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2011-02-01, 19:57   #364
Mr. P-1

Jun 2003

7·167 Posts

Quote:
 Originally Posted by science_man_88 is this what you wanted ?
Not really, though I've had to spend quite a while thinking about what I do "want", why what you've written doesn't satisfy, and how to explain it to you. Plus I've been distracted...

Basically mathematicians try to express their ideas using a very limited vocabulary. The reason for this is so they can define very precisely how these terms interact, and how one can derive new statements (theorems) from previously proved theorems or unproven-but-assumed statements (axioms).

One consequence of this limited vocabulary as it applies to set theory is that it isn't necessary or even really possible to talk about "repeats". Given a set S and an object a, the only thing you can say about the latter's membership of the former is that either a$\in$S or a$\not \in$S. It isn't possible to even discuss how many times a$\in$S. The vocabulary doesn't permit it.

So what I "want" I guess, is for you to define these terms using the following vocabulary only:

Variables: a, b, c, X, Y, Z, etc.
Quantifiers: all, at least one, exactly one, no (as in no set)
Logical operators: and, or, not, neither ... nor, if ... then, iff (short for "if and only if".)
Set membership: member of
Punctuation: brackets and commas.

Other terms already defined using these terms.
Unambiguous synonyms, for example: "each", "every", instead of "all"; "implies" instead of "if ... then", etc.
Any additional words needed to make your sentences into grammatical English, so long as they don't carry substantive meaning.

For example. I can define "subset of" using the above by saying that "A is a subset of B" means the following:

For every x in A, x is in B.

Here "For" and "is" don't carry substantive meaning. (You could omit them and still understand the statement), "every" means "all"; "in" means "member of".

Another example: Define $\empty$:

For every x, x is not in $\empty$.

Binary union: A$\cup$B means:

For every x, x is in A$\cup$B iff (x is in A or x is in B)

Union of a collection: $\cup$C:

For every x, x is in $\cup$C iff x is in at least one X in C.

Do you think you could define binary intersection and intersection of a collection like this?

2011-02-02, 00:13   #365
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by Mr. P-1 Not really, though I've had to spend quite a while thinking about what I do "want", why what you've written doesn't satisfy, and how to explain it to you. Plus I've been distracted... Basically mathematicians try to express their ideas using a very limited vocabulary. The reason for this is so they can define very precisely how these terms interact, and how one can derive new statements (theorems) from previously proved theorems or unproven-but-assumed statements (axioms). One consequence of this limited vocabulary as it applies to set theory is that it isn't necessary or even really possible to talk about "repeats". Given a set S and an object a, the only thing you can say about the latter's membership of the former is that either a$\in$S or a$\not \in$S. It isn't possible to even discuss how many times a$\in$S. The vocabulary doesn't permit it. So what I "want" I guess, is for you to define these terms using the following vocabulary only: Variables: a, b, c, X, Y, Z, etc. Quantifiers: all, at least one, exactly one, no (as in no set) Logical operators: and, or, not, neither ... nor, if ... then, iff (short for "if and only if".) Set membership: member of Punctuation: brackets and commas. Other terms already defined using these terms. Unambiguous synonyms, for example: "each", "every", instead of "all"; "implies" instead of "if ... then", etc. Any additional words needed to make your sentences into grammatical English, so long as they don't carry substantive meaning. For example. I can define "subset of" using the above by saying that "A is a subset of B" means the following: For every x in A, x is in B. Here "For" and "is" don't carry substantive meaning. (You could omit them and still understand the statement), "every" means "all"; "in" means "member of". Another example: Define $\empty$: For every x, x is not in $\empty$. Binary union: A$\cup$B means: For every x, x is in A$\cup$B iff (x is in A or x is in B) Union of a collection: $\cup$C: For every x, x is in $\cup$C iff x is in at least one X in C. Do you think you could define binary intersection and intersection of a collection like this?
thank you for the lesson. I love the breakdown of language like floccinaucinihilipilification broken down it can be defined by the sum worthless + worthless +worthless + the act of. Using your examples as a guide I believe I'm up to the challenge. For every x, x is in A$\cap$B iff (x is in A and x is in B). For every x, x is in $\cap$C iff x is in all X in C . as you've probably seen in the theory on Mersenne primes thread I've been semi distracted as well. Needless to say , I've been trying to apply my understanding to my latest idea. This still hasn't completely helped me prove it and I've already been asked twice for the next Mersenne prime exponent.

Last fiddled with by science_man_88 on 2011-02-02 at 00:26

2011-02-02, 17:56   #366
CRGreathouse

Aug 2006

3·1,993 Posts

Quote:
 Originally Posted by science_man_88 For every x, x is in A$\cap$B iff (x is in A and x is in B).
Yes.

Quote:
 Originally Posted by science_man_88 For every x, x is in $\cap$C iff x is in all X in C .
Yes.

Quote:
 Originally Posted by science_man_88 as you've probably seen in the theory on Mersenne primes thread I've been semi distracted as well. Needless to say , I've been trying to apply my understanding to my latest idea. This still hasn't completely helped me prove it and I've already been asked twice for the next Mersenne prime exponent.
We don't understand what you're saying. You make observations (which are, themselves, hard to follow) and then claim a link to Mersenne exponents without making the precise claim you're making explicit. If you wrote it as you did above (that is, unambiguously) it would go over much better.

2011-02-04, 12:01   #367
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by CRGreathouse Yes. Yes. We don't understand what you're saying. You make observations (which are, themselves, hard to follow) and then claim a link to Mersenne exponents without making the precise claim you're making explicit. If you wrote it as you did above (that is, unambiguously) it would go over much better.
yeah but then I sound real book-like, my latest -> pairwise disjoint: a collection C is pairwise disjoint iff(for all $A,B\in C$ $A\cap B = \empty$)

their's(pasted) ->
Quote:
 A collection {Ci} of sets is called pairwise disjoint if Ci ∩ Cj = ∅ for all i, j with i =6 .
okay their's has one differnece to what is pasted it has the not equal sign and a j not the =6 .

2011-02-04, 12:15   #368
xilman
Bamboozled!

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

22×3×11×83 Posts

Quote:
 Originally Posted by science_man_88 yeah but then I sound real book-like.
That's to be commended!

If your statements are both correct and "book-like" then others will be able to understand you relatively easily. If they are wrong and book-like, it will be much easier to work out where you are going wrong and to provide corrections.

Paul

2011-02-04, 12:55   #369
Wacky

Jun 2003
The Texas Hill Country

32×112 Posts

Quote:
 Originally Posted by science_man_88 yeah but then I sound real book-like, my latest -> pairwise disjoint: a collection C is pairwise disjoint iff(for all $A,B\in C$ $A\cap B = \empty$) their's(pasted) ->
Do you see the difference between your definition and theirs?

2011-02-04, 12:58   #370
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by CRGreathouse Yes. Yes. We don't understand what you're saying. You make observations (which are, themselves, hard to follow) and then claim a link to Mersenne exponents without making the precise claim you're making explicit. If you wrote it as you did above (that is, unambiguously) it would go over much better.

technically I think I could get that second one to align with: For all x, x is in $\cap C$ iff(x is a one element subset of all X in C).

2011-02-04, 13:48   #371
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by Wacky Do you see the difference between your definition and theirs?
they use i and j as indexes where mine looks to use full set names ?

2011-02-04, 13:51   #372
Wacky

Jun 2003
The Texas Hill Country

32×112 Posts

Quote:
 Originally Posted by science_man_88 they use i and j as indexes where mine looks to use full set names ?
No. The difference is that you have specified ANY sets, A and B which are elements of C.

They have excluded certain pairs of sets.

Which pairs have they excluded?
Why is that important?

2011-02-04, 13:54   #373
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

838410 Posts

Quote:
 Originally Posted by Wacky No. The difference is that you have specified ANY sets, A and B which are elements of C. They have excluded certain pairs of sets. Which pairs have they excluded? Why is that important?
oh doh I forgot that A!=B it's important because a set is always equal to itself and so no collection could be called pairwise disjoint under my definition, thanks for catching that.

2011-02-04, 14:18   #374
CRGreathouse

Aug 2006

10111010110112 Posts

Quote:
 Originally Posted by science_man_88 oh doh I forgot that A!=B it's important because a set is always equal to itself and so no collection could be called pairwise disjoint under my definition, thanks for catching that.
Well... just to get nitpicky...

There are sets which are "sm-disjoint", that is, collections C for which all a,b in C are such that a ∩ b = {}. Can you find them?

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