Proof by contradiction:
Assuming that there is a rational solution to the equation $x^3+2x-1=0$.
Let $x=a/b$ where $a$ and $b$ are coprime with $b$ not equal to zero.
Performing a substitution into the equation, it simplifies to $a^3+2ab^2-b^3=0$.
Three cases to consider (since $a$ and $b$ are coprime so they can't be both even):
Case 1: $a$ is even and $b$ is odd, then
$a^3$ is even
$2ab$ is even
$b^3$ is odd
So the LHS of the equation is odd and the RHS of the equation is even. Therefore, there is a contradiction.
Case 2: $a$ is odd and $b$ is even, then
$a^3$ is odd
$2ab$ is even
$b^3$ is even
So the LHS of the equation is odd and the RHS of the equation is even. Therefore, there is a contradiction.
Case 3: $a$ and $b$ are odd, then
$a^3$ is odd
$2ab$ is even
$b^3$ is odd
So the LHS of the equation is even and the RHS of the equation is even.
So does that means that there is a rational solution when $a$ and $b$ are odd? I am stuck with this. Can someone help me out? When we perform proof by contradiction, do we have to perform it for all cases? Thanks in advanced!
Best Answer
By the rational root theorem, only $\pm1$ could possibly be rational roots of your polynomial, but they aren't. Therefore, it has no rational roots.