4題關於集合的證明題 plz

2014-10-01 12:25 am
A={n|n∈ℕ and n=2k+1 for some k∈ℕ},B={n|n∈ℕ and n=4k+1 for some k∈ℕ},C={m∈ℕ|m=2k-1 and k∈ℕ and k>=1}

prove (1).A=C (2)B⊂C

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A={n|n∈ℕ and n=3k+2 for some k∈ℕ},B={n|n∈ℕ and n=5k-1 for some k∈ℕ such that k>=5}

prove (3).A≠B


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Let A,B and C be sets.

(4) prove that if A⊆B and A ⊆/ C then B ⊆/ C


求救T_T

回答 (4)

2014-10-02 2:04 am
✔ 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
2014-10-03 1:02 pm
這家不錯 lv333。cC買幾次啦真的一樣
勬哄儺余侈
2014-10-01 5:44 am
不等於的意思 我找不到那個符號

2014-10-02 21:20:49 補充:
可以在問一下嗎 第一題裡面 A和C裡面的k是不同的?

2014-10-02 21:47:30 補充:
還有 for some k ∈ ℕ 是什麼意思 是代表有些K是自然數,然後有些K不是嗎,那K有可能是負的?
2014-10-01 4:29 am
1-1 & onto


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