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Forbidden Solution

  1. Jul 18, 2010 #1
    I am not sure if the result I get for the equation below is forbidden or not.

    ((x+1)*(x-1))/(x-1)=3
    x^2-1=3x-3
    x^2-3x+2=0
    (x-1)(x-2)=0

    Their for x= 1 and 2
     
  2. jcsd
  3. Jul 18, 2010 #2
    Do both of your final answers work in the original equation?
     
  4. Jul 18, 2010 #3
    putting the value 1 into the equation results in 0/0=3. 0/0= undefined, the problem for me is the process of how it was solved seems mathematically sound.
     
  5. Jul 19, 2010 #4

    HallsofIvy

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    Your first step was multiplying both sides by x- 1. If x- 1= 0 (that is, if x= 1) that is NOT "mathematically sound" because multiplying both sides of any equation, true or false, by 0 results in the true equation 0= 0.
     
  6. Jul 19, 2010 #5
    I suggest you whenever you multiplie with x or with, for exemple (x-1) to put x><0 (wich means x can be 0) or in second exemple x-1><0; x><1 (which means x can't be 1). That way when u get answers you can check are they forbidden solutions very easy.
    Also insted of >< you can writte crossed = which should look something like / on =.
    Sry for my bad english and i hope i helpped
     
  7. Jul 19, 2010 #6

    Gib Z

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    From the starting equation it is implicitly implied that x can not be 1 as division is meaningless otherwise.
     
  8. Jul 19, 2010 #7
    some time ago, i became uncomfortable with the graph this produces. I considered instead of a hole at 1 there should be a vertical line. this was based on the idea that you do not know 0/0= as it could be anything based on AxB=C their for A=C/B. Using other intersecting lines to to determine the co-ordinates, just as in the beginning post, their is a vertical line rather than a hole.

    I just feel a vertical line is a better depiction.
     
  9. Jul 19, 2010 #8

    Gib Z

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    No, in fact its the opposite. The definitely nothing, its undefined. The division operation is meaningless when dividing by zero, and by writing the function like that there is an implicit assumption that x=1 is not in the domain of the function.
     
  10. Jul 22, 2010 #9
    Consider this. When taught in high school about gradients, we are taught that m=rise/run. depicted often is a vertical line for an undefined gradient. The formula for that line is y=(infinity/0)x. When x is equal to any number besides zero it does not occupy that domain as y=(infinity/0) and is undefined, but when x=0 y=(0/0) and depicted is a vertical line, defining it as occupying all values. Does this mean that this high school depiction is false?
     
  11. Jul 22, 2010 #10

    Office_Shredder

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    y=(infinity/0)x is not the equation of a line, so that doesn't really make any sense
     
  12. Jul 29, 2010 #11

    HallsofIvy

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    I cannot speak for all high schools- certainly I did NOT learn that in highschool. But whether it is a "high school depiction" or not, yes, it is false!
     
  13. Jul 30, 2010 #12
    To my understanding the mathematical consensus is that any value divided by zero, is undefined. when I look at 0/0 compared to 1/0 i tend to find that they are different. Are they still both undefined, even if they are different?

    Consider

    X*0=1

    There is no value as to which X could be that will have the result be 1 and hence if X = 1/0 then 1/0 is undefined.

    Consider

    X*0=0

    X could assume any value at all and the equation holds true; hence if X=0/0 then my logic would consider that 0/0 is equal to any value.

    How is this logic, bad logic?
     
  14. Jul 30, 2010 #13

    HallsofIvy

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    Yes, 0/0 is "different" from a/0 for a non-zero. Some text books use the word "undetermined" for a limit, as x goes to a, that, if you just plugged a in for x would give 0/0.

    But your logic is bad when you say "0/0 is equal to any value". The limit of something of the form "0/0" (if is exists) is some specific value- we just don't know which. But, in any case, 0/0 itself is not a number. The graph of [itex]y= (x^2- 2)/(x- 2) is almost a straight line. It is identitical to the graph of y= x+ 2 except at x= 2 where there is a hole. There is not a vertical line at x= 3 because 0/0 is NOT any number- certainly not "all numbers".
     
  15. Jul 31, 2010 #14
    im not sure that im getting you, but you say that 0/0 if it were some value would have to be a specific value. If (y-1)(y-2)=x then if X=0 and Y= 1 and 2, so to my understanding a particular domain can have multiple results. If you consider all those little tricks that can be played with proofs that consider 0/0=1 having 0/0=all values can give a better explanation. you often see the 1=2 proof, but if you were to say that 1*1=2*1 isn't correct, but 1*0/0=2*0/0 is correct as both would simply =0/0

    Im sorry to push this perspective, its just i see it as being more advantageous to have 0/0 = all values, as it highlights that it is different X/0 (X not=0). (-1)^0.5 can't exist either but there is a whole branch of maths dealing with it.

    Did you mean (X^2-4)/(x-2) ?
     
  16. Jul 31, 2010 #15

    Mark44

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    Yes, I'm confident that HallsOfIvy meant (x^2 - 4)/(x - 2).

    In your example, with (y-1)(y-2)=x, and x = 0, y can't simultaneously be 1 and 2.
    If (y-1)(y-2)= 0, then EITHER y = 1 OR y = 2.

    Your statement "so to my understanding a particular domain can have multiple results." makes no sense to me.

    The trouble with an indeterminate expression of the form 0/0 is that if you could conceivably define it to be any value. For example, if you decided to define it to be 5, you would have 0/0 = 5. Now multiply both sides of the equation by 0 to get 0 = 0, a true statement. The same thing would happen if you decided to define 0/0 as -3. The problem is that a quotient should have a unique answer. For that reason it is NOT advantageous to have 0/0 being any arbitrary number.

    In all of the 1 = 2 proofs, one of the steps typically involves division by zero, leading you to a contradictory result.

    (-1)^(.5) doesn't exist in the real numbers, but it does exist in the complex numbers.
     
    Last edited: Jul 31, 2010
  17. Jul 31, 2010 #16
    how about any value rather than all values?
     
  18. Jul 31, 2010 #17

    Mark44

    Staff: Mentor

    If an expression can take on any value, it can take on all values, so you're saying the same thing twice.

    A division problem has to have a single, unique answer, or none at all. It can't have multiple answers.
     
  19. Jul 31, 2010 #18

    Mark44

    Staff: Mentor

    No it isn't. The formula for a vertical line is x = K, where K is some real number. The formula for a vertical line doesn't even involve y!
     
  20. Jul 31, 2010 #19
    Im aware that high school depictions can be mislead. I really just think that this whole thing is more or less a matter of opinion.

    To say a division problem, has to result in a single value, is more or less a rule someone has said. Show me the mathematical proof for this and ill back down from that statement.

    I think that the idea of 0/0= any value should be played with to see if it has any application, because this is just a matter of opinion.
     
  21. Jul 31, 2010 #20

    jgens

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    It's not a matter of opinion. An operation is defined in terms of functions, making it impossible for one input to produce multiple outputs. If you would like to define something differently, you can go ahead and do so, but you're probably not going to get very far.
     
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