0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ You've shown that if there is a rational root, then the numerator and denominator of that root are both odd. Nothing more. $\endgroup$ – Patrick Stevens Feb 1 '18 at 11:07
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks Jose but I am not quite sure how the theorem works. This is our first lesson on proof by contradiction in highschool. $\endgroup$ – user450003 Feb 1 '18 at 11:28
  • $\begingroup$ @user450003 Did you read the statement of the theorem? $\endgroup$ – José Carlos Santos Feb 1 '18 at 11:29
  • $\begingroup$ Yes I did and I am still confused. $\endgroup$ – user450003 Feb 1 '18 at 11:55
  • $\begingroup$ @user450003 What's the problem? It says that if there was a rational root $\frac ab$, with $a\in\mathbb Z$ and $b\in\mathbb N$, then $a\mid-1$ and $b\mid 1$, which means that $\frac ab=\pm1$. $\endgroup$ – José Carlos Santos Feb 1 '18 at 11:59
  • $\begingroup$ I don't understand why a∣−1 and b∣1? $\endgroup$ – user450003 Feb 1 '18 at 12:23
1
$\begingroup$

Much easier using the following theorem: if $f(x) = a_nx^n + \cdots a_1 x + a_0\in\Bbb Z[x]$, then $$f(p/q) = 0 \hbox{ ($p/q$ irreducible)}\implies p\vert a_0,q\vert a_n.$$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Rewrite the equation in the form $x^3+2x=1$. Suppose there is a rational solution $m/n$ with $\gcd(m,n)=1$. Such a solution can be chosen with $ n$ positive integer. Then we get $$ \frac{m^3}{n^3} +2\frac{m}{n} =1. $$

Multiplying by $n^3$ we get, $$m^3+2mn^2= n^3;\quad \mbox{equivalently}\quad m(m^2+2n^2) =n^3\qquad(*)$$

This shows $m$ is a factor of $n^3$. The hypothesis $\gcd(m,n)=1$ forces to conclude $m=1$. Now putting $m=1$, in equation $(*)$ we get $$1+2n^2=n^3 $$ which can be rewritten as $n^2(n-2)=1$. This last equation contradicts the supposition that $n$ is a positive integer. So there are no rational roots to your equation.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

$a^3+2ab^2-b^3=0$ implies $a^3=(-2ab+b^2)b$ and so $b$ divides $a^3$. Since $a$ and $b$ are coprime, we must have $b=\pm 1$. Then $a(a^2+2)=\pm 1$, which cannot happen because $a^2+2\ge2$ cannot divide $1$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for the help but I don't understand why b must be plus or minus 1. $\endgroup$ – user450003 Feb 1 '18 at 11:29
  • 1
    $\begingroup$ @user450003, if $p$ is a prime dividing $b$, then $p$ divides $a$, but this cannot happen because $a$ and $b$ are coprime. $\endgroup$ – lhf Feb 1 '18 at 11:31
  • $\begingroup$ @lhf Why didn't you look at the case $b = -1 $? $\endgroup$ – john Sep 12 '19 at 16:43
  • $\begingroup$ @john, fixed and simplified. Thanks for the push. $\endgroup$ – lhf Sep 12 '19 at 16:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.