## A Hopeless Problem: Can π + e or π – e be rational?

October 28, 2013 § Leave a comment

Today, I thought I would share a problem with you that, much like a lot of the things I will be posting, I heard from my Complexity Theory professor who inspired me to write this blog.

The problem is simple, and requires only a little bit of background information, so I will start with that.

First, a rational number is a number that can be expressed as the quotient *a/b* of two integers *a* and *b* where *b*≠0. We can add to the definition that *a* and *b* do not have common divisors as is useful when demonstrating that the square root of 2 is irrational.

From here, we can say that a real irrational number is a real number that can not be expressed as such. Much to the surprise of many people long ago and very few people now, there exist many of these (in fact, there are many more of these than there are rational numbers). Among these are the famous mathematical constants *e* and *π*.

It has been known for a very long time that each of these is irrational (proofs can be found on Wikipedia as well as many other places on the internet here and here), but shockingly, it remains unknown whether or not their sum or difference is irrational (that is, we do not know if *π+e* is rational, and we do not know if *π-e* is rational) . Peculiar, no?

What is even more interesting about this problem is that we can can conclude that at least one of their sum or difference is irrational rather easily. We simply assume that neither of them are irrational (so both of them are rational). Then, we have, because the rational numbers are closed under addition, that (*π-e)+( π+e)=2π* is rational. Because dividing a rational number by another non-zero rational number produces a rational number, we then get that

*π*is rational. However, this contradicts the theorem that

*π*is irrational, so we can conclude that

*π+e*and

*π-e*can not both be rational. Of course, most people, if not all people, who think about this would probably guess with a high degree of certainty that neither of these numbers are rational, and yet, we find ourselves hopeless to prove it.

Additionally, because *π* and* e *are each transcendental (meaning not algebraic), as well, and algebraic numbers are closed under the same operations we used in our proof, we find that at least one of the sum or difference must be transcendental, as well.

Thanks for reading!

-A Student of Logic