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Who am I?

  1. Mar 10, 2005 #1
    x/1 = x+1/x
     
    Last edited: Mar 11, 2005
  2. jcsd
  3. Mar 11, 2005 #2
    0/0 . . or
    SqrRoot of -0
     
  4. Mar 11, 2005 #3
    My answer is "hedons"
     
  5. Mar 11, 2005 #4
    x/1 = x
    so your eq becomes:

    x = x - 1/x
    ->0=-1/x...
    unless you take the limit you are nobody...
     
  6. Mar 12, 2005 #5
    x=±sqrt(1/(1-1))

    you're the 4 grade student, who just studied what an "equation" is
     
  7. Mar 12, 2005 #6
    Is that written as [tex]\frac {x}{1} = x + \frac{1}{x}[/tex] or is it supposed to be [tex]\frac {x}{1} = \frac{x+1}{x}[/tex]?
     
  8. Mar 14, 2005 #7
    oh yea, cause if its the second one then its the golden ratio. 1.61828 or whatever.
     
  9. Mar 15, 2005 #8
    then if it's the second

    x^2 = x+1
    x^2-x-1 =0
    x=(1(+-) (((-1)^2)-(4)(1)(-1))^0.5) (2(1))^-1
    x= (1(+-)(5^0.5))/2
    interesting ???
     
  10. Mar 26, 2005 #9
    An anomoly, that's what you are!
     
  11. Mar 29, 2005 #10
    i think it should be infinity. putting value of xas infinite we will satisfy the equation.
     
  12. Mar 29, 2005 #11
    Are we ever getting a answer here??? :confused:
     
  13. Apr 11, 2005 #12
    x = 1/x has no solution. For x = 1 + 1/x, x = ~1.618 (the "golden ratio").
     
  14. Apr 11, 2005 #13
    ?! I think x=1 is a solution.
     
  15. Apr 11, 2005 #14
    So is [tex]x = -1[/tex].
     
  16. Apr 11, 2005 #15
    I believe he meant to say 0=1/x.
     
  17. Apr 12, 2005 #16
    Ahhhh....x = 1 is so obvious. but x = -1 doesn't work.
     
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