Prove that there is no rational $x$ such that $x-1 = 1/x$

Question

As said in title,

I need to prove that there doesn't exist a rational number $x$ that satisfies $x-1 = 1/x$.

I remember doing something like this a while back at school but I can't recall how to do it. Am I right in starting out by simplifying it to:

$x^2-x=1$?

Answer 1

By Ruffini's Theorem (or Rational root Theorem), all possible rational solutions of $$x^2-x-1=0$$ are those whose numerator divides 1 (the independent term) and whose denominator divides 1 (the leading coefficient).

So the only possible rational solutions are $\pm1$ and, trivially, they aren't.

Answer 2

Write $x=a/b$ in its lowest terms so$$x=\frac{1}{x-1}=\frac{b}{a-b},$$forcing a contradiction if $|x|>1$. See if you can do the same for $|x|<1$.

Answer 3

The only solutions for x are GoldenRatio and 1-GoldenRatio, which are both irrational.

GoldenRatio =( 1+Sqrt(5) ) /2

Answer 4

Using the quadratic formula on $x^2-x=1$ we have

$$x = \frac {1\pm\sqrt5}2$$

If $x$ can be rational, so can $2x-1 = \pm \sqrt 5$. But the irrationality of $\pm\sqrt 5$ can be proved by infinite descent, for instance.

Answer 5

Let $x=\frac{p}{q}$ be a solution in lowest terms, i.e., $\gcd(p,q)=1$, with $q>0$. By substituting this into the equation for $x$, we get

$$\left(\frac{p}{q}\right)^2-\frac{p}{q}=1\Leftrightarrow p^2-pq=q^2\Leftrightarrow p^2=q(p+q).$$

That is, $q$ divides $p^2$. But we assumed $\gcd(p,q)=1$ and thus $q$ must be $1$. Hence, we get

$$p^2=p+1\Leftrightarrow p(p-1)=1,$$

which means that $p$ divides $1$, i.e., $p=\pm 1$, but obviously neither $p=1$ nor $p=-1$ satisfies $p^2=p+1$. We have thus reached a contradiction and $x$ cannot be rational.