In mathematics, we denote the limit \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x as e, and we can calculate its value as e = 2.7182818284590452353602\dots. This is a number of equal importance to \pi (some books even consider it more important than \pi). Like \pi, this number is also an “irrational number.” Now, let us prove it.
Assume e = p/q is a rational number, where p and q are positive integers. According to the Maclaurin expansion of e, we have: p/q = e = 1 + \frac{1}{2} + \frac{1}{3!} + \dots + \frac{1}{q!} + \frac{\varphi^{q+1}}{(q+1)!} \tag{1} p/q - \left(1 + \frac{1}{2} + \frac{1}{3!} + \dots + \frac{1}{q!}\right) = \frac{\varphi^{q+1}}{(q+1)!} \tag{2} where \varphi \in (0, 1).
Then, by multiplying both sides of (2) by q!, we obtain an integer on the left side, while the right side = \frac{\varphi^{q+1}}{q+1} is a non-integer. Thus, the two sides are not equal. This contradicts the assumption that they are equal! Therefore, the assumption is false, and e is an irrational number.
This proof using the Maclaurin expansion is relatively simple, provided one understands the Maclaurin expansion of e. Compared to the proof provided on Wikipedia, I find this proof to be more concise and intuitive!
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