I was doing some math for a programming project of myself and ran into decimal numbers and how to define them without losing precision while calculating an expression, so I tried writing them down as fractions, which actually worked quite nicely.
I know that irrational numbers exist and that they cannot be expressed as a fraction (hence the name!).
However, there are a lot of irrational numbers that can be expressed as a root, like ${\sqrt2}$.
Now I have been trying to find a way to express ${\pi}$ as a root in combination with fractions.
The rules are simple:

  • All numbers used must be integers
  • Valid operators are adding, subtracting, multiplying, dividing, powers and roots

So an expression like ${\sqrt[4]{2^\frac{4}{5} + 4 -\frac{8}{3}}}$ would be valid for my case.
I just seem to lack the insight to either proof that it is possible or that it is not.

I hope my question is clear, if not please leave a comment.

I changed the question a bit to put more focus on the fact that I want to see the proof why it is/isn't possible.

  • 4
    $\begingroup$ It is transcendental. It is very much not obvious, but it was found in the 19th century that your rules will never give $\pi$. $\endgroup$
    – Empy2
    Commented Oct 19, 2013 at 19:03
  • $\begingroup$ But look into approximations of pi. The thing with programming is that at some point you'll be limited by the machine with regards to precision. en.wikipedia.org/wiki/Approximations_of_%CF%80 If you give the wiki page a read, you'll see that there are simple ways to represent pi with arbitrary precision. So, if you want to abstract away the precision of pi, it could come in handy. (or else, just use 22/7 ;) ) $\endgroup$ Commented Oct 19, 2013 at 19:14
  • $\begingroup$ @Michael: Is this strong enough? I don't think Lindemann-Weierstrass says anything about numbers like $5^\sqrt2$, which his rules allow. $\endgroup$ Commented Oct 19, 2013 at 19:22
  • $\begingroup$ Hm, some further Wikipedia explorations lead me to guess that Baker's theorem is strong enough, because maybe you could iterate away the power towers. But I am not quite sure if I am reading the theorem right. $\endgroup$ Commented Oct 19, 2013 at 19:39
  • $\begingroup$ Good point. Sorry, my initial comment is wrong. $\endgroup$
    – Empy2
    Commented Oct 19, 2013 at 19:39

1 Answer 1


This proof is from The Transcendence of $\pi$ by Steve Mayer, November 2006

I've rewritten it below partly so that I understand it and partly so that if you're too lazy to click on the link or if the link goes away there's a copy here.

Definition: A complex number is algebraic over $\mathbb{Q}$ if it is a root of a polynomial equation with rational coefficients.

i.e: $a$ is algebraic if there are rational numbers $\alpha_0, \alpha_1,...,\alpha_n$ not all 0 such that $\alpha_0a^n + \alpha_1a^{n-1} + ... + \alpha_{n-1}a + \alpha_n = 0$

Definition: A complex number is transcendental if it is not algebraic.

Theorem (Lindemann-Weierstrass): $\pi$ is transcendental over $\mathbb{Q}$

Proof: If $\pi$ satisfies an algebraic equation with coefficients in $\mathbb{Q}$, so does $i\pi$. Let this equation be $\theta_1(x) = 0$ with roots $i\pi = \alpha_1, ..., \alpha_n.$ Now, $e^{i\pi} + 1 = 0$ so

$(e^{\alpha_1} + 1)...(e^{\alpha_n} + 1) = 0$

We now construct an algebraic equation with integer coefficients whose roots are the exponents of $e$ in the expansion of the above product. For example, the exponents in pairs are $\alpha_1 + \alpha_2, \alpha_1 + \alpha_3, ..., \alpha_{n-1} + \alpha_n$. The $\alpha$'s satisfy a polynomial equation over $\mathbb{Q}$ so their elementary symmetrix functions are rational. Hence the elementary symmetric functions of the sums of pairs are symmetric functions of the $\alpha$'s and are also rational. Thus the pairs are the roots of the equation $\theta_2(x) = 0$ with rational coefficients. Similarly, sums of 3 $\alpha$'s are roots of $\alpha_3(x) = 0$, etc...

Then, the equation

$\theta_1(x)\theta_2(x)...\theta_n(x) = 0$

is a polynomial equation over $\mathbb{Q}$ whose roots are all sums of $\alpha$'s. Deleting zero roots from this, if any, we get:

$\theta(x) = 0$

$\theta(x) = cx^r + c_1x^{r-1}+...c_r$

and $c_r \neq 0$ since we have deleted zero roots. The roots of this equation are the non-zero exponents of $e$ in the product when expanded. Call these $\beta_1,...,\beta_r$. The original equation becomes

$e^{\beta_1} + ... e^{\beta_r} + e^0 + ... e^0 = 0$

i.e: $\sum e^{\beta_i} + k = 0$

where k is an integer $> 0$ ($\neq 0$ since the term $1...1$ exists)

Now define

$f(x) = c^sx^{p-1}\frac{[\theta(x)]^p}{(p-1)!}$

where $s = rp-1$ and $p$ will be determined later.


$F(x) = f(x) + f'(x) +...+f^{(s+p)}(x)$


$\frac{d}{dx}[e^{-x}F(x)] = - e^{-x}f(x)$

Hence, we have

$e^{-x}F(x) - F(0) = - \int\limits_{0}^{x}e^{-y}f(y)dy$

Putting $y = \lambda x$ we get

$F(x) - e^xF(0) = -x \int\limits_{0}^{1}e^{(1-\lambda)x}f(\lambda x)d\lambda$

Let $x$ range over the $\beta_i$ and sum. Since $\sum e^{\beta_i} + k = 0$ we get

$\sum\limits_{j = 1}^{r} F(\beta_j) + kF(0) = - \sum\limits_{j=1}^{r} \beta_j \int\limits_{0}^{1} e^{(1-\lambda)\beta_j} f(\lambda \beta_j)d\lambda$

CLAIM For large enough $p$ the left hand size is a non-zero integer.

$\sum\limits_{j=1}^{r} f^{(t)}(\beta_j) = 0$ $(0 < t < p)$ by definition of $f$. Each derivative of order p or more has a factor $p$ and a factor $c^s$, since we must differentiate $[\theta(x)]^p$ enough times to not get $0$. And $f^{(t)}(\beta_j)$ is a polynomial in $\beta_j$ of degree at most $s$. The sume is symmetric, and so in an integer provided each coefficient is divisible by $c^s$, which it is. (symmetric functions are polynomials in coefficients $=$ polynomials in $\frac{c_i}{c}$ of degree $\leq s$). Thus we have,

$\sum\limits_{j=1}^r f^{(t)}(\beta_j) = pk_t$

$t = p,...,p+s$

Thus, the left hand side $LHS=$ (integer) = $kF(0).$ What is $F(0)?$

$f^{(t)}(0) = 0$ $t = 0, ...,p-2$

$f^{(p-1)}(0) = c^sc_r^p$ $(c_r \neq 0)$

$f^{(t)}(0) = p$(some integer) $t = p, p+1,...$

So, $LHS$ is an integer multiple of $p+c^sc_r^pk$. This is not divisible by $p$ if $p > k, c, c_r$. So it is a non-zero integer. But the right hand side tends to $0$ as $p \rightarrow \infty$ and thus we get a contradiction and thus $\pi$ is transcendental.

  • $\begingroup$ Admittedly I haven't gone though the whole thing in detail, but I don't see anything in this post that considers expressions like $x^\sqrt2+1$. $\endgroup$ Commented Oct 19, 2013 at 21:41
  • $\begingroup$ I tried to understand the document, but I'm not familiar with most terms, making it hard for me to understand. On the other hand I do see that through pure algebra it is impossible to achieve an expression that defines ${\pi}$. Non the less I'm very curious about the potential answer on the comment of Eric Stucky. $\endgroup$
    – Mixxiphoid
    Commented Oct 20, 2013 at 10:32
  • 1
    $\begingroup$ I read it again and there is no mention on the use of irrational numbers to express ${\pi}$ (beside one mention of ${i}$). An Irrational number like ${\sqrt2}$ has a lot common with ${\pi}$ (not a real ending, not fractionable, etc) and is (as far as I can understand) not used in any context to disprove the expression of ${\pi}$ in the way as noted in the question. $\endgroup$
    – Mixxiphoid
    Commented Oct 20, 2013 at 12:37

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