Constructive proof of transcendence of $e$ and $\pi$?

Someone asks to me that can we prove the transcendence of $\pi$ without using proof by contradiction. I find some proofs of transcendence of $\pi$ and $e$ and I found that all of proof I found starts from the sentence "Suppose $x$ is algebraic...", which means the proof relies on the proof by contradiction.

Since $\pi$ and $e$ are computable, I suspect that their transcendence can be proven in constructive sense, and I guess that we may eliminate non-constructive nature of a proof of transcendence of them. However I didn't get how to do this. Thanks for any idea.

• Trascendence means something does not exist. One does not usually do that constructively... Sep 6 '15 at 2:47
• @MarianoSuárez-Alvarez But Wikipedia says about the constructive proof of irrationality of $\sqrt{2}$, so I think constructive proof of transcendence of some (irrational) numbers might exist, though it may be hard to find. Sep 6 '15 at 2:50
• That proof is based on the fact that rational numbers can be approximated in a specific way: as that does not work with $\sqrt2$, then it is not rational. I would not call such an argument constructive. Sep 6 '15 at 2:52
• @MarianoSuárez-Alvarez There is a difference between our understanding of constructiveness. The proof gives the difference between $\sqrt{2}$ and given rational constructively, and it does not use proof by contradiction, since it explicitly gives the fact that $\sqrt{2}$ is different from any given rational, so it is irrational. Sep 6 '15 at 2:57

I have not examined the proof that $\pi$ is transcendental to see whether it can be rephrased as a constructive proof. There are many other issues in classical logic that are nonconstructive, such as the way that "or" is handled, and the way that equality of real numbers is handled. But the mere fact that one proves $\pi$ is transcendental by assuming $\pi$ is algebraic is not a real concern in mathematical constructivism.