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Easy Questions, Difficult Answers

  1. Apr 8, 2007 #1
    Hi. I know the following two questions might seem stupid, but I really need the correct answers in order to better understand the way we deal with numbers.


    1. How can we prove that (0! = 1)?
    We all know (4! = 4*3*2*1), (3! = 3*2*1), (2! = 2*1), (1! = 1)
    But for 0, (0! = 1) doesn't make sense. Don't you agree?

    2. Is number 1 a prime? Why? Why not?
    Is it true that some mathematicians refuse to consider number 1 a prime? :confused:


    Thanks in advance
     
  2. jcsd
  3. Apr 8, 2007 #2

    cristo

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    n!=n(n-1)(n-2)...3.2.1, thus n!=n(n-1)!. Consider 1!, which you agree is equal to 1, then this can be expressed as 1=1!=1.0!, and so 0!=1
    A prime number, p, is a number which has exactly two positive integer factors (namely 1 and p). Since 1 has only one factor, it is not a prime number.
     
    Last edited: Apr 8, 2007
  4. Apr 8, 2007 #3
    However,

    the definition
    n!=n(n-1)(n-2)…3.2.1
    means the factorial of any non-negative integer, n, is the product of n and all integers we obtain by decrementing n continuously by 1; and we would stop decrementing when number 1 is reached. Therefore, for any value of n, (n-1) cannot be < 1. Otherwise, the factorial of n is n.
    Now, apply this to 1! :uhh:

    What do you think?


    This isn't fair. You cannot prove using a fictional definition

    :uhh:
     
  5. Apr 8, 2007 #4

    matt grime

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    0! is *defined to be 1*
    1 is *defined not to be a prime*


    They are just definitions. They are useful and completely justifable ones from practical view points. And there is nothing to discuss about them beyond that. So you're better saving you energy and doing something else than attempting to redefine everything in mathematics.

    n! is defined to be n.(n-1)!, for n>0 and 0!=1. A prime is an object which is not a unit and satisfies p|ab implies p divides one of a or b. Since 1 is invertible (i.e. a unit), it cannot be a prime.
     
    Last edited: Apr 8, 2007
  6. Apr 8, 2007 #5

    arildno

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    You seem to believe, Rhythmer, that there exists some ineffable truth about mathematical structures/quantities that our definitions more or less are able to approximate or encapsulate.

    This is a false view of maths, albeit a very common one.

    What is true in maths, is quite simply what we SAY is true, and whatever that is derivable from rules that we SAY are valid rules.
    That's it. Get used to it.
     
    Last edited: Apr 8, 2007
  7. Apr 8, 2007 #6
    A practical example of a case where 0! must be 1 is in the definition of the power rule for taking derivatives, which is

    [tex]
    D_x^n \big[ x^k \big] = \frac{k! \, x^{k-n}}{(k - n)!}
    [/tex]

    This formula will give you the nth derivative of an expression like x^k. You'll notice that if n=1 (we take the first derivative) and k=1 (such as in "x^1") then, if 0! does NOT equal 1, we won't get the right answer for the derivative. For example, if you thought that 0! = 0, the expression would divide by zero since (k-n) = 0 is in the denominator of the expression. However, if we let 0! = 1, we get the correct answer: the derivative of x^1 is 1! * x^(0) / (0!) = 1 * 1 / 1 = 1.

    EDIT: ignore this :uhh:
     
    Last edited: Apr 8, 2007
  8. Apr 8, 2007 #7

    arildno

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    Since that is a subsidiary definition UTILIZING the standard definition of the factorial, the result isn't very surprising, is it?
     
  9. Apr 8, 2007 #8

    HallsofIvy

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    And it isn't fair for you to assume that a definition you do not recognize is "fictional"! That is a perfectly valid and commonly used definition. What do you say is the definition of "prime number"?
     
  10. Apr 8, 2007 #9
    Of course, it's not a proof, but the method used to derive that formula doesn't necessarily assume the part of the definition that states 0! = 1, just the part that n! = n(n-1)(n-2)...3.2.1. In order for it to hold true for the case where n=1 and k=1, we must have defined that 0! = 1. It's just a practical example showing how it is more useful to define the factorial that way.
     
  11. Apr 8, 2007 #10

    matt grime

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    That doesn't make any sense, gabee. You're saying that the theorem holds becuase of the definition of the factorial, and thus, because the theorem holds, the definition of the factorial must mean 0!=1. That is circularity. Your statement doesn't even mention (as it ought to) that you're requiring k=>n, as well. So if we need that restriction, why not the restriction to k>n? Using your logic, the symbol k!/(k-n)! is defined, and is zero whenever n>k.
     
    Last edited: Apr 8, 2007
  12. Apr 8, 2007 #11
    Ah, I understand what you're saying...whoops! That's not a very good example. :X Thanks.
     
  13. Apr 9, 2007 #12
    I think a prime number is a number that has no factors other than itself and 1

    People are not to decide whether a prime number itself can be 1

    Could you give us any example of a case where 1 must not be a prime number?
     
  14. Apr 9, 2007 #13

    arildno

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    You think that? Or do you choose that?

    Ok.
    Who's to decide it, then?
    1. In the formulation of just about every theorem there is concerning primes, for example in the fundamental theorem of arithmetic.

    If we were to use your definition of primes, then we'd need to reformulate these theorems by appending "except fo the number 1" to most of them.

    2. There exists definitions of number sets that can be proven to be equivalent to the set of primes, using the standard definition of primes.
    This would not be the case if we were to use your definition of primes.
     
  15. Apr 9, 2007 #14

    matt grime

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    So people aren't to decide. Except you who can decide what is and isn't prime. Do you see the contradiction in that position?

    A definition is just a definition. It makes more sense to exclude 1 from the list of primes than to include it. Thus we have come to have a definition which explicitly prohibits primes from including 1. There is no conspiracy here. Some centuries ago 1 was considered a prime. Then it was realised that that definition created issues, and better characterisations are used that get round these issues. Primes crop up all the time in number theory, and cases of divisbility. 1 divides every number, and is thus special and will need to be excluded every time, so why not exclude it by definition?
     
  16. Apr 9, 2007 #15
    Why do you think that? Because someone told you so.

    One of the nice things about mathematics is that you can define anything that you like! The standard definition for prime happens not to include 1 (of course, you can also state equivalent definitions without ever explicitly mentioning 1; for example, "a natural n is prime iff n has exactly two distinct natural divisors").

    The reason that 1 is not prime under the standard definition is simply that it makes things more tedious if you include it. That's the same reason that, say, 6 isn't prime under the standard definition (you'd have to exclude 6 from a bunch of results on primes; the same would apply for 1).
     
  17. Apr 9, 2007 #16
    Well guys I'd like to emphasize that my purpose of this thread is just
    as I said before.

    I'm not attempting to redefine the factorial of 0 or to let 1 be a prime but what I'm trying to do is to discuss the topic with 'neutrality'. I do think all of you are right and I'm convinced to all of what you're saying. I really appreciate your posts.

    Let's forget it!
     
  18. Apr 9, 2007 #17

    arildno

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    What do you mean by discuss with "neutrality"?
     
  19. Apr 9, 2007 #18
    I mean I was trying to be "neutral" and not biased against people who don't believe 1 is not a prime, etc.
     
  20. Apr 9, 2007 #19

    arildno

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    What is there to "believe" about it?

    Definitions are all-important, adequate, and arbitrarily chosen.
     
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