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

Question

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)

Answer 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.

Answer 2

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.

Answer 3

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.

Answer 4

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

Answer 5

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.

Answer 6

Before proceeding further, we observe that every $\varphi$ satisfying

$\varphi^2 - \varphi - 1 = 0 \tag 1$

must be irrational, for via the quadratic formula we see that

$\varphi = \dfrac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot -1}}{2} = \dfrac{1 \pm \sqrt{(-1)^2 +4}}{2} = \dfrac{1 \pm \sqrt 5}{2}, \tag 2$

and $\sqrt 5$ is known to be irrational via classic arguments; indeed, we may to a great extent follow the classic argument given for $\sqrt 2$, in which we assume

$\sqrt 2 = \dfrac{p}{q}, \; \gcd(p, q) = 1, \tag 3$

and take

$\sqrt 5 = \dfrac{p}{q}, \; \gcd(p, q) = 1; \tag 4$

(3) yields

$p^2 = 2q^2, \tag 5$

whilst from (4)

$p^2 = 5q^2; \tag 6$

now from (5) it follows that $p^2$ is even, and from this we deduce that $p$ itself is even, for if it were odd we could write

$p = 2t + 1, \tag 7$

whence

$p^2 = 4t^2 + 4t + 1 = 2(2t^2 + 2t) + 1, \tag 8$

clearly odd. With $p$ even we have

$p = 2t, \tag 9$

$2q^2 = p^2 = 4t^2, \tag{10}$

from which

$q^2 = 2t^2, \tag{11}$

at which point we may now repeat the argument just given and conclude that

$q = 2s; \tag{12}$

now (9)-(12) show us that

$2 \mid p, \; 2 \mid q, \tag{13}$

from which we may infer that

$2 \mid \gcd(p, q); \tag{14}$

(14) is in fact an application of the following simple fact concerning the $\gcd$ function: if $a \mid p$ and $a \mid q$ then $a \mid \gcd(p, q)$; this in turn may be seen via Bezout's identity which grants the existence of integers $x$ and $y$ such that

$xp + yq = \gcd(p, q); \tag{15}$

now it is easy to see $a \mid xp + yq$, so indeed $a \mid \gcd(p,q)$; but (14) is in fact prohibited by (3), which affirms that

$\gcd(p, q) = 1; \tag{16}$

this contradiction of course affirms the fact that $\sqrt 2$ is irrational, that

$\sqrt 2 \notin \Bbb Q. \tag{17}$

The case (6) is only slightly more complicated; we have

$5 \mid p^2, \tag{18}$

and we want to show that

$5 \mid p; \tag{19}$

now, $5$ being prime its only divisors are $1$ and $5$ itself, so turning to $\gcd(5, p)$, we see that it too can only take the values $5$ or $1$, since it too is a divisor $5$; if

$\gcd(5, p) = 5, \tag{20}$

then we see that (19) holds without further argument; if on the other hand

$\gcd(5, p) = 1, \tag{21}$

we may again invoke Bezout's and affirm the existence of $a, b \in \Bbb Z$ such that

$5a + bp = 1, \tag{22}$

which upon multipliction by $p$ yields

$5ap + bp^2 = p; \tag{23}$

now since

$5 \mid 5ap \tag{24}$

and (18) we conclude that (19) indeed binds; then writing

$p = 5t \tag{25}$

it follows via (6) that

$25t^2 = p^2 = 5q^2, \tag{26}$

whence

$5t^2 = q^2; \tag{27}$

we now have attained

$5 \mid q^2, \tag{28}$

and so applying the same logic as in (20)-(24) we have

$5 \mid q; \tag{29}$

but now (19) and (29) taken in concert imply (see the remarks ca. (14)-(15))

$5 \mid \gcd(p, q), \tag{30}$

which contradicts our assumption (4) that $\gcd(p, q) = 1$, and in light of this we are forced to conclude that $\sqrt 5$ is irrational:

$\sqrt 5 \notin \Bbb Q. \tag{31}$

This being the case, it follows that $\varphi$ is also irrational, since it may be formed from $\sqrt 5$ by strictly rational operations; indeed, from (2) we may write

$1 + \sqrt 5 = 2\varphi, \tag{32}$

where we have dropped the "$-$" sign since only the positive root of (1) is of interest here; then

$\sqrt 5 = 2\varphi - 1, \tag{33}$

forcing $\sqrt 5$ to be rational if $\varphi$ is; but this is impossible in light of what we have done above.

At this juncture we point out an important and vast generalization of our work so far, viz. that if

$1 < \pi \in \Bbb Z \tag{33.1}$

is any positive prime integer, and

$2 \le n \in \Bbb Z, \tag{33.2}$

then

$\sqrt[n] \pi \notin \Bbb Q; \tag{33.3}$

for indeed, if

$\sqrt[n] \pi = \dfrac{p}{q}, \; \gcd(p, q) = 1, \tag{33.4}$

then evidently

$\pi q^n = p^n, \tag{33.5}$

so as in the above we have

$\pi \mid p^n; \tag{33.6}$

since $\pi$ is prime its only two divisors are $1$ and $\pi$ itself; thus $\gcd(\pi, p)$ is either $\pi$ or $1$; if

$\gcd(\pi, p) = \pi, \tag{33.7}$,

then

$\pi \mid p; \tag{33.8}$

if, on the other hand

$\gcd(\pi, p) = 1, \tag{33.9}$

then in accord with the above we may find integers

$a, b \in \Bbb Z \tag{33.10}$

with

$a\pi + bp = 1; \tag{33.11}$

we multiply this equation by $p^{n - 1}$:

$a\pi p^{n - 1} + bp^n = p^{n - 1}, \tag{33.12}$

and now in light of (33.6) we see that

$\pi \mid p^{n - 1}, \tag{33.13}$

since we also have

$\pi \mid \pi p^{n -1}; \tag{33.14}$

we may repeat steps (33.6)-(33.14) a long as the exponent occurring on $p$ in (33.6) is greater than $2$, finally arriving at

$\pi \mid p^2 \tag{33.15}$

in lieu of (33.6), from which (33.8) follows via one final pass over this "loop".

Note:

Then from (33.8),

$p = \pi t, \tag{33.15}$

$p^n = \pi^n t^n = \pi q^n, \tag{33.16}$

and since $n \ge 2$,

$\pi^{n -1} t^n = q^n, \tag{33.17}$

$\pi \mid q^n, \tag{33.18}$

and as in (33.5)-(33.15) we infer that

$\pi \mid q; \tag{33.19}$

combining (33.8) and (33.19) we find

$\pi \mid \gcd(p, q), \tag{33.20}$

whence

$1 < \pi \le \gcd(p, q), \tag{33.21}$

which contradicts (33.4); hence there are no integers $p$, $q$ such that

$\sqrt[n] \pi = \dfrac{p}{q}, \tag{33.22}$

and so we conclude

$\sqrt[n] \pi \notin \Bbb Q. \tag{33.23}$

End of Note.

Now, amongst other things, we have accomplished up to this point a demonstration of the fact that

$\varphi \notin \Bbb Q, \tag{34}$

effected in perhaps the most basic and obvious way, perhaps at the expense of being overlong and somewhat inelegant.

There are, however, other means to attack such problems; one may, for example, directly assume $\varphi$ is rational,

$\varphi = \dfrac{p}{q}, \; p, q \in \Bbb Z, \; \gcd(p,q) = 1, \tag{35}$

and then (1) becomes

$\left (\dfrac{p}{q} \right )^2 - \dfrac{p}{q} - 1 = 0, \tag{36}$

or

$\left (\dfrac{p}{q} \right )^2 = \dfrac{p}{q} + 1, \tag{36}$

which we multiply through by $q^2$:

$p^2 = pq + q^2 = q(p + q), \tag{37}$

which shows that

$q \mid p^2; \tag{38}$

now an application of Bezout's identity in light of $\gcd(p, q) = 1$ yields $a, b \in \Bbb Z$ with

$ap + bq = 1, \tag{39}$

which we multiply through by $p$:

$ap^2 + bpq = p, \tag{40}$

from which we immediately infer in light of (38) that

$q \mid p; \tag{41}$

but this implies that

$\gcd(p, q) = q, \tag{42}$

and we now conclude, given that we have $\gcd(p, q) = 1$ by hypothesis, that

$q = 1, \tag{43}$

so that (37) becomes

$p^2 = p + 1, \tag{44}$

or

$p(p - 1) = p^2 - p = 1; \tag{45}$

however, no $p \in \Bbb Z$ with

$\vert p \vert \ge 2 \tag{46}$

satisfies (45), implying as it does that

$p \mid 1; \tag{47}$

the only integers with

$\vert p \vert < 2 \tag{48}$

are

$p = -1, \; 0, \; 1, \tag{49}$

none of which fulfill (44)-(45); thus we see that (44)-(45) has no integral solutions, which ultimately contradicts our assumption that

$\varphi = \dfrac{p}{q} \tag{50}$

is rational.

Careful scrutiny of these considerations and results leads one to ask under what circumstances a more general result might be obtained; in order to ascertain just what such a result might be, we observe that the work we have done so far may all be seen as addressing the roots of polynomials with integer coefficients, that is, of polynomials

$f(x) \in \Bbb Z[x]; \tag{51}$

certainly equation (1) is in this class, and if we cast our questions about the rationality of $\sqrt 2$ and $\sqrt 5$ in polynomial form we obtain

$x^2 - 2 = 0, \tag{52}$

$x^2 - 5 = 0; \tag{53}$

furthermore, our assertion that

$\sqrt[n]\pi \notin \Bbb Q \tag{54}$

is equivalent to the assertion that

$x^n - \pi = 0 \tag{55}$

has no rational zeroes as well. We further observe that each of these polynomials is monic, i.e. has leading coefficient $1$, and that this also applies to our original polynomial for $\varphi$, which satisfies

$x^2 - x - 1 = 0. \tag{56}$

These observations invite us to speculate that an even broader generalization may be possible, one which embraces nearly all of the results we have obtained so far. To this end we examine monic polynomials in $\Bbb Z[x]$, which may be written in the form

$f(x) = x^n + \displaystyle \sum_0^{n -1} a_i x^i, \; a_i \in \Bbb Z; \; \tag{57}$

if such a polynomial has a rational root

$x = \dfrac{p}{q}; \; \gcd(p, q) = 1, \tag{58}$

then

$\left ( \dfrac{p}{q} \right )^n + \displaystyle \sum_0^{n -1} a_i \left ( \dfrac{p}{q} \right )^i = f\left (\dfrac{p}{q} \right ) = 0; \tag{59}$

we observe this may be written in the form

$\dfrac{p^n}{q^n} + \displaystyle \sum_0^{n -1} a_i \dfrac{p^i}{q^i} = 0; \tag{60}$

we multiply this through by $q^n$:

$p^n + \displaystyle \sum_0^{n -1} a_i p^iq^{n-i} = 0; \tag{61}$

since

$n - i \ge 1 \tag{62}$

for every $i$ in the range

$0 \le i \le n - 1, \tag{63}$

we may write (61) in the form

$p^n + \displaystyle q\sum_0^{n -1} a_i p^iq^{n-i - 1} = 0, \tag{64}$

or

$p^n = -\displaystyle q\sum_0^{n -1} a_i p^iq^{n-i - 1}, \tag{65}$

which shows that

$q \mid p^n. \tag{66}$

From this we may infer that

$q \mid p \tag{67}$

in a manner similar to our work in the above, to wit:

$\gcd(p, q)= 1 \tag{68}$

implies the existence of

$a, b \in \Bbb Z \tag{69}$

such that

$ap + bq = 1, \tag{70}$

which we multiply through by $p^{n - 1}$ to obtain

$ap^n + bqp^b = p^{n - 1}, \tag{71}$

which shows that

$q \mid p^{n - 1} \tag{72}$

in the light of (66). This sequence of steps may now be repeated until (67) is obtained. In light of (67) we see that

$\gcd(p, q) = q, \tag{73}$

and in concert with (68) this yields

$q = \gcd(p, q) = 1; \tag{74}$

referring to (58), it is seen that

$x = p \in \Bbb Z \tag{75}$

is in fact a root of $f(x)$. A moment's reflection reveals that what we have in fact shown is that any rational zero of $f(x)$ must in fact be an integer.