How can we prove the injectivity of f(x) = x³ + x without using calculus?

[MorallyObtuse]

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

 

[jgens]

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 $a \neq b$ we have that $f(a) = f(b)$. If you can arrive at a contradiction from this, then that will prove that $f$ is injective.

 

[rasmhop]

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 $x^3$ 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:
$$a^3 + a = b^3 + b$$
Move to one side and factor like:
$$\begin{align*} 0 &= a^3 -b^3+ a-b \\<br /> &= (a-b)(a^2-ab+b^2) + (a-b) \\<br /> &= (a-b)(a^2 +b^2 - ab + 1) \end{align*}$$
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).