+3점 · (a) 조건을 만족하는 집합 A \displaystyle A A (3-1) (a) A = { 1 , 2 , 4 , 8 } \displaystyle A=\{1,2,4,8\} A = { 1 , 2 , 4 , 8 } A ~ = { 2 , 3 , 4 , 5 , 6 , 8 , 9 , 10 , 12 , 16 } \displaystyle \widetilde{A}=\{2,3,4,5,6,8,9,10,12,16\} A = { 2 , 3 , 4 , 5 , 6 , 8 , 9 , 10 , 12 , 16 } n ( A ~ ) = 10 \displaystyle n(\widetilde{A})=10 n ( A ) = 10
+7점 · (b) n ( A ~ ) \displaystyle n(\widetilde{A}) n ( A ) (b) A \displaystyle A A A ~ \displaystyle \widetilde{A} A A = { 1 , 2 , 2 2 , ⋯ , 2 k − 1 } \displaystyle A=\{1,2,2^2,\cdots,2^{k-1}\} A = { 1 , 2 , 2 2 , ⋯ , 2 k − 1 } A \displaystyle A A n ( A ~ ) = k 2 + k 2 \displaystyle n(\widetilde{A})=\frac{k^2+k}{2} n ( A ) = 2 k 2 + k 그러므로 n ( A ~ ) \displaystyle n(\widetilde{A}) n ( A ) k 2 + k 2 \displaystyle \frac{k^2+k}{2} 2 k 2 + k
+3점 · (a) 조건을 만족하는 집합 A \displaystyle A A (3-2) (a) A = { 1 , 2 , 3 , 4 , 5 } \displaystyle A=\{1,2,3,4,5\} A = { 1 , 2 , 3 , 4 , 5 } A ~ = { 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 } \displaystyle \widetilde{A}=\{2,3,4,5,6,7,8,9,10\} A = { 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 } n ( A ~ ) = 9 \displaystyle n(\widetilde{A})=9 n ( A ) = 9
+7점 · (b) n ( A ~ ) \displaystyle n(\widetilde{A}) n ( A ) (b) A \displaystyle A A a 1 < a 2 < ⋯ < a k \displaystyle a_1<a_2<\cdots<a_k a 1 < a 2 < ⋯ < a k 그러면 2 a 1 < a 1 + a 2 < 2 a 2 < a 2 + a 3 < ⋯ < a k − 2 + a k − 1 < 2 a k − 1 < a k − 1 + a k < 2 a k \displaystyle 2a_1<a_1+a_2<2a_2<a_2+a_3<\cdots<a_{k-2}+a_{k-1}<2a_{k-1}<a_{k-1}+a_k<2a_k 2 a 1 < a 1 + a 2 < 2 a 2 < a 2 + a 3 < ⋯ < a k − 2 + a k − 1 < 2 a k − 1 < a k − 1 + a k < 2 a k A ~ \displaystyle \widetilde{A} A 2 k − 1 \displaystyle 2k-1 2 k − 1 n ( A ~ ) ≥ 2 k − 1 \displaystyle n(\widetilde{A})\ge2k-1 n ( A ) ≥ 2 k − 1 A = { 1 , 2 , ⋯ , k } \displaystyle A=\{1,2,\cdots,k\} A = { 1 , 2 , ⋯ , k } A ~ = { 2 , 3 , ⋯ , 2 k } \displaystyle \widetilde{A}=\{2,3,\cdots,2k\} A = { 2 , 3 , ⋯ , 2 k } n ( A ~ ) = 2 k − 1 \displaystyle n(\widetilde{A})=2k-1 n ( A ) = 2 k − 1 n ( A ~ ) \displaystyle n(\widetilde{A}) n ( A ) 2 k − 1 \displaystyle 2k-1 2 k − 1
(3-3) (3-1)(b)와 (3-2)(b)에서 n ( A ~ ) \displaystyle n(\widetilde{A}) n ( A ) k 2 + k 2 \displaystyle \frac{k^2+k}{2} 2 k 2 + k 2 k − 1 \displaystyle 2k-1 2 k − 1 n ( A ~ ) \displaystyle n(\widetilde{A}) n ( A ) 2 k − 1 \displaystyle 2k-1 2 k − 1 k 2 + k 2 \displaystyle \frac{k^2+k}{2} 2 k 2 + k
명제 : 임의의 자연수 2 k − 1 ≤ m ≤ k 2 + k 2 \displaystyle 2k-1\le m\le\frac{k^2+k}{2} 2 k − 1 ≤ m ≤ 2 k 2 + k n ( A ) = k \displaystyle n(A)=k n ( A ) = k n ( A ~ ) = m \displaystyle n(\widetilde{A})=m n ( A ) = m A \displaystyle A A
k \displaystyle k k k = 1 \displaystyle k=1 k = 1 2 k − 1 = k 2 + k 2 = 1 \displaystyle 2k-1=\frac{k^2+k}{2}=1 2 k − 1 = 2 k 2 + k = 1 A = { 1 } \displaystyle A=\{1\} A = { 1 } k = 2 \displaystyle k=2 k = 2 2 k − 1 = k 2 + k 2 = 3 \displaystyle 2k-1=\frac{k^2+k}{2}=3 2 k − 1 = 2 k 2 + k = 3 A = { 1 , 2 } \displaystyle A=\{1,2\} A = { 1 , 2 } k > 2 \displaystyle k>2 k > 2 k − 1 \displaystyle k-1 k − 1 2 k − 1 ≤ m ≤ 3 k − 4 \displaystyle 2k-1\le m\le3k-4 2 k − 1 ≤ m ≤ 3 k − 4 3 k − 3 ≤ m ≤ k 2 + k 2 \displaystyle 3k-3\le m\le\frac{k^2+k}{2} 3 k − 3 ≤ m ≤ 2 k 2 + k A \displaystyle A A
+7점 · 2 k − 1 ≤ m ≤ 3 k − 4 \displaystyle 2k-1\le m\le3k-4 2 k − 1 ≤ m ≤ 3 k − 4 m \displaystyle m m n ( A ~ ) = m \displaystyle n(\widetilde{A})=m n ( A ) = m A \displaystyle A A a) 2 k − 1 ≤ m ≤ 3 k − 4 \displaystyle 2k-1\le m\le3k-4 2 k − 1 ≤ m ≤ 3 k − 4 A = { 1 , 2 , ⋯ , k − 1 , m − k + 1 } \displaystyle A=\{1,2,\cdots,k-1,m-k+1\} A = { 1 , 2 , ⋯ , k − 1 , m − k + 1 } 이때, A ~ = { 2 , 3 , ⋯ , 2 k − 2 } ∪ { m − k + 2 , m − k + 3 , ⋯ , m } ∪ { 2 m − 2 k + 2 } \displaystyle \widetilde{A}=\{2,3,\cdots,2k-2\}\cup\{m-k+2,m-k+3,\cdots,m\}\cup\{2m-2k+2\} A = { 2 , 3 , ⋯ , 2 k − 2 } ∪ { m − k + 2 , m − k + 3 , ⋯ , m } ∪ { 2 m − 2 k + 2 } m − k + 2 ≤ 2 k − 2 \displaystyle m-k+2\le2k-2 m − k + 2 ≤ 2 k − 2 m < 2 m − 2 k + 2 \displaystyle m<2m-2k+2 m < 2 m − 2 k + 2 A ~ = { 2 , 3 , ⋯ , m − 1 , m , 2 m − 2 k + 2 } \displaystyle \widetilde{A}=\{2,3,\cdots,m-1,m,2m-2k+2\} A = { 2 , 3 , ⋯ , m − 1 , m , 2 m − 2 k + 2 } n ( A ~ ) = ( m − 1 ) + 1 = m \displaystyle n(\widetilde{A})=(m-1)+1=m n ( A ) = ( m − 1 ) + 1 = m
+8점 · 3 k − 3 ≤ m ≤ k 2 + k 2 \displaystyle 3k-3\le m\le\frac{k^2+k}{2} 3 k − 3 ≤ m ≤ 2 k 2 + k m \displaystyle m m n ( A ~ ) = m \displaystyle n(\widetilde{A})=m n ( A ) = m A \displaystyle A A b) 3 k − 3 ≤ m ≤ k 2 + k 2 \displaystyle 3k-3\le m\le\frac{k^2+k}{2} 3 k − 3 ≤ m ≤ 2 k 2 + k 2 ( k − 1 ) − 1 = 2 k − 3 ≤ m − k ≤ k 2 − k 2 = ( k − 1 ) 2 + ( k − 1 ) 2 \displaystyle 2(k-1)-1=2k-3\le m-k\le\frac{k^2-k}{2}=\frac{(k-1)^2+(k-1)}{2} 2 ( k − 1 ) − 1 = 2 k − 3 ≤ m − k ≤ 2 k 2 − k = 2 ( k − 1 ) 2 + ( k − 1 ) n ( B ) = k − 1 \displaystyle n(B)=k-1 n ( B ) = k − 1 n ( B ~ ) = m − k \displaystyle n(\widetilde{B})=m-k n ( B ) = m − k B \displaystyle B B B \displaystyle B B x \displaystyle x x A = B ∪ { 2 x + 1 } \displaystyle A=B\cup\{2x+1\} A = B ∪ { 2 x + 1 } 그러면 A ~ = B ~ ∪ { 2 x + 1 + b ∣ b ∈ B } ∪ { 4 x + 2 } \displaystyle \widetilde{A}=\widetilde{B}\cup\{2x+1+b\mid b\in B\}\cup\{4x+2\} A = B ∪ { 2 x + 1 + b ∣ b ∈ B } ∪ { 4 x + 2 } B ~ \displaystyle \widetilde{B} B 2 x \displaystyle 2x 2 x { 2 x + 1 + b ∣ b ∈ B } \displaystyle \{2x+1+b\mid b\in B\} { 2 x + 1 + b ∣ b ∈ B } 2 x + 1 \displaystyle 2x+1 2 x + 1 3 x + 1 \displaystyle 3x+1 3 x + 1 B ~ \displaystyle \widetilde{B} B { 2 x + 1 + b ∣ b ∈ B } \displaystyle \{2x+1+b\mid b\in B\} { 2 x + 1 + b ∣ b ∈ B } { 2 x + 2 y + 2 } \displaystyle \{2x+2y+2\} { 2 x + 2 y + 2 } Correction. 이 집합의 2 x + 2 y + 2 \displaystyle 2x+2y+2 2 x + 2 y + 2 4 x + 2 \displaystyle 4x+2 4 x + 2 따라서 n ( A ~ ) = ( m − k ) + ( k − 1 ) + 1 = m \displaystyle n(\widetilde{A})=(m-k)+(k-1)+1=m n ( A ) = ( m − k ) + ( k − 1 ) + 1 = m
수학적 귀납법에 의하여 모든 자연수 k \displaystyle k k
수리논술, 배움에서 논증의 완성까지.