Dividing a number by zero

Discussion in 'Physics & Math' started by chikis, Mar 10, 2013.

  1. chikis Registered Senior Member

    Messages:
    328
    Who can help me solve this problem:

    (0 * P) + 6 = 5
    Make P the subject of the formular. Thank you.
     
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  3. Nehushtan Registered Member

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    That equation is equivalent to \(6=5\) which is false regardless of the value of \(P\).
     
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  5. Prof.Layman totally internally reflected Registered Senior Member

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    982
    I thought it was equal to 0 = -1, in that case there is no solution for P, or it is no reals.
     
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  7. Pete It's not rocket surgery Registered Senior Member

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    10,167
    All false statements are equivalent.

    Starting from anything false (like (0 * P) + 6 = 5), you can conclude anything at all (eg 6=5, 0=-1, or even "Your mama's not fat.")
     
  8. eram Sciengineer Valued Senior Member

    Messages:
    1,877
    Actually, I think P is -∞.

    (0*(-∞))+6 = -1 + 6 = 5

    Of course it can produce any other answer but its the only "value" that fits.
     
  9. Pete It's not rocket surgery Registered Senior Member

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    10,167
    Ew, I need to scrub my eyes.
    .
     
  10. eram Sciengineer Valued Senior Member

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    1,877
    that's if you're trying to solve for P. you get P = -1/0
     
  11. GeoffP Caput gerat lupinum Valued Senior Member

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    22,087
    Many fairies have died from the reading and posting of this thread. Please, stop now.
     
  12. James R Just this guy, you know? Staff Member

    Messages:
    37,345
    \(0 \times P = 5 -6 = -1\)

    \(P = -1/0\)

    ///ERROR Will Robinson. ERROR!

    Division by zero.

    Does not compute.

    Core dump follows: 0394048583743FAB43949CD3940390AAA3485401285405885...

    Dave, my mind is going. I can feel it. I ... can ... feel ... it.

    Would you like me to sing you a song, Dave?

    Daisy, Daisy, give me your answer do...

    xxx Blurk Kerching.
     
  13. eram Sciengineer Valued Senior Member

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    1,877
    and......boom.

    Please Register or Log in to view the hidden image!

     
  14. Nehushtan Registered Member

    Messages:
    29
    And \(\frac{-1}0\) does not exist. It is NOT \(-\infty\); it simply does not exist.
     
  15. Secret Registered Senior Member

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    299
    This thread fully account of the various division by zero issues contributed by many users. It also briefly talk about infinity. The only thing it does not account for is whether there is a proof on every conceivable number system involving division by zero must end with contradictions

    Took ages to find the thread because the search function does not search any threads beyond a year ago (return only my latest two threads), which I am not sure whether its a bug or intended

    Therefore there exist no P in (R or C or Q etc.) such that (0 * P) + 6 = 5 is true
     
  16. eram Sciengineer Valued Senior Member

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    1,877
    What do you mean by "exist"? How do you define it?

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  17. KitemanSA Registered Senior Member

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    624
    Sorry, his equation was correct. Anything except zero, when divided by zero equals infinity with the appropriate sign. 0/0 and infinity*0(and their inverses) are undefined.
     
  18. rpenner Fully Wired Valued Senior Member

    Messages:
    4,833
    "Exist" in this context means "is a number in a logically well-defined system"

    Even in systems with numbers bigger than any finite numbers, transfinite numbers, multiplying any number by zero is still zero.

    Introducing a "reciprocal" of zero breaks arithmetic in many ways.

    \( \color{red} 1 = 0 \times \infty = 0^2 \times \infty = 0 \times ( 0 \times \infty ) = 0 \times 1 = 0 N = ( M - 1 ) + (N + 1 - M ) = [ ( (M - 1) \times \infty ) + ( ( N + 1 - M ) \times \infty ) ] \times 0 = ( ( M - 1 ) \times \infty ) \times 0 + 0 \times ( ( N + 1 - M ) \times \infty ) = ( ( M - 1 ) \times 1 ) + ( 0 \times \infty ) = M\)

    // Edit -- the red in the above mathematics was used to show that the above trains of symbols are Contradictions and reasons why the reciprocal of zero is not a number
     
    Last edited: Mar 11, 2013
  19. KitemanSA Registered Senior Member

    Messages:
    624
    It only breaks if you try to define 0/0 or its inverse varients.
     
  20. GeoffP Caput gerat lupinum Valued Senior Member

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    22,087
    I don't see the point of all this. I could make up any factorial illogical premise - even make the initial assertions exclusive of each other, and then claim they're not, and post a fucked up Venn diagram. So what? Why waste time with it?
     
  21. origin Heading towards oblivion Valued Senior Member

    Messages:
    11,607
    So since you say \(\frac{-1}{0} = -\infty\).

    Does that mean \( -\infty \times 0 = -1\)?
     
    Last edited: Mar 12, 2013
  22. Nehushtan Registered Member

    Messages:
    29
    \(\lim_{x\to0^-}\frac{-1}{x}=+\infty\) whereas \(\lim_{x\to0^+}\frac{-1}{x}=-\infty\) So \(\frac{-1}{x}\) goes in different directions depending on which side of 0 we approach. Hence \(\lim_{x\to0}\frac{-1}{x}\) does not exist.

    You are right in that \(0/0\) is not defined. But \(-1/0\) is not \(-\infty\) and neither is \(1/0\) equal to \(\infty\). This is an all too common mistake – and what’s more, people should stop writing things like \(-1/0\) and \(1/0\) and treating them as numbers (which they’re not).

    Perhaps you’re thinking of \(\lim_{x\to\infty}\frac1x=0\) and \(\lim_{x\to-\infty}\frac{-1}x=0\). These limits exist. However \(\lim_{x\to0}\frac{\pm 1}{x}\) do not exist (see above). If you’d only avoid writing the first two limits as \(1/\infty=0\) and \(-1/-\infty=0\) (which really you shouldn’t) then you wouldn’t be fooled into thinking that \(1/0=\infty\) and \(-1/0=-\infty\).
     
  23. Pete It's not rocket surgery Registered Senior Member

    Messages:
    10,167
    Tex Tip:
    In tex, "x" shows as a cursive letter x, and so shouldn't be used for a multiplication symbol. Use \times instead.
    \(7 \times 6 = 42\)​
     

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