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The Algebraic Closure
Previously, we explored extending fields using irreducible polynomials. One classic result arises from the complex numbers, where

$\mathbb C \cong \mathbb R[x]/\langle x^2 + 1\rangle.$

The fundamental theorem of algebra tells us that any polynomial $f(x) \in \mathbb C[x]$ has a root $\alpha \in \mathbb C$ . In this case, we say that $\mathbb C$ is algebraically closed. The word “closed” broadly translates to “properties are more or less preserved”.

This closure property turns out to work for any field. Let $\mathbb K \supseteq \mathbb F$ .

Definition 1. An element $\alpha \in \mathbb K \supseteq \mathbb F$ is algebraic over $\mathbb F$ if there exists a polynomial $f \in \mathbb F[x]$ such that $f(\alpha) = 0$ . We say that $\mathbb K$ is algebraic over $\mathbb F$ if every element in $\mathbb K$ is algebraic over $\mathbb F$ .

We remark that being “algebraic” is a property of a field and its elements, not the polynomial. That is, given the element $\alpha$ , there exists a sufficiently nice polynomial $f$ such that $f(\alpha) = 0$ .

Lemma 1. The set $\mathbb K_{\mathbb F} := \{ u \in \mathbb K : u\ \text{is algebraic over}\ \mathbb F\}$ forms a sub-field of $\mathbb K$ .

Proof. Fix $u,v \in \mathbb K_{\mathbb F}$ . Recall that $\mathbb F(u, v) \supseteq \mathbb F$ is the smallest field that contains $\mathbb F$ as a subfield. Since the extension is finite, it is algebraic, so that $\mathbb F(u, v) \subseteq \mathbb K_{\mathbb F}$ . In particular, $u-v \in \mathbb K_{\mathbb F}$ and $u \cdot v^{-1} \in \mathbb K_{\mathbb F}$ , so that $\mathbb K_{\mathbb F}$ is closed under the field operations inherited from $\mathbb K$ . Therefore, $\mathbb K_{\mathbb F}$ forms a field.

In particular, $\mathbb K$ is algebraic over $\mathbb F$ if and only if $\mathbb K = \mathbb K_{\mathbb F}$ .

To attain algebraic closure, we ask the reverse question: given the polynomial $f$ , is there a sufficiently nice element $\alpha$ such that $f(\alpha) = 0$ ? Notice that in general, $\mathbb F[x] \supsetneq \mathbb F$ .

Definition 2. We call a field $\mathbb F$ algebraically closed if every polynomial $f \in \mathbb F[x]$ has at least one root $\alpha \in \mathbb F$ such that $f(\alpha) = 0$ .

Lemma 2. If $\mathbb F$ is algebraically closed, then any non-constant polynomial $f \in \mathbb F$ can be decomposed into a product of linear factors (i.e. polynomials with degree $1$ ).

Proof. We prove this result holds by induction: For any $n$ , if $f \in \mathbb F[x]$ has degree $n$ , then it can be written as a product of linear factors. The result is obvious for $n = 1$ . Suppose the result holds for $\deg(f) = k$ . Fix $f \in \mathbb F[x]$ with $\deg(f) = k +1$ . Since $\mathbb F$ is algebraically closed, there exists $\alpha \in \mathbb F$ such that $f(\alpha) = 0$ . By the factor theorem, there exists a polynomial $\tilde{f}$ with $\deg(\tilde{f}) = k$ such that $f(x) = (x-\alpha) \cdot \tilde{f}(x)$ . By the induction hypothesis, $\tilde f$ can be written as a product of linear factors, so that this holds for $f$ as well, as required.

We claim that for any field $\mathbb F$ , there exists an essentially unique field extension $\mathbb F' \supseteq \mathbb F$ that is algebraically closed.

Lemma 3. There exists a field extension $\mathbb K \supseteq \mathbb F$ such that every non-constant polynomial $f \in \mathbb F[x]$ has at least one root $\alpha \in \mathbb K$ . We remark the following caveats:
- $\mathbb K$ may not be algebraically closed. That is, there may exist a non-constant polynomial $g \in \mathbb K[x]$ with no roots in $\mathbb K$ .
- $\mathbb K$ may not be algebraic over $\mathbb F$ . That is, there may exist an element $u \in \mathbb K$ that is not algebraic in $\mathbb F$ .
Proof. For any non-constant $f \in \mathbb F[x]$ , let $x_f$ denote a variable. Then given non-constant polynomials $f,g$ , the finitely many nonzero constants $c_{i,j} \in \mathbb F$ , generate a polynomial $\psi_{f,g}$ defined by

$\displaystyle \psi_{f,g}(x_f, x_g) = \sum_{i=0}^\infty \sum_{j=0}^\infty c_{i,j} \cdot x_f^i \cdot x_g^j \in \mathbb F[x_f, x_g],$

where $c_{i,j} \in \mathbb F$ . This observation can be generalised. For non-constant polynomials $\mathbf f = (f_1,\dots, f_n)$ whose corresponding variables are denoted $\mathbf x = (x_{f_1},\dots, x_{f_n})$ , the finitely many nonzero constants $c_{\mathbf v} \in \mathbb F$ , where $\mathbf v = (v_1,\dots, v_n) \in (\mathbb N^+)^n$ , generate a polynomial $\psi_{\mathbf f}$ defined by

$\displaystyle \psi_{\mathbf f}(\mathbf x) = \sum_{\mathbf v \in (\mathbb N^+)^n} c_{\mathbf v} \cdot x_{f_1}^{v_1} \cdot \cdots \cdot x_{f_n}^{v_n} \in \mathbb F[\mathbf x].$

Define $\mathbf x := \{x_f : f \in \mathbb F[x]^*\}$ and the ring

$\mathbb F[\mathbf x] := \{ \psi_{\mathbf f} : \mathbf f \subseteq \mathbb F[x]^*\}.$

By considering the polynomial $f(x) = x$ , $\mathbb F[x] \subseteq \mathbb F[\mathbf x]$ . Now given a non-constant polynomial $f = \sum_{i=0}^\infty c_i \mu_i \in \mathbb F[x]$ and its corresponding variable $x_f$ ,

$\displaystyle f(x_f) = \sum_{i=0}^\infty c_i \cdot \mu_i(x_f) = \sum_{i=0}^\infty c_i \cdot x_f^i \in \mathbb F[\mathbf x],$

where the right-hand side means that only finitely many $c_i$ is non-zero. Define the ideal

$I := \langle \{ f(x_f) : f \in \mathbb F[x]^* \} \rangle \subseteq \mathbb F[\mathbf x].$

We claim that $I \neq \mathbb F[\mathbf x]$ . Suppose instead that $I = \mathbb F[\mathbf x]$ . Since $1 \in I$ , there exist finitely many $w_i \in \mathbb F[\mathbf x]$ such that

$\displaystyle 1 = \sum_{i=1}^\infty w_i \cdot f_i(x_{f_i}).$

Define $A_i := \{ \alpha_{i,1},\dots,\alpha_{i,\deg(f_i)} \}$ , where $f_i(\alpha_{i,j}) = 0$ . Define

$\displaystyle A := \bigcup_{i=1}^\infty \{\alpha \in \mathbb F : f_i(\alpha) = 0\},$

and $\mathbb K := \mathbb F(A)$ , and the ring homomorphism $\phi : \mathbb F[\mathbf x] \to \mathbb K$ by $\phi|_{\mathbb F} = \mathrm{id}|_{\mathbb F}$ and

$\phi(x_{f_i}) \mapsto \alpha_{i,1}.$

Then

$\displaystyle 1 = \phi(1) = \sum_{i=1}^r \phi(w_i) \cdot f(\phi(x_{f_i})) = \sum_{i=1}^r \phi(w_i) \cdot f(\alpha_{i,1}) = 0,$

a contradiction. Now define

$\mathcal S := \{J \subseteq \mathbb F[\mathbf x] : J \supseteq I\ \text{as proper ideals}\} \ni I.$

Fix any ascending chain $\mathcal C = \{J_t\}$ in $\mathcal S$ . Since the set $K := \bigcup_{t} J_t$ is also an ideal that contains $I$ , it is an upper bound in $\mathcal C$ . By Zorn’s lemma, $\mathcal S$ has a maximal element $M$ .

To show that $M$ is a maximal ideal, we observe that for any ideal $M'$ such that $M \subseteq M' \subseteq \mathbb F[\mathbf x]$ , if $M' \neq \mathbb F[\mathbf x]$ , then $M'$ being a proper ideal of $\mathbb F[x]$ implies that $M' \in \mathcal S$ , so that $M \subseteq M'$ implies $M = M'$ , as required.

Finally, since $M \subseteq \mathbb F[\mathbf x]$ as a maximal ideal, $\mathbb K :=\mathbb F[\mathbf x]/M$ forms a field that contains $\mathbb F$ via the natural inclusion map $f \mapsto f + M$ , so that

$f(x_f + M) = f(x_f) + M = I +M = M = 0.$

Hence any polynomial $f \in \mathbb F[x]$ has a root $x_f + M \in \mathbb K$ .

Theorem 1. For any field $\mathbb F$ , there exists an algebraic extension $\mathbb F' \supseteq \mathbb F$ that is algebraically closed.

Proof. Use Lemma 3 to construct field extensions

$\mathbb F \subseteq \mathbb K_1 \subseteq \mathbb K_2 \subseteq \cdots$

and define the field extension $\mathbb K := \bigcup_{i=1}^\infty \mathbb K_i \supseteq \mathbb F$ . By Lemma 1, the set $\mathbb F' := \mathbb K_{\mathbb F}$ forms a sub-field of $\mathbb K$ . In fact, by definition, $\mathbb F' \supseteq \mathbb F$ is algebraic.

It suffices to show that $\mathbb F'$ is algebraically closed. To do that, fix $f = \sum_{i=0}^n c_i \cdot \mu_i \in \mathbb F'[x]^*$ . By definition, $c_i \in \mathbb K_{m_i}$ for some $n_i$ . Define $m := m_1 + \dots + m_n$ . Then $c_i \in \mathbb K_m$ for each $i$ . In particular, $f$ has a root in $\mathbb K_{m+1} \subseteq \mathbb F'$ .

Furthermore, this field extends uniquely with respect to the base field in the following sense.

Theorem 2. Let $\mathbb F_1, \mathbb F_2$ be field extensions of $\mathbb F$ that are algebraically closed. Then there exists a field isomorphism $\phi : \mathbb F_1 \to \mathbb F_2$ such that $\phi|_{\mathbb F} = \mathrm{id}_{\mathbb F}$ .

Proof. Define the collection of $\mathbb F$ –homomorphisms by

$\mathrm{Hom}_{\mathbb F}(\mathbb F_1, \mathbb F_2) := \{\phi \in \mathrm{Hom}(\mathbb F_1, \mathbb F_2) : \phi|_{\mathbb F} = \mathrm{id}_{\mathbb F}\}.$

Consider the field homomorphism $\mathrm{id}|_{\mathbb F} : \mathbb F \to \mathbb F_2$ . Define the collection $\mathcal P$ of pairs

$\mathcal P := \{ (\mathbb K, \phi_{\mathbb K}) : \mathbb F_1 \supseteq \mathbb K \supseteq \mathbb F, \phi \in \mathrm{Hom}_{\mathbb F}(\mathbb F_1, \mathbb F_2) \}$

with the partial ordering $(\mathbb K_1, \phi_{\mathbb K_1}) \leq (\mathbb K_2, \phi_{\mathbb K_2})$ if and only if $\mathbb K_1 \subseteq \mathbb K_2$ and $\phi_{\mathbb K_2} |_{\mathbb K_1} = \phi_{\mathbb K_1}$ . Fix any chain

$\mathcal C = \{ (\mathbb K_{\alpha}, \phi_{\mathbb K_\alpha}) : \alpha \in I \} \subseteq \mathcal P.$

Define $\mathbb K := \bigcup_{\alpha} \mathbb K_\alpha$ and $\phi_{\mathbb K}$ by $\phi_{\mathbb K}(x) = \phi|_{\mathbb K_\alpha}(x) \cdot \mathbb I_{\mathbb K_\alpha}(x)$ . It is a simple exercise to check that $(\mathbb K, \phi_{\mathbb K}) \in \mathcal P$ is an upper-bound for $\mathcal C$ . By Zorn’s lemma, $\mathcal P$ contains a maximal element $(\mathbb K_{\max}, \phi_{\mathbb K_{\max}})$ .

It is clear that $\mathbb K_{\max} \subseteq \mathbb F_1$ . If $\mathbb K_{\max} \neq \mathbb F_1$ , then there exists some $u \in \mathbb F_1 \backslash \mathbb K_{\max}$ so that $(\mathbb K_{\max}, \phi_{\mathbb K_{\max}})< \mathbb (\mathbb F_1(u), \phi_{\mathbb F_1(u)})$ , contradicting the maximality of $\mathbb K_{\max}$ . Therefore, $\mathbb K_{\max} = \mathbb F_1$ , and the field homomorphism $\phi := \phi_1 : \mathbb F_1 \to \mathbb F_2$ satisfies $\phi|_{\mathbb F} = \mathrm{id}_{\mathbb F}$ . As a field homomorphism, $\phi$ must be injective. Therefore, it suffices to check that $\phi$ is surjective, that is, $\phi(\mathbb F_1) = \mathbb F_2$ .

Now $\phi(\mathbb F_1) \supseteq \mathbb F$ is algebraic, since for any $u \in \mathbb F_1$ ,

$f_{\mathbb F_1}^u(\phi(u)) = \phi(f_{\mathbb F_1}^u(u)) = \phi(0) = 0,$

so that $\phi(u)$ is algebraic over $\mathbb F$ . Since $\phi(\mathbb F_1) \subseteq \mathbb F_2$ , it suffices to check that the reverse inclusion holds.

We claim that $\phi(\mathbb F_1)$ is not just algebraic, but furthermore, algebraically closed. Fix $f = \sum_i \alpha_i x^i \in \phi(\mathbb F_1)[x]$ . Since $\alpha_i \in \phi(\mathbb F_1)$ , there exists $\beta_i \in \mathbb F_1$ such that $\alpha_i = \phi(\beta_i)$ . Since $\mathbb F_1$ is algebraically closed, the polynomial $g := \sum_i \beta_i x^i \in \mathbb F_1[x]$ has a root $\beta \in \mathbb F_1$ . Since $\phi$ is a field homomorphism,

$\begin{aligned} f(\phi(\beta)) &= \sum_i \alpha_i (\phi(\beta))^i \\ &= \sum_i \phi(\beta_i) (\phi(\beta))^i \\ &= \phi \left(\sum_i \beta_i \cdot \beta^i \right) \\ &= \phi(g(\phi(\beta))) \\ &= \phi(g(\beta)) \\ &= \phi(0) = 0. \end{aligned}$

Therefore, $\phi(\mathbb F_1)$ is algebraic closed.

For any $u \in \mathbb F_2$ , $\phi(\mathbb F_1)(u) \supseteq \phi(\mathbb F_1)$ is a finite extension with minimal polynomial $f_{\mathbb F_1}^u \in \phi(\mathbb F_1)[x]$ . Since $\phi(\mathbb F_1)$ is algebraically closed, by induction on Lemma 2, all of the roots of $f_{\mathbb F_1}^u$ belong to $\phi(\mathbb F_1)$ . In particular, $u \in \phi(\mathbb F_1)$ , so that $\phi(\mathbb F_1) = \mathbb F_2)$ .

Therefore, $\phi(\mathbb F_1) = \mathbb F_2$ , and $\phi$ is a field isomorphism, as required.

Corollary 1. Any field $\mathbb F$ has an effectively unique field extension $\overline{\mathbb F} \supseteq \mathbb F$ . We call the field extension $\overline{\mathbb F} \supseteq \mathbb F$ the algebraic closure of $\mathbb F$ , which is unique up to the existence of field isomorphisms according to Theorem 2.

Remark 1. Any algebraic closure of $\mathbb R$ must be isomorphic to $\mathbb C$ . Hence, we could have started with $\mathbb C = \overline{\mathbb R}$ and defined $i$ as a solution to $x^2 + 1 \in \mathbb R[x] \subseteq \mathbb C[x]$ , so that $i = \sqrt{-1}$ in a formal sense, then subsequently derive its algebraically convenient properties. In fact, this is how mathematicians formally “construct” a bona fide $\sqrt{-1}$ from what seems to be thin air.

We remark that Theorem 2 yields a special case of a more general result.

Theorem 3. Let $\phi : \mathbb F_1 \to \mathbb F_2$ be a field isomorphism. Suppose $\mathbb F_i' \supseteq \mathbb F_i$ as algebraic closures according to Theorem 1. Then there exists a field automorphism $\Phi : \mathbb F_1' \to \mathbb F_2'$ such that $\Phi|_{\mathbb F_1} = \phi$ .

Proof Sketch. Adapt the proof of Theorem 2.

Remark 2. In the special case $\mathbb F_1 = \mathbb F_2 = \mathbb F$ , we call $\phi$ a field automorphism on $\mathbb F$ . Then $\Phi : \mathbb F_1' \to \mathbb F_2'$ is a field isomorphism such that $\Phi|_{\mathbb F} = \phi$ . In this case, we say that $\Phi$ is an $\mathbb F$ -automorphism.

Next time, we revisit automorphisms in detail—they will play crucial roles in connecting fields and groups, and in particular, the star of abstract algebra: Galois groups.

—Joel Kindiak, 3 Mar 26, 1226H
May 21, 2026
Who Cares About Binomials?

As part of the school curriculum, students are taught the binomial theorem that they use solve various meaningless problems. I even asked the following question out of pure curiosity: what is the expansion of $(1+x)^5$ ? But why should we even bother with such a question in the first place?

As a pure math student by training, I find no application of the binomial theorem at the secondary school level, except for the proof of one concept: derivatives. Let me use the examples $n = 1,2,3$ to illustrate.

Among other crucial ideas, one key component in differentiation is to evaluate the expression

$\displaystyle \frac{(x+h)^n - x^n}{h}.$

If we consider $n = 1, 2, 3$ , then leave it as an exercise to verify the following simplifications:

$\begin{aligned} \frac{(x+h)^1 - x^1}{h} &= \frac{(x+h)-x}{h} = 1, \\ \frac{(x+h)^2 - x^2}{h} &= \frac{(x^2 + 2xh + h^2) - x^2}{h} = 2x + h, \\ \frac{(x+h)^3 - x^3}{h} &= \frac{ x^3 + 3x^2h + 3xh^2 + h^3 }{ h } = 3x^2 + (3x + h) \cdot h. \end{aligned}$

In the differentiation process, we would set $h = 0$ on the right-hand side and obtain the expressions $1, 2x, 3x^2$ respectively. Continuing the pattern, we would expect that

$\displaystyle \frac{(x+h)^n - x^n}{h} = nx^{n-1} + (\text{finite stuff}) \cdot h,$

so that setting $h = 0$ , we obtain the expression $nx^{n-1}$ . This is better known as the derivative of the function $x^n$ , but how can we know this information for sure?

It boils down to expanding the expression $(x+h)^n$ , how do we do so?

Before we answer this question more completely, we remark that the binomial expansion does get extended to the cases when $n$ is not a positive integer in various ways. To achieve that goal, however, would require us to explore some A-Level calculus, and we will delay that task for now.

The content in this post will mirror the discussion in this one, which ties in the binomial expansion with counting problems in probability, but perhaps with less technical terminology.

Firstly, by observation, the highest power of $x$ in the expansion $(1+x)^n$ is $n$ . We see this observation play out in the expansions

$\begin{aligned} (1+x)^2 &= 1+ 2x + x^2, \\ (1+x)^3 &= 1+ 3x + 3x^2 + x^3. \end{aligned}$

Therefore, we will suppose that for any $k$ , there will be constants $a_{k,0}, a_{k,1},\dots, a_{k,k}$ such that

$(1+x)^k = a_{k,0} + a_{k,1} x + a_{k,2} x^2 + \cdots + a_{k,k} x^k.$

How do we expand $(1+x)^{k+1}$ ? A power of $(k+1)$ denotes multiplying the expression $(1+x)$ a total of $k$ times, so that expanding the expression,

$\begin{aligned} (1+x)^{k+1} &= (1+x)(1+x)^k \\ &= (1+x)(a_{k,0} + a_{k,1} x + a_{k,2} x^2 + \cdots + a_{k,k} x^k) \\ &= (a_{k,0} + a_{k,1} x + a_{k,2} x^2 + \cdots + a_{k,k-1} x^{k-1} + a_{k,k} x^k) \\ &\phantom{====}+ (a_{k,0}x + a_{k,1} x^2 + a_{k,2} x^3 + \cdots + a_{k,k-1}x^k + a_{k,k} x^{k+1}) \\ &= \underbrace{ a_{k,0} }_{ a_{k+1,0} } + \underbrace{ (a_{k,1} + a_{k,0}) }_{ a_{k+1, 1} }x + (\underbrace{ a_{k,2} + a_{k,1} }_{a_{k+1,2}} )x^2 + \cdots + ( \underbrace{ a_{k,k} + a_{k,k-1} }_{ a_{k+1,k} } )x^k + \underbrace{ a_{k,k} }_{ a_{k+1,k+1} } x^{k+1} \\ &= a_{k+1, 0} + a_{k+1, 1} x + a_{k+1, 2} x^2 + \cdots + a_{k+1, k+1} x^{k+1}. \end{aligned}$

Therefore, our expansion of $(1+x)^{k+1}$ is given by the coefficients $a_{k+1, 0} = a_{k, 0}$ , and for any $0 \leq i \leq k$ ,

$a_{k+1, i+1} = a_{k, i+1} + a_{k, i}.$

For the very special case $k = 1$ , $(1+x)^1 = 1+ x = a_{1,0} + a_{1,1}x$ . Comparing coefficients, we set $a_{1,0} = a_{1,1} = 1$ . We will declare $a_{0,0} = 1$ and $a_{k, k+1} = 0$ for convenience / convention / simplicity in formulas, so that for $k = 0$ , $a_{1,0} = a_{0,0} = 1$ , and

$a_{1, 1} = a_{0, 1} + a_{0, 0} = 0 + 1 = 1,$

as expected.

Lemma 1. For any $k$ , $a_{k, 0} = a_{k,k} = 1$ .

Proof. We have $a_{0,0} = 1$ and $a_{k+1, 0} = a_{k,0} = \cdots = a_{0,0} = 1$ . Furthermore,

$a_{k+1,k+1} = a_{k, k+1} + a_{k,k} = 0 + a_{k,k} = a_{k,k},$

so that $a_{k+1,k+1} = a_{k,k} = \cdots = a_{0,0} = 1$ .

Now working on the case $k = 1$ ,

$\begin{aligned} a_{2,0} = 1, \quad a_{2,1} = a_{1,1} + a_{1,0} = 1+1 = 2, \quad a_{2,2} = 1. \end{aligned}$

Hence,

$\begin{aligned} (1+x)^2 &= a_{2,0} + a_{2,1} x + a_{2,2} x^2 = 1 + 2 x + x^2, \end{aligned}$

as expected.

Remark 1. Since the numbers $a_{k,i}$ are coefficients of the binomial $(1+x)^k$ , they are unsurprisingly called the binomial coefficients, and often denoted $\displaystyle a_{k, i} = {k \choose i}$ . It turns out that the formulas that define these numbers generate Pascal’s triangle.

Theorem 1 (Binomial Theorem). For real numbers $a, b$ and any positive integer $n$ ,

$\displaystyle (a+b)^n = a^n + {n \choose 1} a^{n-1} b + {n \choose 2} a^{n-2} b^2 + \cdots + b^n.$

Furthermore, using ideas in counting problems (i.e. Theorem 4 of this post), for any $0 \leq k \leq n$ ,

$\displaystyle {n \choose k} = \frac{ n! }{ k! (n-k)! } = \frac{n \cdot (n-1) \cdot \cdots \cdot (n-k+2) \cdot (n-k+1)}{k \cdot (k-1) \cdot \cdots \cdot 2 \cdot 1},$

where $0! := 1$ for convenience / convention / simplicity.

Proof. The result is trivial if $a = 0$ . Suppose $a \neq 0$ . Setting $a = 1, b = x$ , we have

$\displaystyle (1+x)^n = 1 + {n \choose 1} x + {n \choose 2} x^2 + \cdots + {n \choose n-1} x^{n-1} + x^n.$

For the general case, we first factorise to reduce to the special case:

$\begin{aligned} (a+b)^n &= \left(a \left(1 + \frac ba \right) \right)^n = a^n \left(1 + \frac ba \right)^n . \end{aligned}$

We then use the special case to obtain

$\begin{aligned} a^n \left(1 + \frac ba \right)^n &= a^n \left( 1 + {n \choose 1} \cdot \frac ba + {n \choose 2} \frac {b^2}{a^2} + \cdots + {n \choose n-1} \frac { b^{n-1} }{ a^{n-1} } + \frac { b^n }{ a^n } \right) \\ &= a^n + {n \choose 1} \cdot a^n \cdot \frac ba + {n \choose 2} \cdot a^n \cdot \frac {b^2}{a^2} + \cdots + {n \choose n-1} \cdot a^n \cdot \frac { b^{n-1} }{ a^{n-1} } + a^n \cdot \frac { b^n }{ a^n } \\ &= a^n + {n \choose 1} a^{n-1} b + {n \choose 2} a^{n-2} b^2 + \cdots + {n \choose n-1} a b^{n-1} + b^n, \end{aligned}$

as required.

Theorem 2. Given $n \geq 3$ , setting $h = 0$ in the right-hand side of

$\displaystyle \frac{(x+h)^n - x^n}{h}$

yields $nx^{n-1}$ .

Proof. Using the binomial theorem,

$\displaystyle \frac{(x+h)^n - x^n}{h} = {n \choose 1} x^{n-1} + \left( {n \choose 2} x^{n-2} + \cdots + {n \choose n-1} x h^{n-2} + h^{n-1} \right) h.$

Setting $h = 0$ in the right-hand side yields

$\begin{aligned} & {n \choose 1} x^{n-1} + \left( {n \choose 2} x^{n-2} + \cdots + {n \choose n-1} x h^{n-2} + h^{n-1} \right) h \\ &\phantom{==} = nx^{n-1} + \left( {n \choose 2} x^{n-2} + \cdots + 0 + 0\right) \cdot 0 = nx^{n-1}.\end{aligned}$

A slightly more interesting example arises in the expansion of $(1+1/n)^n$ :

$\displaystyle \begin{aligned} \left(1+ \frac 1n \right)^n &= 1 + {n \choose 1} \frac 1n + {n \choose 2} \frac 1{n^2} + \cdots + {n \choose n-1} \frac{1}{n^{n-1}} + \frac 1{n^n} \\ &= 1 + \frac n1 \cdot \frac 1n + \frac{n \cdot (n-1)}{2 \cdot 1} \cdot \frac 1{n^2} + \cdots + \frac{n \cdot (n-1) \cdot \cdots \cdot 2}{(n-1) \cdot \cdots \cdot 2 \cdot 1} \cdot\frac{1}{n^{n-1}} + \frac 1{n^n} \\ &\leq 1 + 1 + \frac 12 + \cdots + \frac{1}{(n-1) !} + \frac{1}{n!}. \end{aligned}$

Coupling this calculation with several other high-level convergence ideas, we can derive the definition of $e \approx 2.718$ . This number is crucial in helping us understand exponential functions when coupled with ideas in calculus.

We touch base with the idea of exponential functions next time. For a fully rigorous construction of these objects, see this post involving real analysis.

—Joel Kindiak, 27 Oct 25, 1827H

February 17, 2026
Common Continuous Probabilities

Having ventured far into the measure-theoretic world and taking great pains to verify that taking the length $\lambda([a,b]) = b-a$ of a compact interval $[a,b]$ is mathematically legitimate, we turn our attention to modelling continuous data. If the uniform counting measure considered the “basic” distribution in discrete probability, then the uniform continuous distribution would naturally be considered the “basic” distribution in continuous probability.

Definition 1. Let $\Omega$ be a sample space. The uniform continuous distribution on $[a, b]$ is the random variable $X : \Omega \to \mathbb R$ with a distribution defined as follows: for any measurable $K \subseteq [a, b]$ ,

$\displaystyle \begin{aligned} \mathbb P_X(K) &= \frac{1}{b-a} \cdot \int_K \mathbb I_{[a, b]} \, \mathrm d\lambda \equiv \frac{1}{b-a} \cdot \int_a^b \mathbb I_K(x) \, \mathrm dx. \end{aligned}$

Here, the probability density function is given by $\displaystyle f_X = \frac{1}{b-a} \cdot \mathbb I_{[a,b]}$ . Henceforth also we will disregard the sample space unless we need to undertake some theoretic construction.

As usual, $\lambda$ here denotes the Lebesgue measure on $\mathbb R$ , and in this case, we write $X \sim \mathcal U(a, b)$ .

The expected value and variance tends to be useful statistical quantities for our purposes. Nevertheless, there is a special expected value and a special variance that causes the rest of our calculations to become trivial.

Lemma 1. Suppose $\mathbb E[X] = \mu$ and $\mathrm{Var}(X) = \sigma^2$ . Define the standardised random variable $\tilde X$ by

$\displaystyle \tilde X := \frac{X - \mu}{\sigma}.$

Then $\mathbb E[\tilde X] = 0$ and $\mathrm{Var}(\tilde X) = 1$ . Conversely, for any $X = a + b \cdot U$ with $b \neq 0$ , if the p.d.f. $f_U$ of $U$ is continuous, then the p.d.f. $f_X$ of $X$ can be written in terms of the p.d.f. of $U$ :

$\displaystyle f_X(x) = \frac 1b \cdot f_U\left( \frac{x-a}{b} \right).$

Furthermore, $\mathbb E[X] = a + b \cdot \mathbb E[U]$ and $\mathrm{Var}(X) = b^2 \cdot \mathrm{Var}(U)$ .

Proof. For the p.d.f. result, by the fundamental theorem of calculus,

$\begin{aligned} f_X(x) &= \frac{\mathrm d}{\mathrm dx} \int_{-\infty}^x f_X(t)\, \mathrm dt \\ &= \frac{\mathrm d}{\mathrm dx} (\mathbb P(X \leq x)) \\ &= \frac{\mathrm d}{\mathrm dx} (\mathbb P(a + b \cdot U \leq x)) \\ &= \frac{\mathrm d}{\mathrm dx} \left( \mathbb P\left( U \leq \frac{x - a}{b} \right) \right) \\ &= \frac{\mathrm d}{\mathrm dx} \int_{-\infty}^{(x-a)/b} f_U(t)\, \mathrm dt \\ &= \frac 1b \cdot f_U\left( \frac{x-a}{b} \right). \end{aligned}$

The other results follow from expectation and variance properties.

Theorem 1. If $X \sim \mathcal U(a, b)$ , then $\mathbb E[X] = (a+b)/2$ and $\mathrm{Var}(X) = (b-a)^2/12$ .

Proof. Let $U \sim \mathcal U(0, 1)$ for simplicity, so that

$\displaystyle f_X(x) = \frac{1}{b-a}\cdot \mathbb I_{[a, b]}(x) = \mathbb I_{[0, 1]} \left( \frac{x-a}{b-a} \right) \cdot \frac{1}{b-a}$

implies that

$X = a + (b-a) \cdot U.$

By definition,

$\begin{aligned} \mathbb E[U] &= \int_{-\infty}^{\infty} x \cdot \mathbb I_{[0,1]}(x)\, \mathrm dx = \int_{0}^{1} x\, \mathrm dx =\frac 12.\end{aligned}$

Similarly,

$\begin{aligned} \mathbb E[U^2] &= \int_{0}^{1} x^2\, \mathrm dx = \frac 13,\end{aligned}$

so that

$\displaystyle \mathrm{Var}(U) = \mathbb E[U^2] - \mathbb E[U]^2 = \frac 13 - \left(\frac 14\right)^2 = \frac 1{12}.$

Then the general case follows from

$\begin{aligned} \mathbb E[X] &= a + (b-a) \cdot \frac 12 = \frac{a+b}{2}, \\ \mathrm{Var}(X) &= (b-a)^2 \cdot \frac 1{12} = \frac{(b-a)^2}{12}. \end{aligned}$

We can generalise Lemma 1 to any differentiable, invertible transformation of $U$ .

Lemma 2. For any $X = g(U)$ where $g$ is a invertible differentiable transformation, if the p.d.f. $f_U$ of $U$ is continuous, then the p.d.f. $f_X$ of $X$ can be written in terms of the p.d.f. of $U$ :

$\displaystyle f_X(x) = f_U (g^{-1}(x)) \cdot (g^{-1})'(x).$

Proof. Apply the same arguments as in Lemma 1, but apply the chain rule and the inverse function theorem for book-keeping purposes.

The continuous analog of the geometric distribution would be the exponential distribution.

Theorem 2. The random variable $X$ follows an exponential distribution with rate parameter $\lambda > 0$ , denoted $X \sim \mathrm{Exp}(\lambda)$ , if is defined by the probability density function

$f_X(x) = \lambda e^{-\lambda x} \cdot \mathbb I_{[0,\infty)}.$

$\mathbb E[X] = 1/\lambda$ and $\mathrm{Var}(X) = 1/\lambda^2$ .

Proof. Let $U \sim \mathrm{Exp}(1)$ for simplicity, so that

$\displaystyle f_X(x) = f_U(\lambda x) \cdot \lambda \quad \Rightarrow \quad X = \frac 1\lambda \cdot U$

gives the final result. Integrating by parts,

$\begin{aligned}\mathbb E[U] &= \int_0^\infty xe^{-x}\, \mathrm dx \\ &= [-xe^{-x} - e^{-x}]_0^\infty \\ &= (0 - 0) - (0 - 1) = 1, \\ \mathbb E[U^2] &= \int_0^\infty x^2 e^{-x}\, \mathrm dx \\ &= \left[-x^2 e^{-x} \right]_0^\infty + 2\int_0^\infty x \cdot e^{-x}\, \mathrm dx \\ &= (0 - 0) + 2 \cdot \mathbb E[U] = 2,\end{aligned}$

so that $\mathrm{Var}(U) = \mathbb E[U^2] - \mathbb E[U]^2 = 2- 1^2 = 1$ .

Another crucial distribution in probability and statistics is the normal distribution. It is a useful model to represent all kings of continuous data—heights of humans, test scores for exams, and even month-on-month returns on investment. Actually more is true—as long as we have a continuous nonzero function $f$ that is integrable, we obtain a corresponding probability distribution.

Lemma 3. For any continuous and integrable nonzero function $f \geq 0$ , there exists a continuous random variable $X$ with probability density function

$\displaystyle f_X = C_f \cdot f,$

where $C_f := (\int_{-\infty}^{\infty} f(x)\, \mathrm dx)^{-1}$ is called the normalising constant of $f$ .

Theorem 3. The random variable $X$ follows an normal distribution with mean $\mu$ and variance $\sigma^2$ , denoted $X \sim \mathcal{N}(\mu, \sigma^2)$ , if is defined by the probability density function

$\displaystyle f_X(x) = \frac 1{\sigma\sqrt{2\pi}}\exp\left( - \frac{ (x - \mu)^2 }{2\sigma^2} \right).$

Then $\mathbb E[X] = \mu$ and $\mathrm{Var}(X) = \sigma^2$ .

Proof. Denoting $Z \sim \mathcal N(0, 1)$ ,

$\displaystyle f_X(x) = \frac 1{\sigma} \cdot f_Z\left( \frac{x-\mu}{\sigma} \right) \quad \Rightarrow \quad X = \mu + \sigma \cdot Z,$

so it suffices to prove that $\mathbb E[Z] = 0$ and $\mathrm{Var}(Z) = 1$ . We remark that the normalising constant of $e^{-z^2/2}$ is $1/\sqrt{2\pi}$ by the Gaussian integral

$\displaystyle \int_{-\infty}^{\infty} e^{-z^2}\, \mathrm dz = \sqrt{\pi}$

and a change of variables. The former is immediate since the function $ze^{-z^2/2}$ is odd and its integral over $\mathbb R$ converges absolutely, so that

$\displaystyle \sqrt{2\pi} \cdot \mathbb E[Z] = \int_{-\infty}^{\infty} ze^{-z^2/2}\, \mathrm dz = 0.$

The latter requires integration by parts (here, as with the exponential distribution calculation, all integrals converge absolutely so we do not need to worry about the limits):

$\begin{aligned}\sqrt{2\pi} \cdot \mathbb E[Z^2] &= \int_{-\infty}^{\infty} z^2 e^{-z^2/2}\, \mathrm dz \\ &= \int_{-\infty}^{\infty} -z \cdot (-z e^{-z^2/2})\, \mathrm dz \\ &= \left[ -z \cdot e^{-z^2/2} \right]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} (-1) \cdot e^{-z^2/2} \, \mathrm dz \\ &= (0 - 0) - (-1) \cdot \sqrt{2\pi} = \sqrt{2\pi}. \end{aligned}$

Therefore,

$\mathrm{Var}(Z) = \mathbb E[Z^2] - \mathbb E[Z]^2 = 1 - 0^2 = 1.$

What makes the normal distribution so ubiquitous is how it arises from samples of virtually any distribution.

Theorem 4 (Central Limit Theorem). Let $X_1,X_2,\dots$ be i.i.d. random variables with mean $0$ and variance $1$ . Define

$\displaystyle \bar X_n := \frac{1}{n} \sum_{i=1}^n X_i.$

Then $\sqrt{n} \cdot \bar X_n \to Z$ in the following sense: for any $\epsilon > 0$ , there exists $n > 0$ such that for any $z \in \mathbb R$ ,

$\displaystyle |\mathbb P(\sqrt{n} \cdot \bar X_n \leq z) - \mathbb P(Z \leq z)| < \epsilon.$

We say that $\sqrt{n} \cdot \bar X_n$ converges in distribution to the standard normal distribution $Z \sim \mathcal N(0, 1)$ .

Proof. Delayed.

Our main goal in these posts on probability is to prove this statement rigorously. The central limit theorem is responsible for our intuition about probability arising from repeated experiments.

A more immediate application comes from simulation. It turns out that the humble uniform random variable $U \sim \mathcal U(0, 1)$ (implemented using pseudorandom numbers) can be used to generate all other random variables.

Theorem 5. Suppose $U \sim \mathcal U(0, 1)$ . For any random variable $X \sim \mathbb P_0$ , there exists a measurable function $F$ such that $F(U) \sim \mathbb P_0$ .

Proof. Let $F_X := \mathbb P_0((-\infty, \cdot ])$ denote the c.d.f. of $X$ . We observe that $F_X( \cdot ) \in [0, 1]$ by definition. Therefore, the idea is to generate $U \in [0, 1]$ , then define the variate $V := F_X^{-1}(U)$ , so that $F_X(V) = U$ .

To that end, define the measurable map $F$ by

$\displaystyle F(u) := \sup \{x \in \mathbb R : F_X(x) \leq u\}.$

Then define $V := F(U)$ . Observe that $\{V \leq v\} = \{U \leq F_X(v)\}$ . Therefore

$\begin{aligned} \mathbb P(V \leq v) &= \mathbb P(U \leq F_X(v)) = F_X(v) = \mathbb P(X \leq v). \end{aligned}$

Therefore, $F(U) = V \sim \mathbb P_0$ .

But we have a slightly more urgent question to answer: what’s the distribution of $X +Y$ in general? We need to construct the relevant sample space, and prove relevant properties in the multivariable setting, before we can legitimately continue on our quest to prove the central limit theorem. In fact, once we have at least done so for normal distributions, we will be in a sufficiently good place to discuss the principle of statistical hypothesis testing—a quantitative implementation of the scientific method.

—Joel Kindiak. 19 Jul 25, 2258H

November 18, 2025