1. Joined
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    16 Apr '08 20:141 edit
    whether the number 7 is finite or infinite has nothing to do with this thread. If you substitude 7 for Aleph 0 (the cardinality of natural numbers) your proofs are clearly not true. The number is not finite yet you could read the thread in finite time ...
  2. Sigulda, Latvia
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    16 Apr '08 20:16
    Originally posted by 3v1l5w1n
    whether the number 7 is finite or infinite has nothing to do with this thread. If you substitude Aleph 0 (the cardinality of natural numbers), clearly your statement is not true. The number is not finite yet you could read the thread in finite time ...
    What I meant was, that the thread I mentioned has 7 posts. If 7 was infinite, nobody could read the WHOLE thread. However, it isn't that hard. Try it.
  3. Joined
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    16 Apr '08 22:27
    It would seem that the method of moving forward would be to define the terms, then see if the properties match.

    What I'm not sure of is exactly how you define "finite" mathematically, even though the term is easy enough to understand intuitively.

    And perhaps defining "finite" is the difficult part.
  4. Joined
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    17 Apr '08 15:25
    I would define "infinite" as number that, when written in the European Arabic Decimal number system, will never come to an end, and "finite" as one that would.

    The Aleph-Null point mentioned above is incorrect because "Aleph-Null" is not in this system.

    If you wanted pi as the title, you would have to write 3.14159265.... and redhotpawn would not let you continue at some point.
    If you wanted to write 7, then redhotpawn would let you. Any number that can fit in a redhotpawn forum title is not infinite.
  5. Joined
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    17 Apr '08 17:08
    Originally posted by doodinthemood
    I would define "infinite" as number that, when written in the European Arabic Decimal number system, will never come to an end, and "finite" as one that would.
    Irrational numbers have the property whereby their exact value written in decimal form would never end nor repeat, and yet irrationals by and large are considered "finite".

    I have found a good working definition at http://www.merriam-webster.com/dictionary/finite

    The one which seems most relevant is as follows.

    less than an arbitrary positive integer and greater than the negative of that integer

    7 is less than 8 (my arbitrary positive integer), and greater than -8, and hence is finite by definition.
  6. Joined
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    17 Apr '08 17:31
    This discussion is interesting!

    Despite yesterdays promise, I wait another day to giving the answer from the book.
  7. Joined
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    17 Apr '08 17:401 edit
    Originally posted by FabianFnas
    This discussion is interesting!

    Despite yesterdays promise, I wait another day to giving the answer from the book.
    Could you define a "finite number" though?

    My proof would go along the lines of: 7 is a cardinal number, 7 is strictly less than Aleph 0 (as there does not exist and injection from any set of size 7 into the natural numbers). However, Aleph 0 is the smallest infinite cardinal, and so 7 is finite.

    I have a feeling that there is a simpler way - I feel like I'm using a nuclear missile to blow up a single soldier...
  8. Joined
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    18 Apr '08 08:44
    The question in the book was:
    "Can you prove, without circularity, that seven is a finite number?"

    The book gives us the answer:
    "Yes, if you say that seven is three plus four, and if you agree that three and four are finit and a sum of two finite numbers is finite.
    The kind of argument that is notacceptable would be: "Seven is the sum of a finite number of ones: 1+1+1+1+1+1+1." For here you are assuming that the given string of seven ones is finite, and this is just what needs to be proven.
    By the same token, counting up to seven involves seven steps, and cannot be used in a proof of the finitness of seven.
    It is bstractly possible beings that count up to infinite numbers without noticing anything wrong."

    Is this crystal clear, or is there any doubts about it?
  9. Joined
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    18 Apr '08 09:17
    Originally posted by FabianFnas
    The question in the book was:
    "Can you prove, without circularity, that seven is a finite number?"

    The book gives us the answer:
    "Yes, if you say that seven is three plus four, and if you agree that three and four are finit and a sum of two finite numbers is finite.
    The kind of argument that is notacceptable would be: "Seven is the sum of a finite ...[text shortened]... thout noticing anything wrong."

    Is this crystal clear, or is there any doubts about it?
    Since at no point does it define what it means by "finite number", I'd say it's not a very satisfying proof.
  10. Standard memberadam warlock
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    18 Apr '08 10:321 edit
    Originally posted by mtthw
    Since at no point does it define what it means by "finite number", I'd say it's not a very satisfying proof.
    I'd say that the proof that the book gives is complete bollocks. Even the question is complete bollocks. Just like you say what is a finite number is never defined and we don't know what we have to assume that 7 is finite. If we were to go with the usual axioms of the set of real numbers I'd do something like this:

    Let us assume that there is a set of numbers that we'll call the set of the real numbers from now on and denote by R

    Lets us say that there are two operations in this set of numbers and let us call them addition and multiplication.

    Here are the axioms for the addition operation that for every two real numbers (a,b) makes a correspondende to another real number (c) a+b=c where + denotes the addition operation.

    1 - a+b=b+a
    2 - (a+b)+c=a+(b+c)
    3 - for every real number there exists one and only one real number (n) which has the following property a+n=n+a=a
    4 - for each real number exists one other real number that gives a+b=n.

    For purposes of clarification let us call the real number that we denoted as n the number zero and write as 0. And let us call the number that summs up to 0 with another real number as the symmetric of the real number a and denote it as -a.

    The axioms for the multiplication operation are:
    1 - a.b=b.a where . denotes the multiplication operation.
    2 - (a.b).c=a.(b.c)
    3 - a.(b+c)=a.b+a.c
    4 - there exists a real number e with the following property: a.e=e.a=a. Let us write 1 for e from now on and call this number one.
    5 - For every real number there exists another real number with following property: a.b=b.a=1 and this number b will be called from now on as the inverse of the number a and an be denoted as a^(-1)

    Now let us note that the numbers 0 and 1 can be different representations for the same number. But if that would be the case than the set of the real numbers would be composed of one number only and that wouldn't be very interesting. So let us say that indeed we have that 1 and o are different. (1!=0) Now we have the set of real numbers is composed of an infinite number of elements.

    But what about the ordering relationships between 0 and 1? And by the way what are these ordering relationships I speak of? Let us define them by using the usual ordering signs.

    I think that first we have to define positive and negative numbers. A number is said to be positive when a > 0 and negative otherwise. 0 is neither negative nor positive or if you wish he is negative and positive at the same time.

    We have the following property for the any given two real numbers a < b, a > b , a = b and they are mutually exclusive.

    1 - a < b implies a + c < b + c
    2- a < b implies a.c < b.c if c is positive and a.c > b.c if c is negative.

    So we are now able to proof that 1 > 0. Let us do this contradiction. 1 < 0 and then we can multiply each side of the equation by 1 so we have 1.1 >1.0 (remember that we assumed that 1 was a negative number in the beginning and so we are using property 2) which gives 1 > 0. But this is against our first assumption so it must be wrong. So now we have that indeed 1 > 0 and we have lots of real numbers to play around with one other thing that we can do is to define the inverse operations to addition and multiplication. This is easy to do and is just a case of thinking backwards. Let us define 2=1+1. Now finite is taken to be an undefined term in this theory but we know that all real numbers are finite. An infinite number is a number that isn't a real one. This seems highly childlike but I really don't see any other way around. So 2 is finite. By the same token we can define seven as being 1+1+1+1+1+1+1=7 so seven is finite.

    I think this way of approaching things is way better that the one the author of that bok used. It is more coherent and more mathematical so to speak. ut waht really pisses me of is his reasoning is this:so 3+4= 7 and since we agreed that 3 and 4 are finite 7 is too. But I never agreed that 3 and 4 are finite did I? And when was it proved that that a sum of two finite number yields a finite number as a result? But what I think is my most powerful argument is this: well 3=1+1+1 and 4=1+1+1+1 so 7 is indeed what he says we can use to prove 7 is finite. He's just masquerading that fact isn't he?

    Had to edit to correct some text that was eaten out.
  11. Standard memberGatecrasher
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    18 Apr '08 11:203 edits
    Well, assuming that, say 2, is finite.

    Infinity = Infinity/2
    But 7 < > 7/2
    Thus 7 < > Infinity, and is therefore finite.
  12. Joined
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    18 Apr '08 12:543 edits
    I would use the Merriam-Webster definition of "finite" since it seems to be defined in a manner easily tested that fits common understanding of the term.

    A finite number is any number which is less than an arbitrary positive integer and greater than the negative of that integer.

    If I choose 8 as the arbitrary number,then I can state the following true inequality.

    -8 < 7 < 8

    Since 7 is less than the arbitrary positive integer, and also greater than its negative, it is therefore finite by definition.

    I see other good answers here as well, although the book's proof is somewhat lacking, as it fails to define "finite". Also, how do you know 3 and 4 are finite?
  13. Joined
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    18 Apr '08 13:57
    Originally posted by geepamoogle
    I would use the Merriam-Webster definition of "finite" since it seems to be defined in a manner easily tested that fits common understanding of the term.

    A finite number is any number which is less than an arbitrary positive integer and greater than the negative of that integer.

    If I choose 8 as the arbitrary number,then I can state the following t ...[text shortened]... omewhat lacking, as it fails to define "finite". Also, how do you know 3 and 4 are finite?
    But that can be said about any number, surely? Apart from, of course, the Aleph's. Do Aleph's count as numbers? If not, surely every number is infinite...?
  14. Joined
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    18 Apr '08 15:27
    Originally posted by FabianFnas
    The question in the book was:
    "Can you prove, without circularity, that seven is a finite number?"

    The book gives us the answer:
    "Yes, if you say that seven is three plus four, and if you agree that three and four are finit and a sum of two finite numbers is finite.
    The kind of argument that is notacceptable would be: "Seven is the sum of a finite ...[text shortened]... thout noticing anything wrong."

    Is this crystal clear, or is there any doubts about it?
    this assumes that there's closure within the set of finite numbers. in other words, it assumes that a finite number + a finite number = a finite number. but if we can't even assume that 7 is a finite number, who says we can do so for 3 or 4? and who says that closure exists?

    to really "prove" that seven is finite, we not only have to define "finite" but probably go back to axiomatic logic similar to that of adam warlock.

    start with the fact that a=a (identity) and go from there... and this will all be very stringent and hard to do without circular logic unless i pay close attention 🙂 but it can be done.. would take a while to get through the definitions of the hindu-arabic numerals as representing physical ideas, and then the abstraction of the position system. and THEN defining the set of finite integers, integral addition, showing that the associative and commutative laws exist, proving that there is closure within the set of integers, etc.

    my point, though, is that if we can't even assume that 7 is a finite number, then at what point are we allowed to make assumptions at all? why assume 3 and 4 are finite, or even that 3 PLUS 4 is finite?
  15. Joined
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    18 Apr '08 15:33
    Originally posted by doodinthemood
    I would define "infinite" as number that, when written in the European Arabic Decimal number system, will never come to an end, and "finite" as one that would.

    The Aleph-Null point mentioned above is incorrect because "Aleph-Null" is not in this system.

    If you wanted pi as the title, you would have to write 3.14159265.... and redhotpawn would no ...[text shortened]... otpawn would let you. Any number that can fit in a redhotpawn forum title is not infinite.
    but what about 9.99999999999999...... ? it won't come to an end, and so you say is infinite.

    if x = 9.999999999....
    then 10x = 99.999999...

    and by subtraction, 9x = 90

    but then x = 10 and the original number clearly was finite to begin with.

    p.s. this is a cute algebraic method, but not a "trick" - has to do with the finite convergence of an infinite series. akin to 1/2 + 1/4 +1/8 +1/16 + ... = 1
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