2
$\begingroup$

I'm trying to prove that all the solutions to the equation $x^2 = x + 1$ are irrational. This statement is equivalent to: If $x^2 = x + 1$, then $x$ is irrational.

I want to prove this using contraposition. The contrapositive statement is: If $x$ is rational, then $x^2 \neq x + 1$.

A rational number is one that can be described as a ratio. So, $r = \frac{k_1}{k_2}$, where $k_1$ and $k_2$ are integers and $k_2 \neq 0$.

Let $x$ be a rational number.

Using the above definition of a rational number we simplify $x^2 = x + 1$ to $(\frac{k_1}{k_2})^2 = \frac{k_1}{k_2} + 1$, where $k_1$ and $k_2$ are integers and $k_2 \neq 0$. All is left is to show that this equation cannot be satisfied for any combination of $k_1$ and $k_2$ (considering they are integers).

How can we do this? Can someone point me in the right direction?

$\endgroup$
3
$\begingroup$

The polynomial $x^2-x-1$ is irreducible over $\mathbb{F}_2$, hence it is irreducible over $\mathbb{Q}$ and cannot have any rational root. (Not really) Alternative way: assuming that $\frac{p}{q}$ is a root of $x^2-x-1$, we have $$ p^2 = pq + q^2 \tag{1}$$ but since $p$ and $q$ are not both even (due to $\gcd(p,q)=1$), the RHS and LHS of $(1)$ have opposite parity, contradiction. (Really) Alternative way: since the roots of $x^2-x-1$ add to one by Vieta's theorem, it is enough to prove that one of them (say, the greatest) is irrational to have that both of them are irrational. But if a number $x>1$ fulfills $x^2=x+1$, it also fulfills $x=1+\frac{1}{x}$, hence its ordinary continued fraction is given by $x=[1;1,1,1,1,1,\ldots]$. That is not a finite ordinary continued fraction, hence $x\not\in\mathbb{Q}$.

$\endgroup$
  • $\begingroup$ Explain your downvote. $\endgroup$ – Jack D'Aurizio Oct 2 '16 at 20:41
5
$\begingroup$

HINT: if I multiply through by $k_2$ I get $${(k_1)^2\over k_2}=k_1+k_2.$$ This means $(k_1)^2\over k_2$ is an integer. This isn't obviously a contradiction . . .

. . . but what if we assumed at the beginning that $k_1\over k_2$ is in lowest terms?


Note that to get a contradiction from the assumption that $k_1\over k_2$ was in lowest terms, we need to know that $k_2\not=1$ - that is, that there is no integer satisfying $x^2=x+1$. But this has a nice proof! HINT: is $x$ even or odd?

$\endgroup$
  • $\begingroup$ This was a lot of help, thanks! However, if both k1 and k2 are odd, how can I prove that the left side of the expression is a non-integer? I understand the right side of the expression is always an integer. I updated my question with my work so far. $\endgroup$ – Georan Sep 29 '16 at 3:45
  • $\begingroup$ @Georan I think you're misunderstanding my hint re: even/oddness: my point is that $x$ can't be an integer, because for any integer $x$, $x+1$ and $x^2$ have different parities (either odd/even or even/odd). The $k_i$ don't enter into this part; we only look at them once we've determined $x$ isn't an integer, which lets us conclude that $k_2\not=1$ (and it's this that really sets us up for a contradiction as hinted at in the first part of my answer). $\endgroup$ – Noah Schweber Sep 29 '16 at 3:47
  • $\begingroup$ I don't think I'm getting it (brain's running a little slow at this time :P). I understand why 'x' can't be an integer and why k2 /= 1, but what contradiction are we trying to set up? $\endgroup$ – Georan Sep 29 '16 at 4:22
4
$\begingroup$

By the rational roots theorem...

$\endgroup$
3
$\begingroup$

I just want to make a note that this is not the easiest way to do this problem. The best way (in my opinion) would be to actually find the solutions and prove their irrationality.

A hint for your proposed solution method: look at the denominator of each side. What equality would $k_2$ have to satisfy for these rational numbers to be equal?

$\endgroup$
3
$\begingroup$

I'd just rewrite it as $x^2 - x - 1 = 0$, run the quadratic formula, and then see that the roots are irrational, i.e. that $\sqrt{5}$ is irrational.

$\endgroup$
2
$\begingroup$

$4+1=4(x^2-x)+1=(2x-1)^2$ but $5$ isn't a perfect square.

$\endgroup$

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.