## Proof of the Irrationality of e

In a previous post we proved that $\log_{10} 5$ is irrational.  In this post we prove the irrationality of e.

A proof of the irrationality of e must start by defining e.  There are some different ways to do that.  We’ll take e to be the unique positive number b such that $\dfrac{d}{dx} b^x = b^x$.  (This property itself can be proved from other ways to define e.  See, for example, Fitzpatrick’s Advanced Calculus [1], Section 5.2.)  We’ll also make the assumptions that (1) $e < 3$ and (2) $e^x$ is an increasing function.

Under this definition of e, the Lagrange Remainder Theorem says that, given $x > 0$ and $n \in \mathbb{N}$, there exists $c \in (0,x)$ such that

$\displaystyle e^x = 1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!} + \frac{e^c x^{n+1}}{(n+1)!}.$

Taking $x = 1$, we obtain, for some $c \in (0,1)$,

$\displaystyle e = 1 + 1 + \frac{1}{2} + \cdots + \frac{1}{n!} + \frac{e^c}{(n+1)!}$.

$\displaystyle \implies 0 < e - \left[ 2 + \frac{1}{2} + \cdots + \frac{1}{n!}\right] = \frac{e^c }{(n+1)!} \leq \frac{e}{(n+1)!} \leq \frac{3}{(n+1)!}.$

Now, suppose e is rational.  Then there exist $m_0, n_0 \in \mathbb{N}$ with $e = m_0/n_0$.  Therefore,

$\displaystyle 0 < \frac{m_0}{n_0} - \left[ 2 + \frac{1}{2} + \cdots + \frac{1}{n!} \right] \leq \frac{3}{(n+1)!}$.

Multiplying by $n!$, this becomes

$\displaystyle 0 < n! \frac{m_0}{n_0} - n! \left[ 2 + \frac{1}{2} + \cdots + \frac{1}{n!} \right] \leq \frac{3}{n+1}$.

Since the above inequality is true for all $n \in \mathbb{N}$, it is true for $n = \max \{n_0, 3\}$.  For this value of n,  $\displaystyle n!\frac{m_0}{n_0} - n!\left[ 2 + \frac{1}{2} + \cdots + \frac{1}{n!}\right]$ is an integer, and $3/(n+1) \leq 3/4$.

This means that there is an integer in the interval $(0,3/4)$.  However, we know that there is no integer in the interval $(0,1)$.  Contradiction; thus e is irrational.

[1] Patrick M. Fitzpatrick, Advanced Calculus, American Mathematical Society, 2006.

## Some Divisibility Properties of the Binomial Coefficients

Recently I read in Koshy’s book [1] on Catalan numbers some divisibility properties of the binomial coefficients I had not seen before.  Koshy credits them to Hermite.   They are particularly interesting to me because (as Koshy notes) some of the most famous divisibility properties of the binomial coefficients are immediate corollaries.  This post will give proofs of Hermite’s properties (following those in Koshy) and discuss the corollaries.

Divisibility properties.  Let $m, n \geq 1$.  Let $(m,n)$ denote the greatest common divisor of $m$ and $n$.  Then

(1) $\displaystyle \binom{m}{n}$ is divisible by $\displaystyle \frac{m}{(m,n)}$, and

(2) $\displaystyle \binom{m}{n}$ is divisible by $\displaystyle \frac{m-n+1}{(m+1,n)}$.

Proofs

(1)  Let $d = (m,n)$.  By the Euclidean algorithm, there exist integers A and B such that $d = Am + Bn$.  Thus

$\displaystyle d \binom{m}{n} = A m \binom{m}{n} + Bn \binom{m}{n} = A m \binom{m}{n} + B m \binom{m-1}{n-1}$, by the absorption identity.  Then

$\displaystyle d \binom{m}{n} = m \left[ A \binom{m}{n} + B \binom{m-1}{n-1} \right] = mC$, where C is an integer.

Thus $\displaystyle \frac{m}{d}$ divides $\displaystyle \binom{m}{n}$.  In other words, $\displaystyle \frac{m}{(m,n)}$ divides $\displaystyle \binom{m}{n}$.

(2)  Let $d = (m+1,n)$.  Then there exist integers P and Q such that $d = P(m+1) + Qn = (m-n+1)P + n(P+Q)$.  Thus

$\displaystyle d \frac{m!}{n! \, (m-n+1)!} = \binom{m}{n} P + \binom{m}{n-1} (P+Q)$.

Let $\displaystyle R = \binom{m}{n} P + \binom{m}{n-1}(P+Q)$.  Clearly R is an integer.  Then we have

$\displaystyle d \binom{m}{n} = (m-n+1) R$.

Thus $\displaystyle \frac{m-n+1}{d}$ divides $\displaystyle \binom{m}{n}$.  In other words, $\displaystyle \frac{m-n+1}{(m+1,n)}$ divides $\displaystyle \binom{m}{n}$.

Corollaries

1. The central binomial coefficient $\displaystyle \binom{2n}{n}$ is even.
1. Proof:  The greatest common divisor of 2n and n is n.  From Divisibility Property (1), $\displaystyle \binom{2n}{n}$ is divisible by $2n/n = 2$.
2. The Catalan number $C_n = \binom{2n}{n}/(n+1)$ is an integer.
1. Proof:  The greatest common divisor of $2n+1$ and $n$ is 1.  By Divisibility Property (2), $\displaystyle \binom{2n}{n}$ is divisible by $n+1$.
3. For p prime and values of n with $1 \leq n \leq p-1$$\displaystyle \binom{p}{n}$ is divisible by p.
1. Proof:  For values of n with $1 \leq n \leq p-1$, the greatest common divisor of p and n is 1.  By Divisibility Property (1), $\displaystyle \binom{p}{n}$ is divisible by p.

References

1. Thomas Koshy, Catalan Numbers with Applications, Oxford University Press, 2009.

## Log-Convexity of the Bell Numbers

A sequence $\{a_n\}$ is said to be log-convex if, for all $n \geq 1$, $a_n^2 \leq a_{n-1} a_{n+1}$.

In this post I give a short, combinatorial-flavored proof that the Bell numbers are log-convex.  The argument is from my answer here on Math.SE.

The combinatorial interpretation of the Bell number $B_n$ is that it counts the number of ways to partition the set $\{1, 2, \ldots, n\}$ into nonempty subsets.  Let $S_n$ denote the total number of sets over all partitions of $\{1, 2, \ldots, n\}$.  Then $A_n = S_n/B_n$ is the average number of sets in a random partition of $\{1, 2, \ldots, n\}$.

Claim 1: $A_n$ is increasing.

Proof: Each partition of $\{1, 2, \ldots, n+1\}$ can be associated with a partition of $\{1, 2, \ldots, n\}$ by removing the element n+1 from the set containing it.  Under the inverse of this mapping, each partition of $\{1, 2, \ldots, n\}$ consisting of k sets maps to k partitions consisting of k sets (if n+1 is placed in an already-existing set) and one partition consisting of k+1 sets (if n+1 is placed in a set by itself) out of the partitions of $\{1, 2, \ldots, n+1\}$.  Thus partitions of $\{1, 2, \ldots, n\}$ with more sets map to more partitions of $\{1, 2, \ldots, n+1\}$ containing the same number of sets as well as one partition with one more set.  This raises the average number of sets as you move from n elements to n+1 elements.

Claim 2: The log-convexity of $B_n$ is equivalent to the fact that $A_n$ is increasing.

Proof: Separate the partitions counted by $B_{n+1}$ into (1) those that have a set consisting of the single element n+1 and (2) those that don’t.  It should be clear that there are $B_n$ of the former.  Also, there are $S_n$ of the latter because each partition in group (2) is formed by adding n+1 to a set in a partition of $\{1, 2, \ldots, n\}$.  Thus $B_{n+1} = B_n + S_n$.

Therefore,

$\displaystyle B_n^2 \leq B_{n-1} B_{n+1} \iff \frac{B_{n+1}}{B_n} \geq \frac{B_n}{B_{n-1}} \iff 1 + \frac{S_n}{B_n} \geq 1 + \frac{S_{n-1}}{B_{n-1}}$ $\iff A_n \geq A_{n-1}.$

Since we established the last inequality in Claim 1, the Bell numbers are log-convex.

I was not the first to find this proof; Canfield [1] published it in 1995.  Engel [2] was the first to prove that the Bell numbers are log-convex; he used a different method.  See my answer here on Math.SE for references to more proofs.

References

1.  E. Rodney Canfield, “Engel’s Inequality for Bell Numbers,” Journal of Combinatorial Theory, Series A, 72 (1995) 184-187.

2. Konrad Engel, “On the Average Rank of an Element in a Filter of the Partition Lattice,” Journal of Combinatorial Theory, Series A, 65 (1994) 67-78.

## Proving the Existence of Irrational Numbers

The ancient Greeks first proved the existence of irrational numbers by proving that $\sqrt{2}$ is irrational.  The proof is, as modern proofs of irrationality go, fairly simple.  It is often the first example of a proof of irrationality that students see in, say, a course in real analysis.

Yesterday, though, I saw an even simpler proof of irrationality in Morgan’s Real Analysis [1] that I thought worth showing to my students in advanced calculus.

Theorem: The number $\log_{10} 5$ is irrational.

Proof: First, $\log_{10} 5 > 0$ because 5 and 10 are both greater than 1.  Suppose that $\log_{10} 5$ is rational.  Then there exist positive integers p, q such that $10^{p/q} = 5$.  Thus $\displaystyle 10^p = 5^q.$  However, $10^p$ ends in 0, while $5^q$ ends in 5.  This is a contradiction, and so $\log_{10} 5$ is irrational.

References

1. Frank Morgan, Real Analysis, American Mathematical Society, 2005.
Posted in irrational numbers, real analysis | 3 Comments

## Shifts, Finite Differences, and Binomial Transforms

The binomial transform of a sequence $a_n$ is the sequence $b_n$ satisfying $\displaystyle b_n = \sum_{k=0}^n \binom{n}{k} a_k.$

In this post we prove the following:

Theorem: If $b_n$ is the binomial transform of $a_n$, then $\displaystyle \Delta^m b_n = \sum_{k=0}^n \binom{n}{k} a_{k+m}.$

In other words, the mth finite difference of $b_n$ is the binomial transform of the shifted sequence $a_{n+m}$.  This theorem is useful because once we know one binomial coefficient identity in the form of a binomial transform we can use it to generate others.

Proof: First, we have

$\displaystyle \Delta b_n = b_{n+1} - b_n = \sum_{k=0}^{n+1} \binom{n+1}{k} a_k - \sum_{k=0}^n \binom{n}{k} a_k = \sum_{k=0}^{n+1} \binom{n}{k-1} a_k = \sum_{k=0}^n \binom{n}{k} a_{k+1}.$

In the second-to-last step we use Pascal’s recurrence $\binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}$ plus the fact that $\binom{n}{n+1} = 0$.  In the last step we use $\binom{n}{-1} = 0$.

Therefore, $\displaystyle \Delta (\Delta b_n) = \sum_{k=0}^n \binom{n}{k} a_{k+2},$ and, in general, $\displaystyle \Delta^m b_n = \sum_{k=0}^n \binom{n}{k} a_{k+m}.$

Example: Start with the identity $\displaystyle \sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{k+1} = \frac{1}{n+1}.$  (I’ve proved this identity three different ways so far on this blog: here, here, and here.)  Multiplying it by -1, we have $\displaystyle \sum_{k=0}^n \binom{n}{k} \frac{(-1)^{k+1}}{k+1} = - \frac{1}{n+1}.$

Applying the theorem we just proved yields

$\displaystyle \sum_{k=0}^n \binom{n}{k} \frac{(-1)^{k+2}}{k+2} = - \frac{1}{n+2} + \frac{1}{n+1} = \frac{-(n+1) + (n+2)}{(n+1)(n+2)} = \frac{1}{(n+1)(n+2)}.$

Since $(-1)^{k+2} = (-1)^2 (-1)^k = (-1)^k$, this simplifies to

$\displaystyle \sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{k+2} = \frac{1}{(n+1)(n+2)}.$

This identity can be generalized by taking $\displaystyle m-1$ finite differences.  I’ll leave it to the reader to prove the generalization

$\displaystyle \sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{k+m} = \frac{n!}{m(m+1) \cdots (m+n)}.$

## A Combinatorial Proof for the Power Sum

Probably the most common formula for the power sum $\displaystyle \sum_{k=0}^n k^m$ is the one involving binomial coefficients and Bernoulli numbers $B_k$, sometimes called Faulhaber’s formula:

$\displaystyle \sum_{k=0}^n k^m = \frac{1}{m+1} \sum_{k=0}^m \binom{m+1}{k} B_k n^{m+1-k}.$

Historically, this, or a variant of it, was the first general formula for the power sum.

There are other ways to express the power sum, though.  The subject of this post is to give a combinatorial proof for the following one involving Stirling numbers:

$\displaystyle \sum_{k=0}^n k^m = \sum_{k=1}^m \binom{n+1}{k+1} \left\{ m \atop k \right\} k! .$

Proof: How many functions f are there from the set $\{1, 2, \ldots, m+1 \}$ to the set $\{1, 2, \ldots, n+1\}$ such that $f(1) > f(j)$ for any $j \in \{2, \ldots, m+1\}$?

Answer 1: Condition on the value of $f(1)$. If $f(1) = k+1$, then there are k choices for $f(2)$, k choices for $f(3)$, etc., for a total of $k^m$ functions. Summing over all possible values of k gives the left side.

Answer 2: Condition on the number of elements in the range of f. (This is not the codomain, which is $\{1, 2, \ldots, n+1\}$, but the subset consisting of elements from the codomain that have something from $\{1, 2, \ldots, m+1\}$ mapped to them.  The range may be the entire codomain or a proper subset of it.) If the range of f has $k+1$ elements, there are $\binom{n+1}{k+1}$ ways to choose exactly which elements will comprise the range. The largest of these must be $f(1)$, and the remaining m elements in the domain must be mapped to the other k elements in the range. There are $\left\{ m \atop k \right\}$ ways to assign the remaining m elements to k nonempty groups, and then k! ways to assign these groups to the actual k remaining elements in the range. (More generally, the number of onto functions from an m-set to a k-set is $\left\{ m \atop k \right\} k!$.) Summing over all possible values of k gives the right side.

(I first gave this proof as an answer here on Math.SE.)

## Yet Another Nice Proof of the Derangements Formula

A couple of posts from a few years back give proofs of the formula for $D_n$, the number of derangements on n elements.  (A derangement is a permutation with no fixed points.)  Both proofs avoid the use of the principle of inclusion-exclusion, which is the standard method of proving the derangements formula.

This post contains another proof of the derangements formula, and I think this one is my favorite of the three.  It starts with a proof of the following identity:

$\displaystyle n! = \sum_{k=0}^n \binom{n}{k} D_{n-k} = \sum_{k=0}^n \binom{n}{k} D_k.$

Proof.  Both sides of the identity count the number of permutations on n elements.  The left side is straightforward.  For the middle formula, condition on the number of fixed points in the permutation.  To count the number of permutations with k fixed points, first choose which k of the n elements are to be mapped to themselves.  This can be done in $\binom{n}{k}$ ways.  The remaining n-k elements cannot be mapped to themselves; i.e., they must be deranged.  There are $D_{n-k}$ ways to do this.  Multiply these together and sum over all possible values of k to get the middle formula.  The right side just reverses the summation on the middle formula (i.e., replacing k with n-k), while using the binomial coefficient symmetry property $\binom{n}{k} = \binom{n}{n-k}$.

Next, apply the binomial inversion formula.  This comes in different forms.  The one we will use says that for sequences $f(n)$ and $g(n)$ we have

$\displaystyle f(n) = \sum_{k=0}^n \binom{n}{k} g(k) \iff g(n) = \sum_{k=0}^n \binom{n}{k} f(k) (-1)^{n-k}.$

Binomial inversion can be proved multiple ways.  My favorite uses exponential generating functions, but there are more direct proofs, too (see, for example, p. 193 of Concrete Mathematics [1]).

Applying the binomial inversion formula to Equation (1), with $f(n) = n!$ and $g(k) = D_k$, we get

$\displaystyle D_n = \sum_{k=0}^n \binom{n}{k} k! (-1)^{n-k} = \sum_{k=0}^n \binom{n}{k} (n-k)! (-1)^k = n! \sum_{k=0}^n \frac{(-1)^k}{k!},$

where the second step reverses the summation.

References

1.  Ronald L. Graham, Donald E. Knuth, and Oren Patashnik.  Concrete Mathematics (2nd ed.), Addison-Wesley, 1994.