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Why Isn't x/0 = x, Instead of Undefined?

  1. Nov 20, 2014 #1
    Hi everyone,

    I'm Pre-Calc here and have a question about a concept that is troubling me. I know that anything divided by zero is undefined. However, I'm not sure why this is or what we even mean by it.

    To me, the more intuitive result would be that x/0 = x, instead of being undefined. If I imagine, say, a pizza pie and I divide it 0 times, then that would conceptually mean that I didn't divide it at all and I still have exactly what I started with (which is "x"). Am I like the only one who thinks that is more common sense?

    Instead, we're told the answer is "undefined," which I'm not sure of the meaning of. What does it even mean? We're just told in school that you can't divide by zero, but never why or what "undefined" means.

    So, how do we make sense of this? Am I the only one who thinks this is odd?
     
  2. jcsd
  3. Nov 20, 2014 #2

    jedishrfu

    Staff: Mentor

    Think of it like this if I divide a value by 2 I get half the value if I divide by 1 I get the value and if I divide by .5 I get twice the value so as the denominator approaches zero the resultant answer tends toward infinity ie dividing by zero is like multiplying by infinity both of which are really undefined.
     
  4. Nov 20, 2014 #3
    Yes and I would also add that dividing by zero brings up the concept of discontinuity. Define 0+ as a number slightly positive (ex:0.0000000...00001) and define 0- as a number slightly negative (ex: -0.0000000....00001) If you divide 2/0+ you would get infinite positive. But if you divide 2/0-, you get infinite negative. So when you position yourself in between those two - exactly at 0 - it is unclear whether you are infinite positive or infinite negative - there is such an abrupt change, that we call discontinuity. Therefore, we consider this operation 2/0 as ''undefined''. The same apply to x/0. Here is a picture that can help.
     

    Attached Files:

  5. Nov 20, 2014 #4

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    If x/0 = x, then this also implies that x/x = 0. You might want to stop relying on your intuition too much.
     
  6. Nov 20, 2014 #5

    Drakkith

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    Staff: Mentor

    My understanding:

    Let's look at a simple division problem, 10/x=2.
    This means that there is a number, which we can put in place of x, that will divide into the number 10 two times. Obviously the answer is 5.
    Since division is the inverse of multiplication, this equation also means that there is a number which can be multiplied by 2 to equal 10, which of course is 5.

    So what happens if x = 0?
    Well, in that case, 10/0 = Y. Rearranging the equations gives us Y x 0 = 10. This would mean that there is some number for Y that will multiply with zero to equal 10. But wait! Any number multiplied by zero equals zero! (including zero times zero) Since there is no number we can put in for Y to make this equation make sense, we say that dividing by zero is undefined.

    The thing is, 10/5 is not saying take 10 and divide it 5 times, it's saying take 10 and divide it by 5. What does "divide X times" even mean? How am I dividing this pizza up? In halves? Slices?
     
  7. Nov 20, 2014 #6

    Mark44

    Staff: Mentor

    That would be x/1, which conceptually means that you "divided" the pizza into one piece. IOW, you didn't really divide it at all. If you make a cut right through the middle, the pizza is now in two pieces, each of size x/2.
     
  8. Nov 27, 2014 #7
    I agree with jedishrfu's *calculus-based* reasoning.
    Like he said,
    ##\frac{1}{2} = 0.5##
    ##\frac{1}{1} = 1##
    ##\frac{1}{0.5} = 2##
    ##\frac{1}{0.05} = 20##
    ##\frac{1}{0.0005} = 2000##
    So, what's the pattern here? As the denominator gets small, the overall quotient gets large. So, what happens when the denominator is closer and closer to zero? Well, the overall value gets larger and larger until it approaches ##\infty##.
    So,
    ##\lim_{n \to 0^-} \frac{1}{n} = -\infty## and ##\lim_{n \to 0^+} \frac{1}{n} = +\infty##
    Basically, we say "undefined" because we can't actually quantify ## \pm \infty##.
    If you looked at a plot of this function, then at ##n = 0##, there'd be a vertical asymptote.
    And, since you're fond of the "dividing is just cutting into pieces" reasoning, consider the fact that
    ##\frac{1}{\pm \infty} = 0##, because if you divide 1 into infinite pieces, then each piece will be extremely small. Now, cross multiply:
    ##\pm \infty = \frac{1}{0}##
    So, in terms of calculus and limits, it makes sense, but algebraically it also makes sense!
     
    Last edited: Nov 27, 2014
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