Solve these 3 math problems and get 0.1 btc (set theory-not too difficult)

[tarzan2]

As the title claims.. first person to post correct solutions for all three problems will receive 0.1 btc. if someone submits correct answer for any 1 of the three they will receive 0.025 btc from me and 0.66 btc will remain on the remaining 2.. same idea if 2 are solved, such that the remaining problem will have a 0.33 bounty

i need the proof of proposition 6.7 as described here:


I need exercises 5 and 6 from here:


good luck!

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[young3dvard]

you should use imgur and post the images here .

using this paste the link of the image between

[polynesia]

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

[tarzan2]

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

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)?

[polynesia]

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

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 

[tarzan2]

something like that i think would make the trolls happy.. i'm sending btc now

[polynesia]

Thanks. Payment received.
https://blockchain.info/tx/97efdd32c3864b50ad56408830256a544689b576b18e8fe81eb7963fec4d78dc

A big thumbs up to you. 

[tarzan2]

edit: hehe.. you think you could define the bijection? i am literally incapable of doing anymore math today. my brain is reeling

[polynesia]

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

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

[tarzan2]

BEAUTIFUL

[polynesia]

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?

[tarzan2]

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

[polynesia]

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.

[tarzan2]

Nice! I think that could be it.. I'm sending coins now:)
Get the last one and you'll get the full 0.1

[polynesia]

Nice! I think that could be it.. I'm sending coins now:)
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?

[tarzan2]

Yea.. I guess I thought it was a common definition
Heres how they do it:

[tarzan2]

I think these pages are relevant as well :

[polynesia]

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.

[TriggerX]

Has anyone solved it yet? If none, I could give it a shot.

[tarzan2]

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

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.