What is infinity subtracted from infinity?

Discussion in 'The Cesspool' started by Betrayer0fHope, Oct 27, 2008.

Thread Status:
Not open for further replies.
  1. Betrayer0fHope MY COHERENCE! IT'S GOING AWAYY Registered Senior Member

    Messages:
    2,311
    In as many different maths as you care to tell me. Please.

    Edit: And in the physical world
     
  2. Guest Guest Advertisement



    to hide all adverts.
  3. camilus the villain with x-ray glasses Registered Senior Member

    Messages:
    895
    \(\infty - \infty \not= 0\), period.

    German mathematician Georg Cantor proved that infinity can have different sizes, and it could be easily demonstrated using limits.

    \( \lim_{n \rightarrow \infty} n = \infty\)

    now, obviously so does

    \( \lim_{n \rightarrow \infty} n^2 = \infty\)

    but also obviously, n^2 grows faster than n, so even though they both equal infinity, if these two limits were subtracted, it would not equal zero because one infinity is bigger than the other.

    Another way to visualize it is using infinite series.

    Cantor proved that infinity can have different magnitudes, and that it could be easily demonstrated that they have different magnitudes using limits to show that \(\infty - \infty \not= 0\)., or even infinite series. Im not saying Cantor was saying that, its just how my calc instructor made the idea more concrete.

    for example, would you agree that \(\sum_{n=1}^{\infty} n = 1 + 2+3+4+\dots\) diverges to \(\infty\)

    So will \(\sum_{n=1}^{\infty} n^2 = 1+4+9+16+\dots\)

    now since both equal infinity, if we were to subtract the second by the first, we would be subtracting \(\infty - \infty\). But its easy to see that the \(\sum_{n=1}^{\infty} n^2\) grows faster than \(\sum_{n=1}^{\infty} n\) so we would not get \(\sum_{n=1}^{\infty} n^2 - \sum_{n=1}^{\infty} n = \infty - \infty \not= 0\).
     
  4. Guest Guest Advertisement



    to hide all adverts.
  5. camilus the villain with x-ray glasses Registered Senior Member

    Messages:
    895
    in laymans terms, this would look like this,
    \(\sum_{n=1}^{\infty} n^2 - \sum_{n=1}^{\infty} n \not= 0\)

    which is the same as

    \( (1+4+9+16+25+36+49+\dots) - (1 +2+3+4+5+6+7+\dots) \not= 0\)

    The left side, \(n^2\) grows faster, and even if you go canceling terms, the right side will stiill always have larger numbers due to the exponentiation.
     
    Last edited: Oct 27, 2008
  6. Guest Guest Advertisement



    to hide all adverts.
  7. Crunchy Cat F-in' *meow* baby!!! Valued Senior Member

    Messages:
    8,423
    Infinity isn't a number

    Please Register or Log in to view the hidden image!

    and of course camilus is quite correct.
     
  8. Steve100 O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔O Valued Senior Member

    Messages:
    2,346
  9. D H Some other guy Valued Senior Member

    Messages:
    2,257
    Cute, but wrong. The left hand side and right hand side are both undefined. That does not mean they are equal. The expression is ill-formed.
     
    Last edited: Oct 27, 2008
  10. Steve100 O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔O Valued Senior Member

    Messages:
    2,346
    Of course, what a tit I am.
     
  11. D H Some other guy Valued Senior Member

    Messages:
    2,257
    To answer the OP: If you are talking about the point at infinity, the answer is not defined. If you are talking about two transfinite numbers \(\sigma\) and \(\mu\), \(\sigma - \mu = \sigma\) iff \(\mu < \sigma\)
     
  12. Reiku Banned Banned

    Messages:
    11,238
    Though, we can subtract two infinities and hope they equal zero, as in renormalization theory. Though most times, they do not and leave non-zero values.
     
  13. prometheus viva voce! Registered Senior Member

    Messages:
    2,045
    Renormalisation is a very precise way of getting finite answers from QFT. It's not simply "subtracting infinity from infinity and getting zero."
     
  14. Reiku Banned Banned

    Messages:
    11,238
    I know. I was simply pointing out if you can be left with a finite answer, then there is also the possibility of total renormalization. A zero remainder otherwise.
     
  15. AlphaNumeric Fully ionized Registered Senior Member

    Messages:
    6,702
    "Total" renormalisation? Renormalisation is about getting finite, meaningful answers from otherwise divergent quantities. Since zero is finite, there's no distinction between getting a contribution and no contribution when you're talking about renormalisation. Not least because sometimes the contribution you get from particular diagrams in a perturbative expansion are dependent upon your renormalisation scheme. The one I'm thinking of is the gluon self energy one loop diagram with a 4 gluon vertex.

    There is 'super renormalisable', where a particular process has only finitely many graphs which need renormalising, but 'total renormalisation' is a phrase you have concocted in order to make it seem like you have the first clue about renormalisation. Which you don't.
     
  16. amark317 game developer-in-training Registered Senior Member

    Messages:
    252
    infinity is any real number that you can add 1 to.
    assuming that both infinities are the same, then the answer would be zero.
    in any other situation, it could be anything.
     
  17. Enmos Valued Senior Member

    Messages:
    43,184
    It's undefined.
     
  18. Steve100 O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔O Valued Senior Member

    Messages:
    2,346
    So the number 64.22209 is infinity?
     
  19. D H Some other guy Valued Senior Member

    Messages:
    2,257
    1+1=2, 2+1=3, 1002.5+1=1003.5. While 1, 2, and 1002.5 are all real numbers to which I can add 1, none of them is infinite. "Infinity", on the other hand, is not a real number. All real numbers are finite by definition.
     
  20. Steve100 O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔O Valued Senior Member

    Messages:
    2,346
    I doubt it, but does infinity count as an integer?

    Silly question, but I'd just like someone in the know to back me up (or not).
     
  21. D H Some other guy Valued Senior Member

    Messages:
    2,257
    No. The real numbers are the set \((-\infty,\infty)\). The parentheses denote that "infinity" is not part of the set. There is no largest real number, but all real numbers are nonetheless finite. The same pertains to the integers. While there is no largest integer, all integers are finite.
     
  22. Reiku Banned Banned

    Messages:
    11,238
    Thanks again for painting a picture that didn't exist. In no way did i suggest i knew more, simply because i used ''total'' in the phrase.

    Indeed, it was my own words, because i somewhat struggle to find words when there is none to example inasmuch prometheus' point, that renormalization leaves finite answers. I knew he meant non-zero answers, to which i suggested a total renormlization, which was nothing more than my inability to describe a zero value, instead of something left over.

    So no, it had nothing to do with some ego on my behalf, moron.
     
  23. Steve100 O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔O Valued Senior Member

    Messages:
    2,346
    Ta.
     
Thread Status:
Not open for further replies.

Share This Page