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Let's assume that we are given an irrational number such as Euler's number $e$. Is it possible to write $e^x$ (where $x$ is a positive integer), as a sum of integer factors of $e$ to $e^{x-1}$?

In other words, is it possible to have the following equality?

$$e^x = a_0 e^0 + a_1 e + \cdots + a_{x-1}e^{x-1}, ~~~~~~~a_i \in Z^+.$$

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    If this were possible, then we could write $e^x$ as a sum of integers, i.e., as a sum of $a_k\cdot e^0$ (for any given integer $x$). In other words, no, it is not possible. – 2017-01-10
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    No, this is not true. $e$ is known to be transcendental, this is a consequence of that. – 2017-01-10

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This would make $e$ a solution to a polynomial with integer coefficients. We know that this cannot be done because $e$ is transcendental (Lindemann-Weierstrass theorem).

However, this is not because $e$ is an irrational number. $\sqrt{2}$ is an irrational number and it is a solution to $x^2-2=0.$

Numbers that are not solutions to any polynomial equations with integer coefficients are called transcendental. $\pi$ and $e$ are prominant examples.