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Really just what the title says. I have checked this old post Why is it hard to prove whether $\pi+e$ is an irrational number? but I am left none the wiser. I am trying to understand why there is a possibility that $e$ + $π$ could be rational.

lightningjay
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    Hint: both $\sqrt2$ and $-\sqrt2$ are irrational numbers. but what about their sum? – user Jun 05 '24 at 09:07
  • @user I suppose that works, but the relationship between $e$ & $π$ makes me a little confused about why this possibility exists. – lightningjay Jun 05 '24 at 09:09
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    It is just possible that the sum of two irrational numbers is rational. – user Jun 05 '24 at 09:10
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    We simply do not have a proof. And in general , the sum of transcendental numbers can be rational. It is in fact very unlikely that $e+\pi$ is rational, but possible until proven otherwise. – Peter Jun 05 '24 at 09:11
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    @Peter I seem to understand now.

    Transmitting to

     – lightningjay Jun 05 '24 at 09:15
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    We know the surprising fact that $e^{i \pi}$ is rational, so why couldn't this happen again and make $e \pi$ also rational? – Trebor Jun 05 '24 at 09:33
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    @Trebor Each of $e+\pi$ and $e\cdot \pi$ might be rational , but they cannot both be rational. In fact, at least one of them must be transcendental. – Peter Jun 05 '24 at 09:46

2 Answers2

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Just like $e^{i\pi}=-1$ and $i^i=e^{-\pi/2}$, it is also possible that $(e+\pi)$ may yield a rational number. Likewise, $2^\sqrt2$ (Gelfond–Schneider constant) is transcendental even if $2$ and $\sqrt2$ are algebraic. But $(e+\pi)$ is still to be proven to be either irrational or rational (or transcendental or algebraic). However, mathematicians and number theorists believe it is unlikely to be rational.

It is important to observe that an algebraic function of several variables may yield a rational number when applied to irrational numbers. For example, $(1+\sqrt2)$ and $(2-\sqrt2)$ are both irrational numbers, but $1+\sqrt2+2-\sqrt2=3$ is obviously not.

Generally, for any two transcendental numbers, $a$ and $b$, at least one of $(a+b)$ and $ab$ must be transcendental or non-algebraic irrational because $(x-a)(x-b)=x^2-(a+b)x+ab$ can not have all algebraic coefficients. It is known that both $e$ and $\pi$ are transcendental. So, at least one of $(e+\pi)$ and $e\pi$ is transcendental and therefore irrational (non-algebraic). The polynomial does not demand both of them to be transcendental/irrational. Hence, there is still a possibility that one of them could be rational.

Hope it clarifies!

Pustam Raut
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Without a proof, we can't know whether any particular number is rational or irrational. Since rationals are countable and measure 0, the default is to expect that an arbitrary number is irrational, but that expectation isn't a proof. In that way, it is the right instinct to think that a number needs a reason to be rational but it doesn't need a reason to be irrational. However, the fact that we haven't found a reason for $e+\pi$ to be rational doesn't mean that no such reason exists.

For example, consider the following constants:$$ c_1=\sum_{n=1}^\infty\frac{1}{n^2+2n} \hspace{12pt}c_2=\sum_{n=1}^\infty\frac{1}{n^2+2n+1} $$ One of them is rational, the other isn't. In this form the reason "why" might not jump out to you immediately, some work must be put in to see which is which.

I think it is hugely surprising that $$ \frac1{\pi^{k}}\sum_{n=1}^\infty \frac 1 {n^{k}} $$ is rational for even $k$, but (probably) not for odd $k$.

So basically, sometimes numbers can be rational even when there's no apparent reason why. That's because of the gap between "I can't think of a reason why this would be rational" and "no reason exists for this to be rational".

Dark Malthorp
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