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Prove this equation is one to one

  1. Nov 11, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex]y=(x+1)^3[/tex]


    2. Relevant equations
    f(a)=f(b)
    a=b


    3. The attempt at a solution
    [tex](a+1)^3 =(b+1)^3[/tex]
    [tex]1+3 a+3a^2+a^3=1+3b+3b^2+b^3[/tex]

    This is where I get stuck.
     
  2. jcsd
  3. Nov 11, 2011 #2

    SammyS

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    How about: if [itex](a+1)^3 =(b+1)^3\,,[/itex] then [itex](a+1)^3-(b+1)^3=0\,.[/itex]

    Then factor into difference of cubes.
     
  4. Nov 11, 2011 #3

    Dick

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    Do you know any calculus? That's the easy way to prove this. Show y is an always increasing function of x.
     
  5. Nov 11, 2011 #4
    I know what the difference of cubes is but I can't seem to figure out how to apply that to this case. I was under the impression that if I was to apply the difference of cubes that the equation would gabber to be in, the form [tex](x^3)+1[/tex] rather than [tex](x+1)^3[/tex] I'm probably forgetting something here :P

    I haven't taken calculus yet but would it suggestion look like this? f(x)=(x+1)^3 I've already graphed it & seen that it's 1:1 but I'm looking to do this algebraically. Excuse me if I misinterpreted your post.
     
  6. Nov 12, 2011 #5

    SammyS

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    The difference of cubes is:

    p3 - r3 = (p-r)(p2+pr+r2)

    If (p-r)(p2+pr+r2) = 0,

    Then either p=r,

    or [itex]\displaystyle p=\frac{-r\pm\sqrt{r^2-4r^2}}{2}[/itex], which is only a real number if r = 0.
     
    Last edited: Nov 12, 2011
  7. Nov 12, 2011 #6

    Mark44

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    m3 - n3 = (m - n)(m2 + mn + n2)

    m3 + n3 = (m + n)(m2 - mn + n2)
     
  8. Nov 12, 2011 #7

    Deveno

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    one could approach this another way:

    if f:A→A is injective, and g:A→A is injective, then fog:A→A is injective.

    proof:

    suppose fog(x) = fog(y), that is:

    f(g(x)) = f(g(y)).

    since f is injective, g(x) = g(y).

    since g is injective, x = y, QED.

    now prove f(x) = x3 is injective, and

    g(x) = x+1 is injective.
     
  9. Nov 12, 2011 #8
    another path, i think the easiest mechanically but the least intuitive if you're given another definition, is to use the logical contrapositive equivalent of 1-to-1 (perhaps better known as proof by contradiction):

    "x != y implies g o h(x) != g o h(y)".

    here you don't need to show 1-to-1 of anything, only that the functions g and h are defined at x and y as you construct the conclusion f= g o h.

    [edit] of course that ends up needing to show that g and h are indeed functions...
     
    Last edited: Nov 12, 2011
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