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)
Is the proof for the Irrationality of the Golden Ratio correct
elementary-number-theorygolden ratioirrational-numbers
Related Solutions
Theoretically, the concept measure of irrationality of a real number $\alpha$ is, technically speaking, the following specialized notion: it is the infimum of all real $\mu$ for which there is a positive constant $A$ such that for all rational $\frac pq\ne \alpha$ with $q>0$ one has $$|\alpha - \frac pq|>\frac {A}{q^{\mu}}$$ This inequality indicates how “far” of the real $\alpha$ is a rational “close” to $\alpha$; in other words, all rational “near” to $\alpha$ determines its “distance” from $\alpha$; Or even, how rational can not approach $\alpha$.
See as example the “striking inequality” (Baker) discovered by Mahler in 1953 and today improved, $$|\pi-\frac pq|>\frac{1}{q^{42}}$$ valid for every rational $\frac pq;\space q> 1$
There is a whole Epica about this topic of measure of irrationality, beginning with Dirichlet and his approximation theorem (1842), Liouville and his discovery of the first known transcendental number (1844), passing through Thue (1909), Siegel (1929), Dyson (1947) and the gold brooch finisher with Klaus Friedrich Roth (1955) and his deep result which earned him the Fields Medal.
Theorem (Roth).- For all algebraic irrational $\alpha$ and all $\epsilon > 0$ the inequation $$ |\alpha- \frac pq |< \frac{1}{q^{2+\epsilon}}$$ has only a finite number of solutions in irreducible rational $\frac pq$ i.e. for all $\epsilon>0$ there is a positive constant $C(\alpha,\epsilon)$ such that for all rational $\frac pq$; $q>0$ one has $$|\alpha- \frac pq|>\frac {C(\alpha, \epsilon)}{q^{2+\epsilon}}$$ “The achievement is one that speacks for itself: it closes a chapter, and a new chapter is now opened. Roth’s theorem settles a question which is both of a fundamental nature and of extreme difficulty. It will stand as a landmark in mathematics for as long as mathematics is cultivated” (Harold Davenport, in his presentation of Roth to the Fields Medal at the International Congress in Edinburgh,1958).
With Liouville, measure of irrationality of a real algebraic $\alpha$ was equal to its degree $n$ and $n$ was successively decreasing (with the discoveries of the above mentioned authors) till the optimal value $ 2 $ established by Roth.
You do have two options.
That not so great and begging the question option:
Show if $\frac 1\varphi = 1-\varphi$ then $\varphi^2 -\varphi + 1 = 0$ and $\varphi =\frac {1 +\sqrt 5}2$ and show that $\sqrt 5$ is irrational.
That's done the "usual" way. If $a^2 = 5b^2$ for integers $a,b$ then if $a$ isnt a multiple of $5$ then $a^2=5b^2$ either. SO $a$ is a multiple of $5$ and $a =5a'$ and $5a'^2 = b^2$ and by the same argument $b$ is a multiple of $5$ so $\sqrt 5 = \frac ab$ where $a,b$ are integers in lowest terms is impossible.
But I call that "begging the question" because it doesn't really have anything to do with the golden ratio aspect; just something about boring old square roots of integers.
Other option:
Let $\varphi =\frac pq$ where $p,q\in \mathbb Z$ and in lowest terms, and $\frac qp = 1-\frac pq$.
Then $q^2 = p(q- p)$. Now $p,q$ are in lowest terms. So any factor $n$ of $q$ so that $n|q$ can not have any factor with $p$ at all. So $n|q^2$ means $n|p(q-p)$ but $n$ has nothing in common with $p$ so $n|q-p$. But $n|q$ so if $n$ also divides $q-p$ then $n|q -(q-p)=p$ but that's a contradiction.... unless $n=1$ is the only factor of $q$. But that means $q = 1$ and ... that just doesn't work. $1 = p(1-p)$ has no integer solutions.
That requires a fair number of assumptions about numbers and factors; namely that all numbers have indivisible prime factors.
YOu can do a well ordering type argument:
$\frac ab$ where $a=1; b=1$ is not a pair of integers where $\frac ba =1-\frac ab$. Let $p,q$ be such that $q$ is the smallest possible positive integer where there is any integer $p'$ where $\frac q{p'} = 1-\frac {p'}q$ and $p$ is the least possible $p'$ integer where that is true. Then $\frac qp = \frac {q -p}q$ and $\frac q{q-p} = \frac pq = \frac {q-(q-p)}q = 1-\frac {q-p}q$ and $q$ and $q-p$ are also such numbers. But $q-p < q$ and that is a contradiction.
Best Answer
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.