✔ 最佳答案
(4) / C 是什麼?
C的補集C' ?
2014-10-01 18:04:04 補充:
(1)
( i )
x ∈ A
⇒ x = 2p+1 for some p ∈ ℕ
⇒ x = 2(p+1) - 1 for some p ∈ ℕ
⇒ x = 2q - 1 for some q ∈ ℕ and q ≥ 1 ( Let q = p+1 )
⇒ x ∈ C
( ii )
x ∈ C
⇒ x = 2q - 1 for some q ∈ ℕ and q ≥ 1
⇒ x = 2(q-1) + 1 for some q ∈ ℕ and q ≥ 1
⇒ x = 2p+1 for some p ∈ ℕ ( Let p = q-1 )
⇒ x ∈ A
By ( i ) & ( ii ), x ∈ A iff x ∈ C
Hence A = C
(2)
x ∈ B
⇒ x = 4p+1 for some p ∈ ℕ
⇒ x = 4p+2-1 for some p ∈ ℕ
⇒ x = 2(2p+1)-1 for some p ∈ ℕ
⇒ x = 2q - 1 for some q ∈ ℕ and q ≥ 1 ( Let q = 2p+1 )
⇒ x ∈ C
Thus, B ⊆ C ......... ( i )
3 = 2*2-1 , so 3 ∈ C
But 3 is obviously not an element of B, so
B ≠ C ......... ( ii )
By ( i ) & ( ii ), we have B ⊂ C
(3)
Suppose A = B
x ∈ A iff x ∈ B
5 = 3*1+2 , so 5 ∈ A
Then 5 ∈ B, that is, 5 = 5k-1 for some k ∈ ℕ
Thus, 5k = 6 for some k ∈ ℕ , a contradiction.
Hence A ≠ B
(4)
Condition ( i ) : A ⊆ B
Condition ( ii ) : A ⊆/ C ( A is not a subset of C )
Suppose B ⊆ C ......... ( iii )
x ∈ A
⇒ x ∈ B ( By ( i ) )
⇒ x ∈ C ( By ( iii ) )
Thus, A ⊆ C , a contradiction with ( ii )
So B ⊆/ C
2014-10-03 16:27:03 補充:
Q : 第一題裡面, A和C裡面的k是不同的?
A : 因為 " for some k ", 所以A和C裡面的k不一定相同.
我在證明中,刻意不用k,而用p,q,就是為了避免混淆.
2014-10-03 16:33:32 補充:
Q : for some k ∈ ℕ 是什麼意思
A : n=2k+1 for some k∈ℕ ..... (1)
意思即為:
存在一自然數k,使得n=2k+1 ..... (2)
或用數學慣用符號寫成:
∃ k ∈ ℕ s.t. n=2k+1 ..... (3)
(1),(2),(3)意義皆相同