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Proof of injective function

  1. Nov 14, 2009 #1
    Hi,

    How do I prove that this functions is injective?

    a.) f : x --> x³ + x x ∈ R

    f(a) = a³ + a, f(b) = b³ + b
    f(a) = f(b) => a³ + a = b³ + b => a³ = b³
    => a = b
    therefore f is one-to-one
     
  2. jcsd
  3. Nov 14, 2009 #2

    jgens

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    Gold Member

    If this is a homework problem, it should be posted in the homework forums here: https://www.physicsforums.com/forumdisplay.php?f=152

    Now, to answer your question, assume that for some [itex]a \neq b[/itex] we have that [itex]f(a) = f(b)[/itex]. If you can arrive at a contradiction from this, then that will prove that [itex]f[/itex] is injective.
     
  4. Nov 14, 2009 #3
    How do you conclude:
    a³ + a = b³ + b => a³ = b³?
    That is pretty much just as hard as showing f injective. How much math have you had? If you have had calculus simply show that it's continuous and strictly increasing (positive derivative). If you don't know calculus, but know that x is strictly increasing and [itex]x^3[/itex] is increasing, then you can conclude that f(x) is strictly increasing because it's the sum of a strictly increasing and an increasing function, and it's continuous, but this implies that it's injective.

    If you want to brute-force it, do as you did till you reach:
    [tex]a^3 + a = b^3 + b[/tex]
    Move to one side and factor like:
    [tex]\begin{align*} 0 &= a^3 -b^3+ a-b \\
    &= (a-b)(a^2-ab+b^2) + (a-b) \\
    &= (a-b)(a^2 +b^2 - ab + 1) \end{align*}[/tex]
    for this to hold you must have a=b, or
    a^2 +b^2 - ab + 1 = 0
    which you can show is a contradiction for real numbers (consider for instance the discriminant).
     
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