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Ripped off!

  1. Jan 27, 2005 #1
    This is rather silly...

    Imagine you paid $10 for a $5 calculator. Then you paid twice the real amount.
    Imagine you paid $100 for a $1 pencil. Then you paid 100 times the real amount.

    So, if you paid $1 for a free bag, how many times the real amount did you pay?

    :rofl: :tongue: :eek:
  2. jcsd
  3. Jan 27, 2005 #2
    There's no such thing as a free lunch. :tongue: :frown:
  4. Feb 2, 2005 #3
    Infinity, albeit a smaller one than if you had paid $2 (or so my reasoning goes)
  5. Feb 2, 2005 #4


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    ...and what might that reasoning be ?
  6. Feb 2, 2005 #5
    err flawed?

    [itex]lim_{x \rightarrow 0} \frac{2}{x} = \inf_1[/itex]

    hence 0 (call this 01) [itex]0 * \inf_1 = 2[/itex]

    [itex]lim_{x \rightarrow 0} \frac{3}{x} = \inf_2[/itex]

    [itex]0 * \inf_2 = 3[/itex]

    assuming the 0's are the same thing (yes I also make a distinction about the zeroes... I'm weird,) then infinity 2 has to be bigger than infinity 1.

    *wow, I don't know how to use Itex.
  7. Feb 3, 2005 #6


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    Even allowing everything till this point :

    [itex]0 * \inf_1 = 2[/itex]
    [itex]0 * \inf_2 = 3[/itex]

    How do you go from there to ... "infinity 2 has to be bigger than infinity 1" ?
  8. Feb 3, 2005 #7
    Because 0 * inf2 > 0 * inf1

    divide both sides by zero and you get that inequality...

    Yeah I know you can't divide by zero, but if you allow for the definitions of 0 and infinity I have then it's' ok.
  9. Feb 3, 2005 #8


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    I'd've started complaining that he wrote [itex]0 * \infty_1 = 2[/itex].

    That's the problem: what you've said only applies to your invention, not to the arithmetic that everybody else uses.
  10. Feb 3, 2005 #9
    well yeah, I'm trying to translate limits of function to this. if you subsitute 0s and infinities with function and put the expressions inside limits, then.... you catch my drift.

    Anyway, it just seems somewhat logical. What is your answer if any for the question 0 * infinity? I would say you need to know to what degree each is 0 and infinity (expressed in function forms with a limit.) Maybe I should've expressed myself like that before.
  11. Feb 3, 2005 #10


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    That, for the reals it's a nonquestion, since infinity is not a real number.

    For the extended reals, it's undefined, because no choice of value would allow the * operator to be continuous there.

    For cardinal numbers, it's 0.

    It all depends on the definition being used.
  12. Feb 3, 2005 #11

    Well you could describe it with functions that behave a certain way for a given limit.. I think. It's not really important.
  13. Feb 3, 2005 #12


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    Behavior of functions near a point (or near the infinity of the extended reals) is, indeed, an important concept. In fact, one might say that's what calculus is all about. :smile:
  14. Feb 3, 2005 #13
    So could you figure it like this:

    [tex]\lim_{x\rightarrow 0} f(x) * \lim_{y\rightarrow \infty} f(y) = 1[/tex]

    where [tex]\lim_{x\rightarrow 0} f(x) = 0[/tex]
    and [tex]\lim_{y\rightarrow \infty} f(y) = \infty[/tex]

    and then

    [tex]\lim_{x\rightarrow 0} f(x) * \lim_{y\rightarrow \infty} g(y) = 2[/tex]

    setting the two in an inequality, couldn't it be said that the limit of g(y) approaches infinity more quickly than f(y). I personally take this to mean that g(y)'s infinity is greater than f(y)'s... It's not really math at that point, more of philosophy really. No?
  15. Feb 4, 2005 #14


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    You haven't gotten rid of the problem: you're still trying to do [itex]0 * \infty[/itex], you just replaced the two symbols with some other equal expression.

    Now, you could say this (and be perfectly right):


    (all of the following are as x --> infinity)

    f(x) --> 0
    g(x) --> infinity
    h(x) --> infinity
    f(x) g(x) --> 1
    f(x) h(x) --> 2

    Then, (I'm pretty sure that) you can conclude that

    h(x)/g(x) --> 2
    |h(x)| - |g(x)| --> infinity

    Which is a more precise statement of what you're trying to say.

    In particular, notice that the arithmetic operations are inside the limit.

    (BTW, f(x) --> 0 as x --> infinity means the same thing as [itex]\lim_{x \rightarrow \infty} f(x) = 0[/itex])
  16. Feb 4, 2005 #15


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    Hahaha, rubbish mathematics about infinities, every mathematicians nightmare to try and explain why the person who wrote it isn't being logical at all. Anyway here is the equivalent I've heard:

    "A mathematician goes fishing, on the way he passes a fish bait show, on the sign outside it says 'No matter how much fish bait, $1'. So the mathematician walks in and asks for $2 worth"
  17. Feb 4, 2005 #16
    Rubbish math for a less than serious question... Honestly, stop being so critical. Had the question been asked in earnest, then fine.
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