Dividing a number by zero

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

  1. Pete It's not rocket surgery Registered Senior Member

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    1+0 is odd, therefore 0 is even.
    Is there any sense in which zero can be considered an odd number?
     
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  3. arfa brane call me arf Valued Senior Member

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    I guess not.
    But that doesn't mean you can't define it as having no parity, since then odd + no_parity = odd, etc.

    Why or whether it's a good idea is another matter.
     
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  5. Emil Valued Senior Member

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  7. Prof.Layman totally internally reflected Registered Senior Member

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    \(\lim_{a \to 0} \frac{0}{a} = 0\)

    \(\lim_{a \to 0} \frac{a}{0} = undefined\)

    \(\lim_{a \to 0} \frac{2a}{a} = 2\)

    In the first example, I think 0 divided by any number is 0, so then the answer to that one would be zero. This seems to show that the zero comes from a different variable than "a", so then it would seem that zero's coming from different sources cannot be canceled. If it came from the same variable it could be canceled since it would still be an "a".

    In the secound example, any number divided by zero is undefined, so then I don't think it would have an answer since the limit would never say "a" is actually 0. That does seem odd that you would get a different answer in this limit even though it seems to really be saying that it is searching for the limit of the same division problem \( \frac {0}{0} \). In this case it is undefined and in the other it is 1. But since this example is undefined, I think the real solution for \( \frac {0}{0} \) would be 1. The problem has only been rearranged so that it is undefined. In the example \(\lim_{a \to 0} \frac {a}{a} \) you never actually have to divide by zero.

    In the third, any number divided by itself would be one, so every answer would then be 2. It is as simple as canceling out the "a" in the equation even if "a" approuches zero. You shouldn't allow yourself to think that just because "a" approuches zero, that it would give you a different answer rather than just canceling them out, you can cancel \( \frac {0}{0} \) in limits and that should always be 1, except in the first case where the zero comes from a constant. If you had a constant of zero and was trying to find the value of a variable that is also zero, then there would be a good chance that the whole thing is just zero.
     
    Last edited: Mar 13, 2013
  8. origin Heading towards oblivion Valued Senior Member

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    I think the problem is you are treating \(a^2 - b^2 = 0 \) like it is a quadradic equation. A quadradic equation is in the form \(ax^2 + bx + c \). The quadradic can of course also be in the form \(ax^2 - b^2 = 0\).

    But \(a^2 - b^2 = 0 \) is not in the right form it is essentially \(b^2 - b^2 = 0 \) or \(x^2 - x^2 = 0 \) so yu cannot expand it using the rules for quadradics.
     
  9. Prof.Layman totally internally reflected Registered Senior Member

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    They do always say that quadradic's cannot be equal to zero, even if there is not division in it by zero. Never got why they cared to do that until now, it seemed kind of redundant in everything. I just thought that they really didn't like the number zero.

    Or they will usually say that "a" is not equal to "b".
     
  10. RJBeery Natural Philosopher Valued Senior Member

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    The "trick" is in moving from step 4 to step 5. Division by (a-b) is equivalent to division by zero, and the point is that dividing by zero leads to contradictions.
     
  11. origin Heading towards oblivion Valued Senior Member

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    Painfully obvious, now that you pointed it out.
     
  12. KitemanSA Registered Senior Member

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    Almost. It is the dividing of zero by zero that leads to the contradiction.
     
  13. Pete It's not rocket surgery Registered Senior Member

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    No, Kiteman. In the given example, there is no division of zero by zero.
     
  14. Emil Valued Senior Member

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    :worship:
     
  15. KitemanSA Registered Senior Member

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    624
    Surely you jest!

    (a+b)(a-b)=b(a-b)

    Then the actual next step, which was tricked past is

    (a+b)(a-b)/(a-b)=b(a-b)/(a-b)

    Where (a-b)/(a-b) in any other case would be 1 but in this case is undefined. Duh!
     
  16. Pete It's not rocket surgery Registered Senior Member

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    Must have had my 'stupid' hat on.
     
  17. funkstar ratsknuf Valued Senior Member

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    So, in other words, your comment amounts to: "if we take this well-defined concept, and redefine it to mean something else, then it can mean something else."

    Such insight.
     
  18. Prof.Layman totally internally reflected Registered Senior Member

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    The only time \( a^{2} - b^{2} = ab - b^{2} \) is when \( a = b \), I would be willing to bet that in almost every math book that teaches these concepts, says "a" does not equal "b". \( a^{2} - b^{2} \) could never equal \( ab - b^{2} \), so then the statement has already gone wrong before there was even actual division by zero. You would be saying that you are factoring out a zero out of each side of the equation, before you even divided by zero. The division by zero is just another error on top of errors already made. If you assumed that "a" and "b" is actually zero, then it would be the only time it is correct, 0+0=0. How do you factor out a zero out of something that is non-zero?

    a=0

    b=0

    The only way you could factor out a zero and it still be a true statement would be if the value of it was already zero. So then "a" and "b" are equal to zero and equal to each other, so then these mathmatical rules do not even apply.
     
  19. Pete It's not rocket surgery Registered Senior Member

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    Right. We're talking about Emil's post, which starts from a=b:
     
  20. Prof.Layman totally internally reflected Registered Senior Member

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    Well ya, and that would have been the first error in the equations.
     
  21. RJBeery Natural Philosopher Valued Senior Member

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    Prof.Layman, there's nothing wrong with proclaiming that a=b. The subsequent equations are derived from that fact.
     
  22. Prof.Layman totally internally reflected Registered Senior Member

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    I would say that it is mathmatical proof that there is something wrong with claiming a=b. I never seen any algebra books that claimed that \( a^{2} = ab \) from then on it is just nonsense.

    When you substitute equations into other equations, then it makes the variable have to agree with all of those equations. So then by saying that a=b, and then factoring out a zero from both sides, the solution to "a" and "b" would then have to be zero. It is the only number that you could factor out a zero and not change its value. Everything multiplied by zero is zero, so if it had a zero multiplied in it and you factored that out, then that number would have to be zero.

    I think this is why they say "a" does not equal "b", and they cannot be equal to zero. The algebra just wouldn't work out anymore, as shown. I think this is a good example of that. It is never said that "a" can equal "b" in any fundemental math principals.
     
  23. arfa brane call me arf Valued Senior Member

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    If you write a program that defines 0 as having no parity, then 0 doesn't 'behave' arithmetically. That doesn't mean the program can't do arithmetic.

    If a = b, then a[sup]2[/sup] - b[sup]2[/sup] = 0. The rest of emil's equations skip over this; the equation a + b = b is true if a and b are both 0.
     

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