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Fitz Watson
Fitz Watson
04:28
Let S = {1, 2, 3} be a sample space. Let P be a probability measure defined
on 2S
(the collection of all subsets of S) such that

P(i) = xi
, for i = 1, 2, 3.

Select the correct statement from the following and complete it.
1. (x1, x2, x3) can be any point in a sphere centered at the origin, having
radius R = .
2. (x1, x2, x3) can be any point in a triangle whose vertices are v1 =
, v2 = , and v3 = .
3. (x1, x2, x3) can be any point in a square whose vertices are v1 = ,
v2 = , v3 = , and v4 = .
Can someone tell how this question should be approached?
 
5 hours later…
user21820
user21820
09:01
@F.Zer Not sure why you cannot get it since you essentially did it before.
@FitzWatson 2S = {2,4,6}.
Fitz Watson
Fitz Watson
10:01
It's actually 2^S, representing the powerset of S
 
6 hours later…
F. Zer
F. Zer
15:56
If k > m:
  Let z ∈ ℕ such that m + z = k
  k = m·1 + z
  ∃t,u∈ℕ ( k = m·t+u ).
  Let r∈ℕ be minimum such that ∃t∈ℕ ( k = m·t+r ).
  Let y∈ℕ such that k = m·y+r.
  If r ≥ m:
    If r = 0:
      k = m·y
      ⊥
    If r = 1:
      k = m·y + 1
      ...
    If r > 1:
      ...
    ...
  r < m.
  r < k.
  r+m < k+m = n.
  Q(r+m).
  ...
  ¬∃d∈ℕ ( d > 1 ∧ d | r ∧ d | m ).
  Let p,q∈ℕ such that r·p = m·q+1.
  k·p = m·(y·p)+r·p = m·(y·p)+m·q+1 = m·(y·p+q)+1.
  ...
@user21820, I don't have a clue, yet.
user21820
user21820
16:29
Jul 6 at 16:04, by F. Zer
@user21820 k · x ≥ k + m
As I said, you did it before. You're getting stuck in exactly the same way as you got stuck before...
@FitzWatson What do you think?
Prithu biswas
Prithu biswas
17:00
@user21820 Can you check my last attempt ?
user21820
user21820
17:23
@Prithubiswas Yes, after fixing the errors it is correct. Note that extra brackets are not considered errors.
Prithu biswas
Prithu biswas 
If (A⇒nB)∧B
   A⇒nB
   B
   If A
      A⇒nB
      nB
      B
      ⊥
   A⇒⊥
   nA
(A⇒nB)∧B⇒nA
@user21820 attempt for P3. Inform me if there are any errors.
Fitz Watson
Fitz Watson
18:00
@user21820 It has something to do with P(Sample Space)=1 and that 0<=P(X)<=1
F. Zer
F. Zer
18:21
If k > m:
  Let z ∈ ℕ such that m + z = k
  k = m·1 + z
  ∃t,u∈ℕ ( k = m·t+u ).
  Let r∈ℕ be minimum such that ∃t∈ℕ ( k = m·t+r ).
  Let y∈ℕ such that k = m·y+r.
  If r = m:
    k = m·y + m
    k = m·(y + 1)
    m | k
    ⊥
  If r > m:
    r + z > m + z
    r + z > k
    r + z > m
    ...
    ...
  r < m.
  r < k.
  r+m < k+m = n.
  Q(r+m).
  ...
  ¬∃d∈ℕ ( d > 1 ∧ d | r ∧ d | m ).
  Let p,q∈ℕ such that r·p = m·q+1.
  k·p = m·(y·p)+r·p = m·(y·p)+m·q+1 = m·(y·p+q)+1.
  ...
@user21820 I looked at my previous attempt but couldn't connect the dots. All I can see are symbols. Could you tell me a bit, in English, how do you search for the proof ? Is it like some statement do you fell it will work ? I try and try variants by brute-force. However, that is a very inefficient way of working. I add a certain variable to both sides, apply distributivity...I should find meaning to my search, I think.
 
4 hours later…
F. Zer
F. Zer
22:15
Essentialy, I should arrive at the statement r' < r ∧ ∃ t ∈ ℕ ( k = m·t + r' ) where r' ∈ ℕ.
F. Zer
F. Zer
22:54
I've also tried one small case: "6 = 5·1 + 1".
From r ≥ m and m > 1, I could derive r > 1. 
Trying all combinations of additions and multiplications on both sides of "r ≥ m" using variables k and y, I get:

r + k ≥ m + k
r + y ≥ m + y
r·k ≥ m·k
r·y ≥ m·y

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