: 1. P, Q, R (P Q) (Q R), Q (R P ) (a) P Q R (P Q) (Q R) Q (R P ) T T T T T F T F T T F F F T T F T F F F T F F F (b) 2. A, B, C Venn X, Y X Y = X Y = {x (x X) (x Y )} A (B C) = ((A B) C) (A C)
11 Division: ID#: Name: 3. R A a, b, c A R (a), (b), (c) R (a) ara, (b) arb bra, (c) (arb brc) arc. a A [a] = {x A xra} (a), (b), (c) [a] [b] [a] [b] =.
12 Division: ID#: Name: 4. a, b Z 2a 2 + 5b 2 0 (mod 7) arb (a) R Z (b)
13 Division: ID#: Name: 5. f A B g B C h = g f : A C (a g(f(a))) A C h = g f (a) g h = g f (b) f h = g f Message : Midterm Exam [ ]
14 Solutions to Quiz 5 May 18, P, Q, R (P Q) (Q R), Q (R P ) (a) P Q R (P Q) (Q R) Q (R P ) T T T T T F F T F F T F T F F T T F T T F T T T T T T F T F T F T T T T F F F F F F T F F T F F T T T F F F T F F F T F F T T F F F T T F F T T T T T F T F F F F T T T T T T F T T F F T F T T F F F F F F T T T F F F F T T F F F T F F F T T (b). ( Q) Q P Q Q P P Q Q P (P Q) (Q R) (P Q) (Q R) ( P Q) (Q R) Q ( R P ) Q (R P ). 2. A, B, C Venn X, Y X Y = X Y = {x (x X) (x Y )} A (B C) = ((A B) C) (A C). ( ) x A x C a C A (A C) (A C) A A B A = (A C) (A C) ((A B) C) (A C) B C (A B) C A (B C) ((A B) C) (A C) ( ) A C A A C A (B C) x (A B) C x A x B x C x A x A (B C) x B x C x B C. x A (B C) A (B C) ((A B) C) (A C) A (B C) = ((A B) C) (A C) A = (A C) (A C) A ((A B) C) (A C) = ((A B) C) (A C) = (A C) (B C) (A C) = (A C) (A C) (B C) = A (B C).
15 3. R A a, b, c A R (a), (b), (c) R (a) ara, (b) arb bra, (c) (arb brc) arc. a A [a] = {x A xra} (a), (b), (c) [a] [b] [a] [b] =.. [a] [b] [a] = [b] [a] [b] [a] [b] a b [a] [b] [b] [a] [a] = [b] [a] [b] c [a] [b] [a], [b] cra crb (b) arc x [a] xra arc (c) xrc crb (c) xrb x [b] x [a] [a] [b] 4. a, b Z 2a 2 + 5b 2 0 (mod 7) arb (a) R Z. 2a 2 + 5b 2 0 (mod 7) 2b 2 7b 2 0 (mod 7) 2a 2 2b 2 (mod 7) 4 a 2 b 2 (mod 7) a 2 b 2 (mod 7) 2 2a 2 2b 2 (mod 7) 5b 2 arb a 2 b 2 (mod 7) (a) (b) (c) (b). a b (mod 7) a 2 b (mod 7), (mod 7) (mod 7) 0, 1, 4, [a] = {x Z x a (mod 7)} [0], [1] [6], [2] [5], [3] [4] 5. f A B g B C h = g f : A C (a g(f(a))) A C h = g f (a) g h = g f. A = {1}, B = C = {1, 2}, f(1) = 1, g(1) = 1, g(2) = 2 h(1) = 1 A = {1} h(a) = 2 a A g h (b) f h = g f. h f f(a) = f(a ) h(a) = g(f(a)) = g(f(a )) = h(a ) h a = a f(a) = f(a ) a = a f
16 Quiz 6 (Due at 10:00 a.m. on Fri. May. 25, 2007) Division: ID#: Name: 1. X = {1, 2, 3}, Y = {a, b, c, d} (a) X Y (b) Y X 2. f X Y g Y Z (a) g f : X Z g (b) g g f : X Z Message HP
17 Solutions to Quiz 6 (May 25, 2007) 1. X = {1, 2, 3}, Y = {a, b, c, d} (a) X Y. f : X Y f(x) 3 Y f(x) 4 f(x) f n n! ) 4 3! = 24 (b) Y X. epi(n, m) n m A = {1, 2,..., n, n + 1}, C = {1, 2,..., n}, B = {1, 2,..., m} f : A B f(n + 1) f(c) f(n + 1) f(c) f C : C B f C : C D, D = B {f(n + 1)} f(n + 1) m epi(n + 1, m) = m(epi(n, m) + epi(n, m 1)) epi(l, l) l! epi(l, 1) = 1 epi(4, 3) = 3(epi(3, 3) + epi(3, 2)) = 3(3! + 2(epi(2, 2) + epi(2, 1))) = 3(6 + 2(2 + 1)) = f X Y g Y Z (a) g f : X Z g. z Z g f x X (g f)(x) = z f(x) = y g(y) = g(f(x)) = (g f)(x) = z z Z g(y) = z y Y g (b) g g f : X Z. X = {1}, Y = Z = {1, 2}, f(1) = 1, g(1) = 1, g(2) = 2 g f(x) = {1} g f(x) = 2 x X g f
18 Quiz 7 (Due at 10:00 a.m. on Fri. June 1, 2007) Division: ID#: Name: 1. A = {x Q x 1} A S 2. a n+2 = (α + β)a n+1 α βa n a 0, a 1, a 2,... α, β (a) a, b a 0 = a + b, a 1 = aα + bβ a n = aα n + bβ n (b) a n+2 = a n+1 + 2a n (n = 0, 1,...) a 0 = 1, a 1 = 5 a n Message HP
19 Solutions to Quiz 7 (June 1, 2007) 1. A = {x Q x 1} A S. S = {1 + 1 n N} n n 1 < = n+1 Q n n S A S a a S a = 1 + 1, m N m m + 1 < m = a, m + 1 S a S S A 2. a n+2 = (α + β)a n+1 α βa n a 0, a 1, a 2,... α, β (a) a, b a 0 = a + b, a 1 = aα + bβ a n = aα n + bβ n. n = 0, 1 a n = aα n + bβ n k 1 n = 0, 1,..., k a n = aα n +bβ n n = k +1 a k 1 = aα k 1 +bβ k 1, a k = aα k +bβ k a k+1 = (α + β)a k α βa k 1 = (α + β)(aα k + bβ k ) α β(aα k 1 + bβ k 1 ) = aα k+1 + bβ k+1 + α β(aα k 1 + bβ k 1 ) α β(aα k 1 + bβ k 1 ) = aα k+1 + bβ k+1. a n = aα n + bβ n n k n a n = aα n + bβ n (b) a n+2 = a n+1 + 2a n (n = 0, 1,...) a 0 = 1, a 1 = 5 a n. α = 2, β = 1 α + β = 1, α β = 2 a n+2 = (α + β)a n+1 α βa n a + b = 1, 2a b = 5 a = 2, b = 1 a n = 2 2 n ( 1) n a n+2 = pa n+1 qa n x 2 px + q = 0 a n+2 = 2pa n+1 p 2 a n a n+2 = 2a n+1 a n a n+2 = pa n+1 qa n
20 Quiz 8 (Due at 10:00 a.m. on Fri. June 8, 2007) Division: ID#: Name: a, b a < b 3, 4, A B 2. f : ( 1, 1) R ( x x 1 x 2 )
21 3. ( 1, 1) (a, b) 4. (a, b) R 5. ( 1, 1) ( 1, 1) (a, b) R Message HP
22 Solutions to Quiz 8 (June 8, 2007) a, b a < b 3, 4, A B. A B f : A B A B ( ) 2. f : ( 1, 1) R x x 1 x 2. f(x) = f(y) x = y x/(1 x 2 ) = f(x) = f(y) = y/(1 y 2 ) 0 = x(1 y 2 ) y(1 x 2 ) = x xy 2 y + x 2 y = (x y)(1 + xy) f x < 1 y < 1 xy < xy > 0 x = y f y R f(x) = y x ( 1, 1) φ(x) = yx 2 +x y φ(1) = 1, φ( 1) = 1 φ(x) = 0 x ( 1, 1) x y = x/(1 x 2 ) = f(x) f ( 1, 1) y R f(x) = y x ( 1, 1) f(x) = x 1 x 2 = 1 ( x 1 ) 1 + x lim f(x) = +, lim x 1 0 f(x) = x 1+0 f(x) ( 1, 1) x ( 1, 1) f(x) = y 3. ( 1, 1) (a, b). g(x) = b a 2 (x + 1) + a g( 1) = a, g(1) = b a < b g(x) g ( 1, 1) g g : ( 1, 1) (a, b) (x b a 2 (x + 1) + a) ( 1, 1) (a, b) 4. (a, b) R. f g 1 : (a, b) R (x f(g 1 (x))) f g g g f (a, b) R 5. ( 1, 1) ( 1, 1) (a, b) R. h : ( 1, 1) ( 1, 1) (a, b) R ((x 1, x 2 ) (g(x 1 ), f(x 2 ))) f(x), g(x) 2, 3 h(x 1, x 2 ) = h(x 1, x 2 ) h h(x 1, x 2 ) = (g(x 1 ), f(x 2 )), h(x 1, x 2 ) = (g(x 1 ), f(x 2 ) g(x 1 ) = g(x 1 ) f(x 2 ) = f(x 2 ) g, f x 1 = x 1 x 2 = x 2 (x 1, x 2 ) = (x 1, x 2 ) h y 1 (a, b), y 2 R g, f g(x 1 ) = y 1, f(x 2 ) = y 2 x 1, x 2 ( 1, 1) h(x 1, x 2 ) = (g(x 1 ), f(x 2 )) = (y 1, y 2 ) h ( 1, 1) ( 1, 1) (a, b) R
23 Quiz 9 (Due at 10:00 a.m. on Fri. June 15, 2007) Division: ID#: Name: a, b a < b 3, 4, A B A B 2. f : ( π, π ) R (x tan x) 2 2 ( 3. [a, b] π 2, π ) 2
24 ( 4. π 2, π ) [a, b] 2 5. [a, b] = R Message ICU ICU
25 Solutions to Quiz 9 (June 15, 2007) a, b a < b 3, 4, A B A B. A B f : A B A B A = C, B = D, A B C D g : C A h : B D h f g : C D 2. f : ( π, π ) R (x tan x) 2 2. f f (x) = sec 2 x 0 f(x) = tan x lim x π/2+0 f(x) =, lim x π/2 0 f(x) = + f(x) = tan x ( 3. [a, b] π 2, π ) 2. g g : [a, b] ( π 2, π ) ( x 1 ) 2 b a (x a) g([a, b]) = [0, 1] ( π/2, π/2) g (x) = 1/(b a) > 0 g(x) [a, b] ( π/2, π/2). Quiz 8 j : [a, b] (a 1, b + 1)(x x) [a, b] (a 1, b + 1) = ( π/2, π/2) 4. ( π 2, π ) [a, b] 2. h h : ( π 2, π ) ( [a, b] x b a ( x + π ) + 2a + b ). 2 3π 2 3 h(( π/2, π/2)) = ((2a + b)/3, (a + 2b)/3) [a, b] h(x) ( π/2, π/2) [a, b] 5. [a, b] = R. X Y Y X Cantor-Bernstein X = Y 3, 4 [a, b] = ( π/2, π/2) 2 ( π/2, π/2) = R [a, b] = R
Solutions to Quiz 1 (April 17, 2008) 1. P, Q, R (P Q) R (P R) (Q R). P Q R (P Q) R (P R) (Q R) X T T T T T T T T T T T T T T T F T T F T T T F F T F F
Quiz 1 Due at 9:00 p.m. on Thursday, April 17, 2008 Division: ID#: Name: 1. P, Q, R (P Q) R (P R) (Q R). P Q R (P Q) R (P R) (Q R) X T T T F T T F F T F T T T F F F F T T F F T F T F F T F F F F F 2. 1.1
1990 IMO 1990/1/15 1:00-4:00 1 N N N 1, N 1 N 2, N 2 N 3 N 3 2 x x + 52 = 3 x x , A, B, C 3,, A B, C 2,,,, 7, A, B, C
0 9 (1990 1999 ) 10 (2000 ) 1900 1994 1995 1999 2 SAT ACT 1 1990 IMO 1990/1/15 1:00-4:00 1 N 1990 9 N N 1, N 1 N 2, N 2 N 3 N 3 2 x 2 + 25x + 52 = 3 x 2 + 25x + 80 3 2, 3 0 4 A, B, C 3,, A B, C 2,,,, 7,
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17 ( ) II III A B C(100 ) 1, 2, 6, 7 II A B (100 ) 2, 5, 6 II A B (80 ) 8 10 I II III A B C(80 ) 1 a 1 = 1 2 a n+1 = a n + 2n + 1 (n = 1,
17 ( ) 17 5 1 4 II III A B C(1 ) 1,, 6, 7 II A B (1 ), 5, 6 II A B (8 ) 8 1 I II III A B C(8 ) 1 a 1 1 a n+1 a n + n + 1 (n 1,,, ) {a n+1 n } (1) a 4 () a n OA OB AOB 6 OAB AB : 1 P OB Q OP AQ R (1) PQ
x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x
[ ] IC. f(x) = e x () f(x) f (x) () lim f(x) lim f(x) x + x (3) lim f(x) lim f(x) x + x (4) y = f(x) ( ) ( s46). < a < () a () lim a log xdx a log xdx ( ) n (3) lim log k log n n n k=.3 z = log(x + y ),
x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y)
x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 1 1977 x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y) ( x 2 y + xy 2 x 2 2xy y 2) = 15 (x y) (x + y) (xy
取扱説明書 -詳細版- 液晶プロジェクター CP-AW3019WNJ
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ALGEBRA II Hiroshi SUZUKI Department of Mathematics International Christian University 2004 1 1 1 2 2 1 3 3 1 4 4 1 5 5 1 6 6 1 7 7 1 7.1....................... 7 1 7.2........................... 7 4 8
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1/68 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量一覧 平成 31 年 3 月 6 日現在 < 留意事項 > (1) 空容量は目安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 特に記載
1/68 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量一覧 平成 31 年 3 月 6 日現在 < 留意事項 > (1) 空容量は目安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 特に記載のない限り 熱容量を考慮した空き容量を記載しております その他の要因 ( 電圧や系統安定度など ) で連系制約が発生する場合があります
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