Topic: Solve these 3 math problems and get 0.1 btc (set theory-not too difficult) (Read 1010 times)
May 03, 2015, 06:56:11 PM
woops.. wouldnt that be something hahaha. i think these problems would have been solved long ago
May 03, 2015, 06:54:40 PM
Payment received as promised.
Thanks. Positive feedback left.
On a lighter note, you should take care of your decimals.
The OP talks about 0.66 BTC for the last 2 problems (instead of 0.066)
Thanks. Positive feedback left.
On a lighter note, you should take care of your decimals.
The OP talks about 0.66 BTC for the last 2 problems (instead of 0.066)
May 03, 2015, 06:48:24 PM
hmm... i think this is correct, but i'd need to spend more time to be sure... anyways its close enough if it isnt.. im sending polynesia the last of the bounty
good job sir and thank you kindly
good job sir and thank you kindly
Thanks! That was helpful
(a)
(5,0) - Xo
(17,0) - Xo
(18,0) - Xe
To check the solution, we can list down the antecedents of (5,0)
(4,1), (3,0), (2,1), (1,0), (0,1)
So (5,0) has 5 antecedents and our answer is correct.
So the element (x,0) can be defined to fall under
Xe: x is even
Xo: x is odd
Xi is a null set
(b) Similarly (y,1) can be define to fall under
Ye: y is even
Yo: y is odd
Yi is a null set
(c) The bijection is defined as
F(x,0) = (x+1,1), if x is even
F(x,0) = (x-1,1), if x is odd.
(a)
(5,0) - Xo
(17,0) - Xo
(18,0) - Xe
To check the solution, we can list down the antecedents of (5,0)
(4,1), (3,0), (2,1), (1,0), (0,1)
So (5,0) has 5 antecedents and our answer is correct.
So the element (x,0) can be defined to fall under
Xe: x is even
Xo: x is odd
Xi is a null set
(b) Similarly (y,1) can be define to fall under
Ye: y is even
Yo: y is odd
Yi is a null set
(c) The bijection is defined as
F(x,0) = (x+1,1), if x is even
F(x,0) = (x-1,1), if x is odd.
May 03, 2015, 06:44:56 PM
Has anyone solved it yet? If none, I could give it a shot.
May 03, 2015, 06:34:52 PM
Thanks! That was helpful
(a)
(5,0) - Xo
(17,0) - Xo
(18,0) - Xe
To check the solution, we can list down the antecedents of (5,0)
(4,1), (3,0), (2,1), (1,0), (0,1)
So (5,0) has 5 antecedents and our answer is correct.
So the element (x,0) can be defined to fall under
Xe: x is even
Xo: x is odd
Xi is a null set
(b) Similarly (y,1) can be define to fall under
Ye: y is even
Yo: y is odd
Yi is a null set
(c) The bijection is defined as
F(x,0) = (x+1,1), if x is even
F(x,0) = (x-1,1), if x is odd.
(a)
(5,0) - Xo
(17,0) - Xo
(18,0) - Xe
To check the solution, we can list down the antecedents of (5,0)
(4,1), (3,0), (2,1), (1,0), (0,1)
So (5,0) has 5 antecedents and our answer is correct.
So the element (x,0) can be defined to fall under
Xe: x is even
Xo: x is odd
Xi is a null set
(b) Similarly (y,1) can be define to fall under
Ye: y is even
Yo: y is odd
Yi is a null set
(c) The bijection is defined as
F(x,0) = (x+1,1), if x is even
F(x,0) = (x-1,1), if x is odd.
May 03, 2015, 11:41:12 AM
I think these pages are relevant as well :
May 03, 2015, 11:24:36 AM
Yea.. I guess I thought it was a common definition
Heres how they do it:
Heres how they do it:
May 03, 2015, 11:14:31 AM
Nice! I think that could be it.. I'm sending coins now:)
Get the last one and you'll get the full 0.1
Get the last one and you'll get the full 0.1
Payment received. Thank you.
Could you help me with the notation for the third question.
What is Xe, Xo, Xi?
Have they been defined elsewhere?
May 03, 2015, 10:26:01 AM
Nice! I think that could be it.. I'm sending coins now:)
Get the last one and you'll get the full 0.1
Get the last one and you'll get the full 0.1
May 03, 2015, 10:01:29 AM
Here goes my attempt at proposition 6.7
Proof that the function is injective
Let F(S) = F(T) = A (which belongs to P(X))
Given the definition of F, we note that S is defined by
S(x) = 1, for all x belongs to A
S(x) = 0, for all x does not belong to A
We also note that since F(T) = A, T is defined by
T(x) = 1, for all x belongs to A
T(x) = 0, for all x does not belong to A
Since the definitions of S and T are identical, we conclude that S = T
Proof that the function is surjective
Let A belong to P(X)
Define the function J from X into |-2-| such that
J (x) = 1, for all x belongs to A
J (x) = 0, for all x does not belong to A
We note that F(J) = A, for any A
Hence F is surjective.
Since F is injective and surjective, it is bijective.
Proof that the function is injective
Let F(S) = F(T) = A (which belongs to P(X))
Given the definition of F, we note that S is defined by
S(x) = 1, for all x belongs to A
S(x) = 0, for all x does not belong to A
We also note that since F(T) = A, T is defined by
T(x) = 1, for all x belongs to A
T(x) = 0, for all x does not belong to A
Since the definitions of S and T are identical, we conclude that S = T
Proof that the function is surjective
Let A belong to P(X)
Define the function J from X into |-2-| such that
J (x) = 1, for all x belongs to A
J (x) = 0, for all x does not belong to A
We note that F(J) = A, for any A
Hence F is surjective.
Since F is injective and surjective, it is bijective.
May 02, 2015, 08:03:00 AM
I'm glad your giving that one a shot because thats the real turd in the punchbowl..
so Y^X means the set of all functions with domain X and codomain Y.
|2| notation thing is like you said, just {0,1}-so basically its talking about X->{0,1} for |2|^X
P(X) is the power series of X, and it is what it is defined to be in the book.. i have a vague idea of what their talking about, the (xsi?-squiggly)X is the characteristic set of P(X) ie it is a set that defines where P(X) holds..
I hope this helps u squash this sucka
so Y^X means the set of all functions with domain X and codomain Y.
|2| notation thing is like you said, just {0,1}-so basically its talking about X->{0,1} for |2|^X
P(X) is the power series of X, and it is what it is defined to be in the book.. i have a vague idea of what their talking about, the (xsi?-squiggly)X is the characteristic set of P(X) ie it is a set that defines where P(X) holds..
I hope this helps u squash this sucka
May 02, 2015, 07:57:13 AM
Some small clarifications with respect to the first question
1) What is under your fingertips in the scanned image?
2) Does |-2-| indicate the set with just 2 elements {0,1}?
3) Is P(X) defined elsewhere?
1) What is under your fingertips in the scanned image?
2) Does |-2-| indicate the set with just 2 elements {0,1}?
3) Is P(X) defined elsewhere?
May 02, 2015, 03:16:30 AM
BEAUTIFUL
May 02, 2015, 03:14:27 AM
0.025btc awarded to polynesia
97efdd32c3864b50ad56408830256a544689b576b18e8fe81eb7963fec4d78dc
edit: hehe.. you think you could define the bijection? i am literally incapable of doing anymore math today. my brain is reeling
97efdd32c3864b50ad56408830256a544689b576b18e8fe81eb7963fec4d78dc
edit: hehe.. you think you could define the bijection? i am literally incapable of doing anymore math today. my brain is reeling
Unfortunately, I don't get the notation required.
Let me try to define it
X = [m] x [n]
Y = [mn]
For each i (1<= i <= m) and j ( 1 <= j <= n), let us define a function f (i, j) such that f(i,j) = (i−1)n+j
f is bijective from X to Y
May 02, 2015, 02:58:48 AM
edit: hehe.. you think you could define the bijection? i am literally incapable of doing anymore math today. my brain is reeling
May 02, 2015, 02:56:39 AM
Thanks. Payment received.
https://blockchain.info/tx/97efdd32c3864b50ad56408830256a544689b576b18e8fe81eb7963fec4d78dc
A big thumbs up to you.
https://blockchain.info/tx/97efdd32c3864b50ad56408830256a544689b576b18e8fe81eb7963fec4d78dc
A big thumbs up to you.
May 02, 2015, 02:50:48 AM
something like that i think would make the trolls happy.. i'm sending btc now
May 02, 2015, 02:45:54 AM
For problem 5, does this work?
f(i,j)=(i−1)n+j
where i goes from 1 to m and j goes from 1 to n
f(i,j)=(i−1)n+j
where i goes from 1 to m and j goes from 1 to n
yea i think thats exactly right as a matter of fact.. could you just be a little more explicit for defining the bijection of m and n with respect to f(i,j)?
Thanks. Let me know a sample of the way you want to define the bijective function and I will help define it for you.
My address: 1Nri6bYYB5jjVSQxum9jGkYNghqZHzUAfu
May 02, 2015, 02:39:16 AM
For problem 5, does this work?
f(i,j)=(i−1)n+j
where i goes from 1 to m and j goes from 1 to n
f(i,j)=(i−1)n+j
where i goes from 1 to m and j goes from 1 to n
yea i think thats exactly right as a matter of fact.. could you just be a little more explicit for defining the bijection of m and n with respect to f(i,j)?
May 02, 2015, 02:32:03 AM
For problem 5, does this work?
f(i,j)=(i−1)n+j
where i goes from 1 to m and j goes from 1 to n
f(i,j)=(i−1)n+j
where i goes from 1 to m and j goes from 1 to n
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