Why is it so hard to prove a number is transcendental?

Why is it so hard and difficult to prove that $e +\pi, \pi - e, \pi e, \frac{\pi}{e}$ etc. are transcendental numbers?

It is generally more difficult to prove a number is transcendental than to prove it is not transcendental, i.e. that it is algebraic. Showing a number x is algebraic amounts to proving it is the root of a polynomial with rational coefficients, and so one often can just exhibit the polynomial and show by computation that a particular x is its root. Proving number x is transcendental amounts to proving no rational polynomial exists with that number as a root, and this requires more work (because we will be "proving the negative", i.e. exhausting all possible polynomials).

We know that $\pi$ and $e$ are transcendental numbers, why we can't deduce that $e + \pi \approx 5.859874482048838473822930854632165381954416493075065395941912...$ or $\pi - e \approx 0.423310825130748003102355911926840386439922305675146246007976...$ are also transcendental numbers?

For example, $\pi$ and $1−\pi$ are transcendental, but $\pi+(1−\pi)=1$ is not.

Since both of you gave imho a great answer, i don't know whom of you should become the credits...

• It is generally more difficult to prove a number is transcendental than to prove it is not transcendental, i.e. that it is algebraic. Showing a number $x$ is algebraic amounts to proving it is the root of a polynomial with rational coefficients, and so one often can just exhibit the polynomial and show by computation that a particular $x$ is its root. Proving number $x$ is transcendental amounts to proving no rational polynomial exists with that number as a root, and this requires more work (because we will be "proving the negative", i.e. exhausting all possible polynomials). May 16 '15 at 14:59
• Thx. This is a good point!
– aGer
May 16 '15 at 15:04

For example, $\pi$ and $1-\pi$ are transcendental, but $\pi+(1-\pi)=1$ is not.

• Thx! This answers my second question.
– aGer
May 16 '15 at 14:55

It is mainly because transcedental numbers behave so weird. For example, would you think that a transcedental number raised to an irrational is an integer? Well, it is possible:

$$(2^{\sqrt{2}})^{\sqrt{2}} = 2^{(\sqrt{2})^2}=4$$

• This is a variant of one of my favorite examples of a nonconstructive proof, namely that there exists an irrational that, when raised to the power of another irrational, yields a rational. Either $\sqrt{2}^{\sqrt{2}}$ is rational, and we're done, or else it is irrational, and $(\sqrt{2}^\sqrt{2})^\sqrt{2} = 2$ is our example. In your example, you have not shown that $2^{\sqrt{2}}$ is irrational, much less transcendental. At the level of this question, I think it is also maybe not helpful to appeal to Gelfond-Schneider, which is presumably what you had in mind. May 16 '15 at 21:59
• I do not find it a surprise at all. A positive transcedental number $t$ raised to a non-zero rational power $\frac{p}{q}$ cannot be a integer $n \gt 1$ (otherwise $n^q - t^p =0$ and $t$ would be algebraic), But clearly $\log_t n$ is a number and $t^{\log_t n} = n$. May 17 '15 at 10:51

A difficulty of showing those numbers transcendental lies in the fact that they are not numbers that seem to occur in a natural way.

Sure, they are the sum, product and so on of two of the most common constants, but in "actual mathematics" they rarely appear in this form, and as illustrated in other answers summing and multiplying might not preserve transcendence.

But on the good side $e^{\pi}$ is known transcendental!

(Also $e+\pi$ or $e\pi$ is transcendental as otherwise $\pi$ and $e$ being roots of the polynomial $x^2 - (e + \pi )x + e \pi= (x-e)(x - \pi)$ yields a contradiction to them being transcendent.)

• Also, two of $e+\pi$, $e-\pi$ and $e\pi$ are transcedental. May 17 '15 at 11:04