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UnkleRhaukus

\[\infty\times0=\]

  • one year ago
  • one year ago

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  1. Fox375
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    0!!!!

    • one year ago
  2. UnkleRhaukus
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    @lgbasallote

    • one year ago
  3. theredhead1617
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    0

    • one year ago
  4. satellite73
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    definitely \(\pi\)

    • one year ago
  5. Fox375
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    @satellite73 IKR

    • one year ago
  6. UnkleRhaukus
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    what is your evidence @satellite73

    • one year ago
  7. satellite73
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    oh wait, i am wrong it is \(e^2\) sorry

    • one year ago
  8. lgbasallote
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    you can rewrite \[\infty \times 0\] as \[\infty \times \frac 1\infty\] right? then it becomes l'hospital...

    • one year ago
  9. lgbasallote
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    assuming of course these are functions...

    • one year ago
  10. UnkleRhaukus
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    there is another way

    • one year ago
  11. lgbasallote
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    really? hmm

    • one year ago
  12. lgbasallote
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    change infinity to 1/0 to make it 0/0?

    • one year ago
  13. lgbasallote
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    either way...it's indeterminate....

    • one year ago
  14. satellite73
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    if this is a serious question, presumably it is about limits, i.e. if \(\lim_{x\to\infty}f(x)=\infty\) and \(\lim_{x\to \infty}g(x)=0\) then what is \[\lim_{x\to \infty}f(x)g(x)\] the answer is it could be anything, it depends on \(f\) and \(g\) the form is not determined

    • one year ago
  15. CliffSedge
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    \[\large a \times b = c\] \[\large a=\frac{c}{b}\] \[\large b=\frac{c}{a}\] Let a = ∞, b=0 \[\large ∞=\frac{c}{0}\] \[\large 0=\frac{c}{∞}\] But c/0 is undefined and so is c/∞, right? What if c were positive? What if c were negative?

    • one year ago
  16. satellite73
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    so it could be \(\pi\) or it could be \(e^2\) or it could be anything

    • one year ago
  17. lgbasallote
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    an interesting question though is... \[\frac 1 \infty = 0\] therefore... \[0 \times \infty = 1\] that should be right?

    • one year ago
  18. lgbasallote
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    just expressing how weird math is

    • one year ago
  19. satellite73
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    not at all

    • one year ago
  20. satellite73
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    infinity is not a number, and so \(\frac{1}{\infty}\) is not a number either

    • one year ago
  21. CliffSedge
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    Agree with @satellite73 To say that 1/∞ = 0, you have to take the limit of 1/x as x-->∞ and then 0 × (x is really really big) still equals 0.

    • one year ago
  22. UnkleRhaukus
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    |dw:1349968485543:dw|

    • one year ago
  23. lgbasallote
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    math is still weird....

    • one year ago
  24. CliffSedge
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    (I'm going to have to go get coffee, then come back for this. Unkle is about to get all Twilight Zone on us, I can feel it.)

    • one year ago
  25. satellite73
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    you are making short cut statements about limits namely if \[\lim_{x\to\infty}f(x)=\infty\] then \[\lim_{x\to \infty}\frac{1}{f(x)}=0\]

    • one year ago
  26. lgbasallote
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    i never really understood the function of limits...

    • one year ago
  27. lgbasallote
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    math is too ambiguous for me

    • one year ago
  28. satellite73
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    \(\infty\) is not a number \(\infty\times 0\) is not a number \[\frac{5}{\infty}\] is not a number

    • one year ago
  29. CliffSedge
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    (aside: Anybody hear the full treatment of 'Hilbert's Hotel?')

    • one year ago
  30. CliffSedge
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    Is the slope of the vertical line +∞ or -∞ @UnkleRhaukus ?

    • one year ago
  31. UnkleRhaukus
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    i dont know what 'Hilbert's Hotel?' is

    • one year ago
  32. CliffSedge
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    Check it out some time. David Deutsch gives a good telling of it in his book, 'The Beginning of Infinity.' (Fantastic book on the philosophy of science; I recommend it to everyone.)

    • one year ago
  33. UnkleRhaukus
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    the product of the slopes of perpendicular lines is

    • one year ago
  34. CliffSedge
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    Yes, but what is the slope of the vertical line?

    • one year ago
  35. UnkleRhaukus
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    \[\pm\infty\]

    • one year ago
  36. helder_edwin
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    this might help

    • one year ago
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  37. JamesWolf
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    infinity is stupid. \[\infty + 1 = \infty \] \[1 = \infty - \infty = 0\]

    • one year ago
  38. UnkleRhaukus
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    \[\infty\times 0=-1\]

    • one year ago
  39. JamesWolf
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    all the even numbers = \[\infty\] which is smaller than all the numbers = \[\infty\] so \[\infty < \infty\]

    • one year ago
  40. CliffSedge
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    \[\large 2 \times ∞ \times 0 = -2?\]

    • one year ago
  41. CliffSedge
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    @JamesWolf there are different cardinalities of infinity. Research "Aleph Numbers" for more info. The infinity of integers is less than the infinity of real numbers, etc.

    • one year ago
  42. Jemurray3
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    The bastardization of mathematics that has taken place in this thread is like salt rubbed into a fresh wound of my soul. Is there a legitimate question, or are you guys just playing around? :)

    • one year ago
  43. 03453660
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    this is definitely an undefined case because any number multiplied with zero become zero and any number multiplied with infinity become infinity that's why both of these cases are possible here therefore we cannot consider one these case separately. thus this is an undefined case.

    • one year ago
  44. CliffSedge
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    @Jemurray3 I don't think there's anything serious going on. Unkle came up with a cute link to slopes of parallel lines, but made the error in saying that vertical lines have a slope, m=∞, which, of course, is false.

    • one year ago
  45. Coolsector
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    if the question is something like lim x->inf of 0*x then it is 0 otherwise we should be more specific about the problem i guess

    • one year ago
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