mathematical mind fuck

Discussion in 'Mind Games' started by s0ma, Apr 23, 2007.

  1. s0ma

    s0ma Member

    Below is a proof I found somewhere proving that 1 is equal to 0.999999999... (a.k.a. 0.9repeating).

    x = 0.999...
    10x = 9.999...
    10x - x = 9
    9x = 9
    x = 1
    1 = 0.999...

    SO CRAZY.
     
  2. prismatism

    prismatism loves you

    this has been posted so many times :D
     
  3. BudBill

    BudBill Dark Helmet

    You can't just randomly assign variables then substitute them back in.

    x = 0.999...
    10x = 9.999... <- this is meaningless
    x = 0.99999....999

    Next equation:

    10y - y = 9
    y = 1


    does x = y no, close but no. They will approximate each other but never equaling. Two separate equations, two separate variables so you cannot just pop them together. :)

    Nice try though.
     
  4. KParker730

    KParker730 Member

    10x = 9.999...
    10x - x = 9

    how did you jump to this?! i have seen it proved that 1 is equal to .9 repeating but i think you missed it
     
  5. I Hate Maths, It's Fuckin Whack!!!
     
  6. s0ma

    s0ma Member

    Why not?
     
  7. Nalencer

    Nalencer Dig Yourself

    It's just common sense. You have to be able to subtract and get a difference, or it's the same number. Since the nines repeat forever, there is no point where they stop and 1 starts. Therefore, same number.
     
  8. s0ma

    s0ma Member

    I still don't get what you're saying. The way I see it, you can assign a variable to the number you start out with (0.999...) and apply the rules of math to the variable (e.g. 10x= 9.999... and 10x - x= 9x). How is 10x = 9.999... meaningless?

    I don't know, this whole proof-thing has intrigued me since I found it. :)
     
  9. it's like saying 1=1
    1-1=1 <=====incorrect
    0=1
     
  10. sourdiesel06

    sourdiesel06 Member

    No it isn't... there's nothing wrong with what soma posted
    there's only 1 variable, x.
    if x=.999...
    then I think we can all agree that 10x would equal 9.999...
    have I lost any of you yet?
    Now, and this is where that grade school education REALLY comes in handy.
    If you subtract x from 10x, you get 9x, which in this case is equal to 9(9.999...-0.999...=9).
    Now, what do you get when you divide 9x(in this case 9) by 9?
    Thats right, you get x=1.

    Budbill please feel free to explain to me what is wrong about that.
     
  11. bluflame

    bluflame Member

    u never passed algebra did you?
     
  12. Peace Child

    Peace Child Member

    I really don't get how this works. I've seen it so many times on the internet. But I just never understand how it works. Probably because I'm not this far in algebra yet. I'm not even in high school yet. Oh well. :p
     
  13. Posthumous

    Posthumous Resident Smartass

    x cannot act as a constant and a variable at the same time...duh!
     
  14. Willy_Wonka_27

    Willy_Wonka_27 Surrender to the Flow

    computed for those who don't understand. and to clarify, there is no variable, x is a constant!

    0.999... = 0.999...
    10 * 0.999... = 9.999...
    (10 * 0.999...) - 0.999... = 9.999... - 0.999...
    9 * 0.999... = 9
    0.999... = 1
    1 = 0.999...
     
  15. The problem with the equation seems to be in this step:

    If x =0.9999 recurring

    THEN

    10x = 9.9999 recurring

    As I see it, this is not possible. With a number that is a 'fixed' number (sorry, my maths terminology sucks), such as "3" or "1001" or "4.27", yes you can multiply it by other numbers. There is no problem here.

    The problem occurs when we treat something with a recurring decimal (such as 0.999 recurring) as a number that can be multiplied by another number. This is NOT the same as multiplying 2 x 3 or 5 x 6 or 254 x 48432.

    Think about it this way. We can multiply 3 x 7, because we COULD actually get three groups of seven apples and ask "how many?". But how the hell do you get 0.999 recurring of something? A recurring number is a theoretical construct used to express something infinite. Multiplication is an action for finite, certain numbers. Hence, it doesn't make sense to say that

    let x = 0.999 recurring
    therefore, 10x = 9.999 recurring

    the equation should simply be that 10x = impossible.

    Now that's the problem as I see it...but I don't pretend to be a maths whizz or anything...so if someone else has another explanation let's hear it! :)
     
  16. Posthumous

    Posthumous Resident Smartass

    Oh really?

    9 * 0.999... = 9

    Approximate much?
     
  17. Willy_Wonka_27

    Willy_Wonka_27 Surrender to the Flow

    not at all.

    it computes out exactly.
    10 * 0.999... - 0.999... = 9
    9 * 0.999... = 9

    think of it this way:
    10 * .6 - .6 = 5.4
    9 * .6 = 5.4

    or

    10 * .234 - .234 = 2.106
    9 * .234 = 2.106

    or

    10 * .333333 - .333333 = 2.999997
    9 * .333333 = 2.999997

    you see?
     
  18. Willy_Wonka_27

    Willy_Wonka_27 Surrender to the Flow

    dude, in that step all your doing is moving the decimal one place to the right. its not rocket science.
     
  19. Posthumous

    Posthumous Resident Smartass

    Treating x as a variable

    x = unknown
    10x = 9.999
    10x - x = 9.999 -x <-can't subtract variables from constants
    10x = 9.999
    x = 0.999

    Yep x= 0.999 [​IMG]

    Treating x as a constant

    x = 0.999
    10x = 9.999
    10x - x = 9.999 -x
    9 = 9

    Yep 9=9 [​IMG]


    again x cannot act as a constant and a variable at the same time.
     

Share This Page


  1. This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
    By continuing to use this site, you are consenting to our use of cookies.
    Dismiss Notice