Is my proof for the Irrationality of the Golden Ratio correct? I am trying to make a proof about the irrationality of the golden ratio by contradiction. I substituted the fraction $\frac{p}{q}$ for $\varphi$, where $p$ and $q$ are coprime integers. I used the following property: $$\varphi^2 - \varphi - 1 = 0$$ and then put $\frac{p}{q}$ in place of $\varphi$, getting $$\frac{p^2}{q^2} - \frac{p}{q} - 1 = 0$$ I combined the terms on the left into one fraction, getting $$\frac{p^2-pq-q^2}{q^2}=0$$ multiplying $q^2$ on both sides $$p^2-pq-q^2=0$$ I added $q^2$ on both sides $$p^2-pq=q^2$$ I factored the left side by $p$ $$p(p-q) = q^2$$ finally,I found the square root for both sides $${\sqrt p}{\sqrt{p-q}} = q$$ I stated that $q$ and $p$ share the factor $\sqrt p$, and since ${\sqrt{p}}{\sqrt{p-q}}$ is an integer, and $p-q \neq p$ then ${\sqrt p}$ must also be an integer, so our first assumption that $p$ and $q$ don't share any factors is wrong, so $\varphi$ is irrational. So, is my proof correct? Thanks for reading the whole thing and helping. (solved)
 A: As already posted in comments, to conclude from
$$\sqrt p \sqrt{ p-q} = q$$
that  $\sqrt p$ is a factor does not make sense. $\sqrt p$ is not an integer number, so it makes no sense do call it a factor of an integer.
But from
$$p(p-q) = q^2 \tag 1 $$ you can conclude $p=1$.
Because if $f$ is a prime number such that $$f|p$$
then by $(1)$ we have
$$f|q \operatorname    {\cdot} q$$
But if a prime divides a product is divides one of its factors, so $$f|q$$
and further $$f|\gcd(p,q)=1$$
so $f=1$. This means $p$ has no prime factors and $p=1$. From $(1)$ follows now
$$1-q=q^2$$
But this quadratic equation has not integer solution.
A: I just noticed that I can actually finish up everything from the first equation only. I get rid of the fraction and replace it with $x$. Then, using the rational root theorem, I find all the rational roots (which is $\pm1$). Since none of the possibilites work, I can conclude that $\varphi$ couldn't be rational since all the possible rational roots aren't solutions
A: There's a geometric description of the golden ratio: If a rectangle's sides $p>q$ are in the golden ratio (i.e., $\frac pq=\phi$) and you chop off a $q$ by $q$ square from one end, the part that remains (a $q$ by $p-q$ rectangle) also has its sides in the golden ratio, i.e., $\frac q{p-q}=\phi$.  (You can verify this using the definition of $\phi$.)
What does this have to do with the irrationality that you want to prove? Well, if $\phi$ were rational, then there would be a rectangle with integer sides in golden ratio. Then the geometric fact above would give a strictly smaller rectangle whose sides are still integers and still in golden ratio. Repeat the construction to get smaller and smaller such rectangles. That's a contradiction, because you can't keep getting smaller and smaller positive integers (the sides of your rectangles).
It's possible, of course, to eliminate the geometry in favor of algebraic computation. Just check that any positive integer solution of $p^2-pq-q^2$ gives rise to a smaller positive integer solution: $p^*=q$ and $q^*=p-q$. But for me, eliminating geometry here spoils much of the fun.
A: Expanding on Stephen Donovan’s comment: If two integers are coprime (share no common factors except 1), then any of their (integer) powers are also coprime. This is because powers don’t introduce any new factors:
$$\begin{align}
n &= p_1^{a_1} \times p_2^{a_2} \times\cdots\times p_i^{a_i}\\
\implies n^m &= p_1^{ma_1} \times p_2^{ma_2} \times\cdots\times p_i^{ma_i}
\end{align}$$
So if $p$ and $q$ are coprime, then there does not exist an integer $r$ such that $pr=q^a$. This can be applied to your second-last line.
A: Although some of the more obvious/practical ways to proceed have already been pointed out, here's another involving modular arithmetic which I found kind of cute.
By a similar argument to yours, we can yied $p^2 = q(p+q).$ Because $p^2$ is an integer multiple of $q,$ we can say that $p^2 \equiv 0 \pmod{q}.$
Now, recall from Bézout's identity that there must exist some integers $a$ and $b$ such that $ap + bq = 1,$ since $p$ and $q$ are coprime. Taking both sides mod $q$ we get that $ap \equiv 1 \pmod{q},$ so $p$ has a modular inverse mod $q.$ (this is a well-known fact about coprime integers but I figured I should motivate it in case you weren't familiar)
So, going back to $p^2 \equiv 0 \pmod{q},$ we can proceed by multiplying both sides by $a,$ so $ap^2 \equiv 0.$ Now notice $ap^2 = a(pp) = (ap)p \equiv p \equiv 0,$ so we have that $p$ must be divisible by $q.$ This means $\gcd(p, q) = q,$ and since $p$ and $q$ are coprime this means $q = 1,$ so we would have that $\varphi$ must be an integer.
From here you can proceed by proving that our original equation has no integer solutions because $\sqrt{5}$ is not an integer, but we can continue more simply by proceeding with a similar argument to show that $q$ is divisible by $p$: $q^2 = p(p-q)$ implies that $q^2 \equiv 0 \pmod{p},$ so $bq^2 \equiv q \equiv 0 \pmod{p}.$
Now consider that because $p$ is divisible by $q,$ there exists some integer $k$ such that $p = kq,$ and similarly there is some $n$ such that $q = np.$ So, $q = nkq,$ meaning either $q = 0$ or $nk = 1.$ $q = 0$ would result in division by zero, so we have $nk = 1,$ and clearly either $n = k = 1$ or $n = k = -1.$ So, either $p = q = 1$ or $p = -q = -1,$ and either $\varphi = 1$ or $\varphi = -1.$ Simply checking our original equation for either choice shows that neither of these can be true, and we can conclude by contradiction that $\varphi$ is not rational.
