Is my proof for the Irrationality of the Golden Ratio correct?

Solution 1:

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.

Solution 2:

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

Solution 3:

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.

Solution 4:

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.