What are some numbers that seem irrational/transcendental because of the way they look but actually aren't?

Question

It is not known whether $\pi+e, \pi-e, \frac{\pi}{e}, etc$ are irrational, it has also been shown that at least one of the numbers $\zeta(5), \zeta(7), \zeta(9), \zeta(11)$ is irrational. From my point of view, it looks "obvious" that all these numbers are irrational however proving them is a different story. Do you know of any numbers that seem irrational/transcendental at first glance but actually aren't? (except $e^{i\pi}$)

Answer 1

The problem is that any rational number can be written, by defintion, under the simple closed form $\frac{a}{b}$ with $a$ and $b$ integers. What you actually look for, I guess, is mathematical expression which look irrational but actually equal a rational number.

You can find convoluted expressions with roots like the one shared by J.-E. Pin, $$ \sqrt{4 + 2\sqrt{3}} - \sqrt{3} = 1. $$ A good way to generate them is to use Cardano's formula for cubic equations. If $p,q$ are real numbers with $-\Delta = \frac{q^2}{4} + \frac{p^3}{27} > 0$, then the equation $t^3 + pt + q = 0$ has one unique real solution, $$ t = \sqrt[3]{-\frac{q}{2} + \sqrt{-\Delta}} + \sqrt[3]{-\frac{q}{2} - \sqrt{-\Delta}}. $$ Similar results exist to find the complex roots when $-\Delta > 0$ and all three real roots when $-\Delta \leqslant 0$ but it doesn't matter here. Now, take a rational number, $r$. Take a complex number $z$ and develop the following real degree $3$ polynomial, \begin{align*} (t - r)(t - z)(t - \overline{z}) & = (t - r)(t^2 - 2\Re(z)t + |z|^2)\\ & = t^3 - (r + 2\Re(z))t^2 + (2r\Re(z) + |z|^2)t - r|z|^2. \end{align*} We want the term in $t^2$ to vanish so $\Re(z) = -\frac{r}{2}$. The condition $-\Delta > 0$ is automatically fulfilled if $\Im(z) \neq 0$ because it is equivalent of having only one real root. All the equalities that follow remain true when $\Im(z) = 0$ by continuity.

We have $p = 2r\Re(z) + |z|^2 = -r^2 + |z|^2$ and $q = -r|z|^2$ so, $$ -\Delta = \frac{q^2}{4} + \frac{p^3}{27} = \frac{r^2|z|^4}{4} + \frac{(|z|^2 - r^2)^3}{27} = \frac{r^2|z|^4}{4} + \frac{|\Im(z)|^6}{27}. $$ It is, as expected, positive, and by Cardano's formula, $$ r = \sqrt[3]{\frac{r|z|^2}{2} + \sqrt{\frac{r^2|z|^4}{4} + \frac{|\Im(z)|^6}{27}}} + \sqrt[3]{\frac{r|z|^2}{2} - \sqrt{\frac{r^2|z|^4}{4} + \frac{|\Im(z)|^6}{27}}}. $$ For example, with $r = 2$ and $z = 1 - i$, $$ 2 = \sqrt[3]{2 + \sqrt{\frac{100}{27}}} + \sqrt[3]{2 - \sqrt{\frac{100}{27}}} = \sqrt[3]{2 + \frac{10}{9}\sqrt{3}} + \sqrt[3]{2 - \frac{10}{9}\sqrt{3}} $$ Or with $r = \frac{2}{3}$ and $z = -\frac{1}{3} + \sqrt[4]{5}$, \begin{align*} \frac{2}{3} = & \sqrt[3]{\frac{1}{3}\left(\frac{1}{9} + \sqrt{5}\right) + \sqrt{\frac{1}{9}\left(\frac{1}{9} + \sqrt{5}\right)^2 + \frac{\sqrt{125}}{27}}} + (\cdots)\\ & = \sqrt[3]{\frac{1}{27} + \frac{1}{3}\sqrt{5} + \sqrt{\frac{406}{729} + \frac{17}{81}\sqrt{5}}} + \sqrt[3]{\frac{1}{27} + \frac{1}{3}\sqrt{5} - \sqrt{\frac{406}{729} + \frac{17}{81}\sqrt{5}}}. \end{align*} It is not related but you also have the famous, $$ \left(\sqrt{2}^\sqrt{2}\right)^{\sqrt{2}} = 2. $$

Answer 2

One example might be the number $$1 / 9899 = .00010102030508132135\ldots$$ given by concatenating the Fibonacci numbers $1,2,3,5,8,13,21,35, \ldots$ (although the pattern breaks down slightly when the Fibonacci numbers require more than two digits.)

Answer 3

The followings of hypergeometric function might be such examples.

$${}_2F_1\bigg(\frac 18,\frac 18;\frac 12;-2400\bigg)=\sum_{n=0}^{\infty}\frac{(\frac 18)_n(\frac 18)_n}{(\frac 12)_n}\frac{(-2400)^n}{n!}=\frac 23$$

$${}_2F_1\bigg(\frac 38,\frac 38;\frac 12;-2400\bigg)=\sum_{n=0}^{\infty}\frac{(\frac 38)_n(\frac 38)_n}{(\frac 12)_n}\frac{(-2400)^n}{n!}=\frac{2}{21}$$

where ${}_2F_1(a,b;c;z)$ is the hypergeometric function and $(q)_n$ is the (rising) Pochhammer symbol whose definition is $$(q)_n=\begin{cases}1&\text{if $n=0$}\\ q(q+1)\cdots (q+n-1)&\text{if $n\gt 0$}\end{cases}$$

So, for example, we have $$\bigg(\frac 18\bigg)_n=\frac 18\bigg(\frac 18+1\bigg)\cdots \bigg(\frac 18+n-1\bigg)=\frac{\Gamma(n+\frac 18)}{\Gamma(\frac 18)}$$

You can see the other examples here.

(You might also want to see (18)(19) here which are written in this answer to the question Jair Taylor wrote in a comment.)

---

Added :

The followings also might be such examples.

$$\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^2}\bigg(\frac 14\bigg)^n\frac{1}{n+1}=2$$ $$\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^2}\bigg(\frac 14\bigg)^n\frac{1}{n+2}=\frac 43$$ $$\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^2}\bigg(\frac 14\bigg)^n\frac{1}{n+3}=\frac{16}{15}$$ $$\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^2}\bigg(\frac 14\bigg)^n\frac{1}{n+4}=\frac{32}{35}$$

By the way, the followings are similar but not rational.

$$\sum_{n=0}^∞\left(\frac{(n!)^2}{(2n)!}\right)^5(-256)^{n}\frac{2(10n^2+14n+5)}{7(2n+1)^5}=ζ(3)$$

$$\sum_{n=0}^∞\left(\frac{(2n)!}{(n!)^2}\right)^3\bigg(\frac 1{256}\bigg)^n\frac{6n+1}{4}=\frac{1}{\pi}$$

$$\sum_{n=0}^∞\left(\frac{(n!)^2}{(2n)!}\right)^7(256)^n\frac{16(21n^3+41n^2+27n+6)}{(2n+1)^7}=π^4$$