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Multiplication=addition 2*2=2+2=4

  1. Feb 12, 2004 #1
    i think everyone is familiar to this but i wonder if someone has made his research to study it.
    what i mean is the namubers that when the result of multiplying them together is the same as adding them:
    for example:
    2*2=2+2=4 (trivial, i know (-: ).
    3*1.5=3+1.5=4.5

    my question has someone prooved that the number of these pairs are finite or infinite in the real numbers?
     
  2. jcsd
  3. Feb 12, 2004 #2

    matt grime

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    xy=x+y

    rearrange and solve for R.

    for N (well, this is the number theory thread)

    xy-x-y+1=1

    (x-1)(y-1)=1

    in N x=y=2
     
  4. Feb 12, 2004 #3
    Perhaps I'm misunderstanding something... You want real solutions to this equation:

    x + y = xy
    =>
    y = x / (x - 1)

    I.e, for any x != 1 we can chose a y so that x + y = xy, namely y = x / (x - 1). Any real x (except for x = 1) will give a real y, so there are an infinite number of such pairs...

    A more interesting question would be to consider only integers...
     
  5. Feb 12, 2004 #4
    i didnt think about it very much so perhaps from this had risen the misunderstanding.
    i guess in the real numbers is really trivial.
    so lets say just prooving for integers how do you go around prooving this?
    [x,y]=[Z] or [x,x/(x-1)]=[Z]
     
  6. Feb 12, 2004 #5
    I have no idea what this means. matt grime provided an answer for what happens if you only consider the natural numbers (i.e, x = y = 2 is the only solution).
     
    Last edited: Feb 12, 2004
  7. Feb 12, 2004 #6
    z doesnt represent the integers?
     
  8. Feb 12, 2004 #7
    Aha, yes it does, but you used the equals sign (and non-curly braces), and I've never seen = used to specify membership in a set ([​IMG] seems to be more commonly used).
     
    Last edited: Feb 12, 2004
  9. Feb 12, 2004 #8

    NateTG

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    For [tex]\frac{x}{x-1}[/tex] to be an integer, [tex]x-1[/tex] must divide[tex]x[/tex]. Since the difference between the two is 1, their greatest common factor is 1, so the only solutions are [tex]x-1=1[/tex] and [tex]x-1=-1[/tex] (2+2=2*2 and 0+0=0*0).
     
  10. Feb 12, 2004 #9

    matt grime

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    NateG, you should state you are only working in N there, by saying integer you implying Z. In Z, there are other answers. All we know is that here (x-1) is a unit, hence x-1=-1 or 1, yielding the other integer answer of x=y=0 as well as x=y=2
     
  11. Feb 12, 2004 #10

    NateTG

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    For [tex]\frac{x}{x-1}[/tex] to be an integer |x-1| must divide |x|. The GCF of the two absolute values is 1 (by Euclid's algorithm). Therefore [tex]|x-1| \leq 1[/tex] since it divides |x| and obviously divides itself. For [tex]x \in \mathbb{Z}[/tex] that leaves three solutions: 0,1, and 2, but 1 leads to division by zero, so the only solutions are x=0 and x=2.

    P.S. My apologies for the mixed formatting
     
    Last edited: Feb 12, 2004
  12. Feb 12, 2004 #11

    matt grime

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    forget that last post if you saw it, i absolutely apoligize, i misread your post.
     
    Last edited: Feb 12, 2004
  13. Feb 12, 2004 #12

    NateTG

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    Don't sweat it, I've posted some beauties myself. I should be less adverserial in my response though. ;)
     
  14. Feb 19, 2004 #13
    Integer solutions of [tex] xy = x + y [/tex] where [tex] x,y \epsilon Z [/tex]

    [tex] xy - x - y = 0 [/tex]

    [tex] (x-1)(y-1) = 1 [/tex]

    let [tex] x' = x -1[/tex] and [tex] y' = y - 1 [/tex] so [tex] x',y' \epsilon Z [/tex]

    So [tex] x' = 1 / y' [/tex]

    The only integers whose reciprocals are also integers are 1 and -1

    So [tex] y' = 1 [/tex] and [tex] x' = 1 [/tex]

    So [tex] y = 2 [/tex] and [tex] x = 2 [/tex]

    AND

    So [tex] y' = -1 [/tex] and [tex] x' = -1 [/tex]

    So [tex] y = 0 [/tex] and [tex] x = 0 [/tex]

    Therefore there are only two solutions to this diophantine equation [tex] x = y = 2 [/tex] and [tex] x = y = 0 [/tex]
     
    Last edited by a moderator: Feb 19, 2004
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