Here's a list of irrational numbers which almost fulfill your criteria.
Each were chosen to be accurate to within ±0.1, so that no two of them implicitly express the same mathematical approximation, and so that none of them "cheats" in order to fudge an exact result involving integers to obtain a slightly inexact, irrational result. Only the last number fails to meet your criteria, as it is slightly larger than 12.
- $\ln(3)$
- $7\pi/11$
- $\sqrt 2 + \frac\pi2$
- $7/{\sqrt[3]5}$
- $\mathrm e^\phi$
- $\sec^2(20)$
- $4\sqrt3$
- $5\phi$
- $2\pi+\mathrm e$
- $\sinh(3)$
- $\pi^3-20$
- $\csc^2(16)$
I would prefer not to use cosecant, integers greater than 12, or more than two additions/subtractions — it is too easy to get results if you rely on these — but I think I've spent enough time on this diversion for now. :-)
[EDIT: revised the formula for 4 two times now: this first to change the formula for 4 from √3 + √5 — which is too close to √4 + √4 — and the second time to correct the formula as I somehow copied a result which was not approximately 4.]
Added: Since it seems that I can't sleep tonight, here is a list of approximate values:
$\begin{align*}
\ln(3) & \approx 1.0986 \\
7\pi/11 & \approx 1.9992 \\
\sqrt 2 + \tfrac\pi2 & \approx 2.9850 \\
7 / \sqrt[3]5 & \approx 4.0936 \\
\mathrm e^\phi & \approx 5.0432 \\
\sec^2(20) & \approx 6.0049 \\
4\sqrt3 & \approx 6.9282 \\
5\phi & \approx 8.0902 \\
2\pi+\mathrm e & \approx 9.0015 \\
\sinh(3) & \approx 10.018 \\
\pi^3-20 & \approx 11.006 \\
\csc^2(16) & \approx 12.064
\end{align*}$
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.
Define:
$F(x) = f(x) + f'(x) +...+f^{(s+p)}(x)$
Then,
$\frac{d}{dx}[e^{-x}F(x)] = - e^{-x}f(x)$
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.
Best Answer
Short answer: You're looking at the wrong sets. It's not about differentiating the irrational numbers from the trancendental numbers. It's about differentiating the rational numbers from the algebraic numbers. Both the rational and algebraic numbers are so important that even their complements have special names, and you seem to be focusing on those.