e is irrational
Tonight's idea
You met \(e\) as a series (grade 10, b9): \(e = \sum_{n=0}^{\infty} \frac{1}{n!} = 1 + 1 + \tfrac12 + \tfrac16 + \tfrac1{24} + \cdots\). Here is Fourier's proof by contradiction that \(e\) is not a fraction. Suppose \(e = p/q\) with \(p\) and \(q\) positive integers. Multiply by \(q!\). The left side, \(q!\,e = p \cdot (q-1)!\), is an integer. On the right, cut the series after \(n = q\):
\[q!\,e = \underbrace{\sum_{n=0}^{q} \frac{q!}{n!}}_{N} \;+\; \underbrace{\sum_{n=q+1}^{\infty} \frac{q!}{n!}}_{T}.\]
Every term of \(N\) is a whole number, because \(n!\) divides \(q!\) when \(n \le q\). So the tail \(T = q!\,e - N\) is an integer too. But \(T\) is a sum of positive terms, and
\(T = \dfrac{1}{q+1} + \dfrac{1}{(q+1)(q+2)} + \cdots \lt \dfrac{1}{q+1} + \dfrac{1}{(q+1)^2} + \cdots = \dfrac{1}{q}.\)
The inequality holds term by term: the denominators \((q+1)(q+2)\), \((q+1)(q+2)(q+3)\), … each exceed the matching power of \(q+1\). The geometric series on the right has first term \(\tfrac{1}{q+1}\) and ratio \(\tfrac{1}{q+1}\), so it sums to \(\tfrac{1}{q}\) (grade 10, b12). Hence \(0 \lt T \lt 1/q \le 1\). No integer lies strictly between \(0\) and \(1\). The assumption was false: \(e\) is irrational.
Watch one: the tail for \(q = 5\)
Take \(q = 5\). The front sum is \(N = 5!\left(1 + 1 + \tfrac12 + \tfrac16 + \tfrac1{24} + \tfrac1{120}\right) = 120 + 120 + 60 + 20 + 5 + 1 = 326\), so \(T = 120\,e - 326\). With \(e = 2.718281828\ldots\), \(120\,e = 326.1938\ldots\), giving \(T = 0.1938\ldots\): positive, and below \(1/5 = 0.2\), as the bound promised. The series agrees: \(T = \tfrac16 + \tfrac1{42} + \tfrac1{336} + \tfrac1{3024} + \cdots = 0.1667 + 0.0238 + 0.0030 + 0.0003 + \cdots\).
The contradiction in miniature: if \(e\) were \(p/5\), then \(120\,e\) would be an integer, so \(T\) would be too. It is \(0.19\). The proof does this for every \(q\) at once.
Your turn
- Compute \(N\) for \(q = 3\), that is \(3!\,(1 + 1 + \tfrac12 + \tfrac16)\). Then compute \(T = 6e - N\) to three decimal places and check it lies between \(0\) and \(1/3\).
- Use the series to show \(2 \lt e \lt 3\). For the upper bound, show \(n! \ge 2^{n-1}\) for \(n \ge 1\) and compare with a geometric series. Why does \(2 \lt e \lt 3\) on its own settle the case \(q = 1\) of the proof?
- The proof needs \(T \lt 1/q\), strictly. Explain exactly where the strictness comes from (which term of \(T\) is strictly smaller than its geometric partner?). What would go wrong for \(q = 1\) if we only knew \(T \le 1\)?
- The number \(\sum_{n=0}^{\infty} 2^{-n} = 2\) is rational, so the same argument must fail for it. Imitate the proof: suppose the sum is \(p/q\), multiply by \(2^q\) instead of \(q!\), and compute the tail \(2^q \sum_{n > q} 2^{-n}\) exactly. Where does the argument break?
Stretch: \(1/e\) too
Also \(1/e = \sum_{n=0}^{\infty} (-1)^n/n! = 1 - 1 + \tfrac12 - \tfrac16 + \cdots\). Adapt the proof to show \(1/e\) is irrational. The tail now alternates in sign. Show it is not zero and has absolute value less than \(\tfrac{1}{q+1}\): group the terms in pairs.
Answers
- \(N = 6 + 6 + 3 + 1 = 16\). \(6e = 16.3097\ldots\), so \(T = 0.310\), which is between \(0\) and \(1/3 = 0.333\). (Series check: \(\tfrac14 + \tfrac1{20} + \tfrac1{120} + \cdots = 0.25 + 0.05 + 0.0083 + \cdots\).)
- Lower: the first two terms alone give \(1 + 1 = 2\), and the rest are positive, so \(e > 2\). Upper: \(n! = 1 \cdot 2 \cdot 3 \cdots n \ge 1 \cdot 2 \cdot 2 \cdots 2 = 2^{n-1}\) for \(n \ge 1\), so \(e \le 1 + \left(1 + \tfrac12 + \tfrac14 + \cdots\right) = 1 + 2 = 3\), and the inequality is strict from \(n = 3\) on (\(6 > 4\)), so \(e \lt 3\). If \(q = 1\) then \(e = p\) is an integer; but no integer lies strictly between \(2\) and \(3\).
- The first terms agree: both are \(\tfrac{1}{q+1}\). The second term of \(T\) is \(\tfrac{1}{(q+1)(q+2)}\), strictly less than \(\tfrac{1}{(q+1)^2}\) because \(q + 2 > q + 1\). One strict term makes the whole sum strict. For \(q = 1\) the bound is \(T \lt 1\); with only \(T \le 1\), the value \(T = 1\) would be an integer and there would be no contradiction.
- Suppose \(\sum 2^{-n} = p/q\). Multiply by \(2^q\): the left side need not be an integer (unless \(q\) is a power of \(2\)), which is one break already. Worse, the tail is \(2^q\left(2^{-(q+1)} + 2^{-(q+2)} + \cdots\right) = \tfrac12 + \tfrac14 + \cdots = 1\) exactly: an integer. The factorial proof worked because the tail terms \(q!/n!\) shrink faster than any geometric series, pushing the tail strictly below \(1/q\). Powers of \(2\) shrink only geometrically, and the tail lands exactly on an integer.
Stretch. Suppose \(1/e = p/q\) and multiply by \(q!\). The front sum \(\sum_{n \le q} (-1)^n q!/n!\) is an integer, so the tail \(T = q!\sum_{n > q} (-1)^n/n!\) must be one too. Up to sign, \(T = \tfrac{1}{q+1} - \tfrac{1}{(q+1)(q+2)} + \tfrac{1}{(q+1)(q+2)(q+3)} - \cdots\). The terms strictly decrease in size, so pairing them as \(\left(\tfrac{1}{q+1} - \tfrac{1}{(q+1)(q+2)}\right) + \cdots\) shows \(|T| > 0\), and pairing from the second term shows \(|T| \lt \tfrac{1}{q+1} \le \tfrac12\). An integer with \(0 \lt |T| \lt \tfrac12\) does not exist. For \(q = 2\): \(2/e = 0.7358\), front sum \(2 - 2 + 1 = 1\), tail \(-0.264\), of size below \(1/3\).