Posted June 18, 2015 By Presh Talwalkar. Read about me , or email me .

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We are taught in school that numbers like π = 3.141… and e = 2.718… are irrational numbers. But actually proving this is not something most of us learn. I went through a math degree at Stanford and never learned the proofs. So let’s fill this gap in mathematical knowledge! Quite nicely there are elementary proofs that can be understood for someone that’s taken high school calculus.

I have previously written about a proof that π is irrational.

This post is about 3 methods to show that Euler’s number, e = 2.718…, is irrational. I have prepared a video that explains the proofs.

Let’s Prove e Is An Irrational Number!

Below I’ll present the proofs in text.

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"All will be well if you use your mind for your decisions, and mind only your decisions." Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon. .

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Proof 1: Contradiction

The most well-known proof comes from Fourier. This is a variation of that proof.

Assume that e is rational, so that e = a/b for two integers a and b.

If e is rational, then its reciprocal is also, because 1/e = b/a.

Now we’ll show this is not possible. The intuitive problem is the denominator a will have to be some fixed number, and the infinite series for 1/e will continue to keep adding smaller terms with denominators n factorial.

Recall the infinite series for 1/e (which can be obtained by putting x = -1 in the power series for ex).

1/e = 1/2! – 1/3! + 1/4! + …

Let’s group this alternating series into terms up to a and terms that have denominator larger than a.

1/e = [1/2! – 1/3! + 1/4! + … + (-1)a/a!] + [(-1)a+1/(a + 1)! + … ]

Now we substitute 1/e = b/a and put all the terms up to a! on the left hand side.

b/a – [1/2! – 1/3! + 1/4! + … + (-1)a/a!] = [(-1)a+1/(a + 1)! + … ]

Now we multiply the entire equation by (-1)a+1a!

(-1)a+1b(a – 1)! – (-1)a+1[a!/2! – a!/3! + a!/4! + … + (-1)aa!/a!] = [1/(a + 1) – 1/[(a + 1)(a + 2)] … ]

This equation is a problem. Note that every single term on the left hand side is an integer. The very first term is an integer, and every term in the alternating series is a! divided by some factorial of a number smaller than a!, which means every term is an integer. Since the sum of many integers is also an integer, we conclude the entire left-hand side is an integer.

integer = [1/(a + 1) – 1/[(a + 1)(a + 2)] … ]

The right-hand side is an alternating series. It is bounded above by the first term, 1/(a + 1), which is strictly less than 1 because a is larger than 1 (we know 1/e is not equal to an integer). The sequence is bounded below by the sum of the first two terms, which will be a positive number. Therefore, the right-hand side is strictly between 0 and 1, so it is a fractional number.

integer = fractional value between 0 to 1

This is a contradiction!

Therefore, e cannot be a rational number and it is an irrational number.

Source of proof: “Elementary Proof that e is Irrational” by L. L. Pennisi. The American Mathematical Monthly, Vol. 60, No. 7 (Aug. – Sep., 1953), p. 474. Published by: Mathematical Association of America. http://www.jstor.org/stable/2308411

Proof 2: Geometric

This is a very cool proof from Jonathan Sondow.

Recall the infinite series representation of e.

e = 1 + 1 + 1/2! + 1/3! + 1/4! + …

We’ll construct e geometrically from line segments. Let the first segment be the interval I 1 = [2, 3] to correspond to the term 1 + 1.

The next segment is found by adding 1/2!. We can do this in the following method. We will split the line segment [2, 3] into two segments, and then keep the second half. So the next interval is I 2 = [5/2, 3] = [5/2!, 6/2!].

The next segment is found by adding 1/3!. We will now split the previous line segment [3/2, 3] into three segments, and then keep the second half. So the next interval is I 3 = [16/3!, 17/3!].

We can iteratively define each line segment I n and then conclude e = ∩ I n for all the indices of n from 1 to ∞.

Note that e is in between the endpoints of each interval. Furthermore, each interval is between two fractions of the form a/n! and (a + 1)/n! That is each interval is a fraction with numerators of two consecutive whole numbers and a denominator of n factorial.

So we can conclude this: e is NEVER a fraction for any denominator n factorial.

On the other hand, every rational number p/q does have this property: it IS a fraction for some factorial denominator. To see why, let us multiply the numerator and denominator by (q – 1)!. Then we have p/q = p(q – 1)!/q! which is a denominator of some factorial.

Therefore, e is an irrational number.

Source of proof: “A geometric proof that e is irrational and a new measure of its irrationality” by Jonathan Sondlow. The American Mathematical Monthly 2006 and 2007, Volume 113, 637-641 and Volume 114, 659. http://arxiv.org/abs/0704.1282

Proof 3: Continued Fraction

I will sketch one final proof which is how Euler proved it.

The number e can be written as a continued fraction.

The continued fraction is some number plus 1 divided by another fraction. It is common to focus on the leading numbers to write the continued fraction more easily.

The continued fraction is e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12, …]

Starting from the third number (2), there is a pattern. Every third term increases by 2 and in between in are two numbers of 1. This continues indefinitely, so e has an infinite continued fraction representation.

(Here is a reference for proving the continued fraction for e: “A Short Proof of the Simple Continued Fraction Expansion of e” by Henry Cohn. The American Mathematical Monthly 2006, Volume 113, 57-62. http://arxiv.org/abs/math/0601660v3)

On the other hand, every rational number has a finite continued fraction. For example, let’s do the continued fraction for 8/3. We first have:

8/3 = 2 + 2/3

We want to get that 2/3 to be a fraction of 1 over something. So we take the reciprocal and divide.

3/2 = 1 + 1/2

Now the last term is 1 over something, so we’re done. So we have the following continued fraction.

8/3 = [2; 1, 2] = 2 + 1/[1 + 1/2]

Why does this always terminate for rational numbers? The reason is the continued fraction is based on the Euclidean division algorithm. When you divide two numbers, you always get a whole number plus some remainder that is smaller. You can repeat this again on the remainder. Since the remainders keep getting smaller, this process always terminates in finite steps.

Since e has an infinite continued fraction, we conclude it must be irrational.

**Update: I got an email pointing out a clarification. It is also necessary to point out that irrational numbers have a unique simple continued fraction representation.

More to read

If you like this proof, you might enjoy my post about a proof that π is irrational. It is more complicated and involves calculus, but it is something I think should be shown to high school students.

You can also check out my video for this post.

Let’s Prove e Is An Irrational Number!

You may also like some of the 150+ videos on my YouTube channel.

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