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UnkleRhaukus

  • 2 years ago

\[\infty\times0=\]

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  1. Fox375
    • 2 years ago
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    0!!!!

  2. UnkleRhaukus
    • 2 years ago
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    @lgbasallote

  3. theredhead1617
    • 2 years ago
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    0

  4. satellite73
    • 2 years ago
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    definitely \(\pi\)

  5. Fox375
    • 2 years ago
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    @satellite73 IKR

  6. UnkleRhaukus
    • 2 years ago
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    what is your evidence @satellite73

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

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

  9. lgbasallote
    • 2 years ago
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    assuming of course these are functions...

  10. UnkleRhaukus
    • 2 years ago
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    there is another way

  11. lgbasallote
    • 2 years ago
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    really? hmm

  12. lgbasallote
    • 2 years ago
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    change infinity to 1/0 to make it 0/0?

  13. lgbasallote
    • 2 years ago
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    either way...it's indeterminate....

  14. satellite73
    • 2 years ago
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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

  15. CliffSedge
    • 2 years ago
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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?

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

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

  18. lgbasallote
    • 2 years ago
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    just expressing how weird math is

  19. satellite73
    • 2 years ago
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    not at all

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

  21. CliffSedge
    • 2 years ago
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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.

  22. UnkleRhaukus
    • 2 years ago
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    |dw:1349968485543:dw|

  23. lgbasallote
    • 2 years ago
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    math is still weird....

  24. CliffSedge
    • 2 years ago
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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.)

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

  26. lgbasallote
    • 2 years ago
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    i never really understood the function of limits...

  27. lgbasallote
    • 2 years ago
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    math is too ambiguous for me

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

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

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

  31. UnkleRhaukus
    • 2 years ago
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    i dont know what 'Hilbert's Hotel?' is

  32. CliffSedge
    • 2 years ago
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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.)

  33. UnkleRhaukus
    • 2 years ago
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    the product of the slopes of perpendicular lines is

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

  35. UnkleRhaukus
    • 2 years ago
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    \[\pm\infty\]

  36. helder_edwin
    • 2 years ago
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    this might help

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

  38. UnkleRhaukus
    • 2 years ago
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    \[\infty\times 0=-1\]

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

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

  41. CliffSedge
    • 2 years ago
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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.

  42. Jemurray3
    • 2 years ago
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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? :)

  43. 03453660
    • 2 years ago
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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.

  44. CliffSedge
    • 2 years ago
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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.

  45. Coolsector
    • 2 years ago
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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

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