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Prove (100-100)/(100-100)=2

  1. Jan 4, 2012 #1
    I got this question from my friend and he solve it like this:

    Code (Text):

              100-100         10^2-10^2     (10-10)(10+10)        (10+10)
            -----------  = -------------=  ---------------   = ----------- =2
              100-100         10(10-10)        10(10-10)                10
     
    Please tell me how is it...what is the error in it according to the rules of mathematics..
     

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    Last edited by a moderator: Jan 6, 2012
  2. jcsd
  3. Jan 4, 2012 #2

    Deveno

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    (100-100)/(100-100) doesn't equal anything, sorry.

    the denominator is 0, which leads to an undefined quantity.
     
  4. Jan 4, 2012 #3

    pwsnafu

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    To expand on Deveno's answer: the phrase
    "divide a by b"
    means:
    "find the number x such that xb = a".

    So the (100-100)/(100-100) means:
    "find x such that (100-100) x = 100-100"
    Clearly, anything can be x. So we leave it undefined because x is not unique.

    Whenever you write down [itex]\frac{a}{b}[/itex] there is always an implicit "b is non-zero", otherwise the expression is invalid.
     
  5. Jan 4, 2012 #4

    Mentallic

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    In a similar fashion you can "prove" that it's equal to 1/2 by just swapping the operations in the numerator and denominator. Also, it's equal to 1 if you just cancel the like terms in the beginning.

    Essentially dividing by zero causes these problems and only until you begin studying limits in calculus do they make a little more sense beyond the typical answer of being undefined.
     
  6. Jan 4, 2012 #5
    The error is in when he goes from the second to last step to the last step. How does he get rid of the (10-10)? He divides both sides by 10-10=0, but you cannot divide by zero.

    It is like saying 6*0=5*0 then if you
    divide out the zeroes you get 6=5
    But you cannot divide by zero and get a meaningful answer.
     
  7. Jan 6, 2012 #6
    Thanks every body..
    I got some more stuff like this which results 2=1..
    i will post it
     
  8. Jan 6, 2012 #7
  9. Jan 6, 2012 #8
    Strictly speaking, that original equation does equal two. Of course, it also equals all other numbers through infinity.
     
  10. Jan 6, 2012 #9
    Alright, so say that we have some f:RxR -> R, and

    h(x,y) = f(x,y) + g(x,y)
    f*(x,y) = h(x,y) - g(x,y)

    so is f(0,0) = f*(0,0) or not? I can argue that for any g:RxR -> R

    g(x,y) - g(x,y) = 0

    Therefore by substitution of the definition of h(x,y) we have f*(x,y) = f(x,y), including f*(0,0) = f(0,0); this is valid because substitution of a function definition is not an algebraic operation.

    So I choose g(x,y) = x/y... but now g(0,0) is indeterminate, and so is h(0,0), and as f*(x,y) is defined as a subtraction over from h(x,y), f*(0,0) is now indeterminate too. Therefore I can also argue that f*(0,0) [itex]\neq[/itex] f(0,0)...

    Both seem to be valid arguments (for me old, lowly brain at least), but one of them is certainly false.
     
  11. Jan 6, 2012 #10

    Hurkyl

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    No, strictly speaking it is undefined, so it doesn't make sense to ask if it equals anything at all.

    Strictly speaking, if something has that property, then all numbers are the same.


    There is a (very) loose sense in which what you said is true -- but you're being so loose with all of the terms involved that the statement becomes almost entirely devoid of content.
     
  12. Jan 6, 2012 #11

    pwsnafu

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    You can't. You previously stated
    Every function is defined on its domain, your choice of g(x,y) does not satisfy that.
     
  13. Jan 6, 2012 #12

    Yes, but I'm trying to get in the mood of the discussion here, which has been around "what value 0/0 has". Of course a well-defined, continuous function f(x,y) = x/y is defined on a domain R x (R - {0}), so the issue of 0/0 is just avoided. That's my comfort zone.

    But yet people always discuss... "value of 0/0 is... etc...". Sounds impossible that thing to have a value, but then sqrt(-1) and ∞ don't exist either, but are referenced all the time.

    Say we assume 0/0 exists, and that R* = R U { 0/0 }. Then I can define all my functions as

    f:R* x R* -> R*
    g:R* x R* -> R*
    h:R* x R* -> R*
    f*:R* x R* -> R*

    So if R* exists then f(0,0) = f*(0,0), which is a pretty casual result.

    This is definitely not in my comfort zone, but with statement like "0/0 is equal to all numbers to infinity", it seems like people just assume that R* exists, and that R = R*.
     
  14. Jan 7, 2012 #13
    The equality relation is transitive; if the above expression equals every number, then all numbers are equal. The expression is undefined, so, in a sense, the equation was flawed before it began.
     
  15. Jan 7, 2012 #14
    As I understand it, 0/0 is indeterminate, not undefined. Unless all my math books and teachers have been wrong.

    fbs7, I don't understand your notation.
     
  16. Jan 7, 2012 #15
    Indeterminate? We're not talking about limits here.
     
  17. Jan 7, 2012 #16

    Pardon me, English is not my first language. In my native language we call 0/0 undefined, as opposed to a well-defined value. Whatever it is, it's 0/0...

    If that's anything, that is. I'm more comfortable thinking that 0/0 is neither undefined nor indeterminate.. it just doesn't exist. But then again I have never been comfortable with i either.

    Let me ask the other way around... what is the true nature of 0/0? What "indeterminate" means in mathematical terms? Which equation/expression defines indeterminate?

    I mean, when one says 0 * x = 0, that just means a multiplication and it holds true for any x [itex]\in[/itex] R. It doesn't imply that x = 0/0, because from 0 * x = 0 one obtains ( 0 * x ) / 0 = 0 / 0, therefore (0/0) * x = 0/0... not x = 0/0.

    So I could think as "indeterminate" meaning that if z = f(x) / g(x), and f(0) = 0 and g(0) = 0, then the value of z, if it exists, depends on the functions f and g. But once the functions are known, z is known too... so it's not that indeterminate, it just depends on f and g.

    Or, let me ask even another way around. If F = [{0} -> {0}] is the set of all continuous functions f such that f(0) = 0, and then I define 0/0 as being the set of all limits limx->0 f(x)/g(x) calculated per the l'Hopital rule, then certainly there is a mapping between F x F and 0/0.

    That way, although I still think that 0/0 doesn't exist (in the same way that i doesn't exist), I can say that if I know f and g, then I can define 0/0 = f(0)/g(0), I can say that it is a member of 0/0, it is defined through that mapping, and for a given (f,g) it has one specific, well-defined value in R. Doesn't seem very indeterminate.

    As you see, the whole thing is just very confusing to me.
     
    Last edited: Jan 7, 2012
  18. Aug 5, 2012 #17
    I will tell you your actual mistake.......you have solved the numerator by the formula 'a^2-b^2'...bt bt there is not 'b' in the numerator only 'a' as,a=10 then how can you use that formula?????
    MATHS CAN'T FAIL BUDDYYYY
     
  19. Aug 5, 2012 #18

    pwsnafu

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    1. Welcome to Physics Forums. Here we value proper spelling, clarity of expression and punctuation.
    2. Why did you feel the need to resurrect an eight month old thread that was already resolved?
    3. The formula a2-b2 = (a+b)(a-b) is perfectly valid when a=b.
     
  20. Aug 5, 2012 #19
    The OPs friend factors out zero. For example:

    [itex]\frac{1*0}{1*0}[/itex] would appear to be [itex]\frac{1}{1} = 1[/itex] if you factor out zero.

    Generally written:
    [itex]\frac{n*0}{n*0} \neq \frac{n}{n}[/itex]

    That is the problem.
    The friend arranged the problem to
    [itex]\frac{0*20}{10*0} \rightarrow \frac{20}{10} \rightarrow \frac{2}{1} \rightarrow 2[/itex]
     
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