courseshub.world · Lives in Mathematics and Physics

George Boole

George Boole, FRS

A course in two strands: the life of a shoemaker’s son from Lincoln who taught himself mathematics while running schools, and the algebra of logic, probability and operators that he built and that now runs inside every digital circuit.

Born 2 November 1815, Lincoln Died 8 December 1864, Ballintemple, Cork Laws of Thought 1854 Level advanced undergraduate
Three classes split the universe 1 into eight constituents. Shaded: x(1−y), the X that are not Y.
Lesson 00

How to use this course

George Boole never took a degree. He learned calculus from French textbooks while teaching in small schools to support his parents, published in the Cambridge Mathematical Journal from the provinces, won a Royal Society gold medal at 28, and became the first professor of mathematics at Queen’s College, Cork. His name survives in Boolean algebra, but his work also includes operator methods for differential equations, finite differences, probability, and a paper that helped start invariant theory.

Prerequisites for Part II
  • Elementary set theory and propositional logic (truth tables).
  • A first course in rings (ideals, quotients) for Lesson 06.
  • Elementary probability for Lesson 08.
  • Linear ordinary differential equations with constant coefficients, and Taylor series, for Lessons 09 and 10.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, proofs or proof sketches, and exercises with folded solutions. Lessons 04 and 05 present Boole’s own system, read through the modern interpretation that makes it rigorous. Lessons 06 and 07 follow what later mathematicians and engineers made of it. Lessons 08 to 10 cover his other mathematics.

Reference

Timeline

1815Born 2 November in Lincoln, eldest child of John Boole, a shoemaker with a strong interest in science and instruments, and Mary Ann Joyce.
182810 September: enters Bainbridge’s Commercial Academy in Lincoln. Teaches himself Greek, French and German.
c. 1831His father’s business fails. At 16 he becomes assistant teacher at Heigham’s School, Doncaster.
1833Teaching in Liverpool, then at Hall’s Academy, Waddington.
1834Opens his own school in Lincoln, aged 19. Active in the new Lincoln Mechanics’ Institute.
1838Takes over Hall’s Academy at Waddington after Robert Hall’s death. Studies Laplace and Lagrange.
1840Opens a boarding school in Lincoln. Begins publishing in the Cambridge Mathematical Journal, encouraged by its editor Duncan Gregory.
1841“Exposition of a general theory of linear transformations”: invariance of discriminants.
1842Begins a lifelong correspondence with Augustus De Morgan.
1844“On a general method in analysis” in the Philosophical Transactions; Royal Medal of the Royal Society, November.
1847The Mathematical Analysis of Logic.
1849Appointed first Professor of Mathematics at Queen’s College, Cork (announced August, takes up post November).
1851Elected Dean of Science at Cork.
1854An Investigation of the Laws of Thought, on Which are Founded the Mathematical Theories of Logic and Probabilities.
185511 September: marries Mary Everest at Wickwar, Gloucestershire. Keith Medal of the Royal Society of Edinburgh.
1857Elected Fellow of the Royal Society. Paper on definite integrals containing Boole’s identity.
1859A Treatise on Differential Equations.
1860A Treatise on the Calculus of Finite Differences.
1864Dies 8 December at Ballintemple, Cork, aged 49, after a fever that followed a soaking in heavy rain.
Part I · Formation
Lesson 01

Lincoln, 1815–1834

George Boole was born in Lincoln on 2 November 1815 and christened the next day. His father John Boole kept a cobbler’s shop at 34 Silver Street, but his real enthusiasm was for science and mathematics, which he passed on to his son. His mother, Mary Ann Joyce, had been a lady’s maid. Three younger children followed within five years: Mary Ann, William and Charles.

George’s schooling was modest and practical. He attended a school for tradesmen’s children, then a commercial school, then a primary school, and on 10 September 1828 entered Bainbridge’s Commercial Academy. His father taught him mathematics and the making of instruments. A local bookseller arranged Latin lessons, and Boole taught himself Greek well enough to translate a poem by Meleager at 14. He went on to teach himself French and German.

Teaching to support a family

Around 1831 his father’s business collapsed, and George, at 16, became the family’s main support. He took a post as assistant teacher at Heigham’s School in Doncaster, then taught for about six months in Liverpool in 1833, then at Hall’s Academy in Waddington, a village four miles from Lincoln. In 1834, aged 19, he opened his own school in Lincoln. He was involved with the Lincoln Mechanics’ Institute, founded in 1833.

His reading in these years was in languages and literature as much as mathematics. The shift came in the next few years, when he started to read the great French analysts.

Lesson 02

The schoolmaster mathematician, 1834–1847

Boole’s first advanced mathematics book was Sylvestre Lacroix’s treatise on the differential and integral calculus. By 1838 he was working through Laplace’s celestial mechanics and Lagrange’s analytical mechanics, making notes that turned into his first research. That year Robert Hall of Waddington died, and Boole took over his academy; the whole Boole family moved there to help run it. In 1840 he moved back to Lincoln and opened a boarding school.

Duncan Gregory and the Cambridge Mathematical Journal

The young Scottish mathematician Duncan Gregory had founded the Cambridge Mathematical Journal and was a champion of the “symbolical” school of algebra, which treated operators such as \(d/dx\) as algebraic quantities obeying formal laws. Gregory encouraged Boole, published his papers from 1840 onward, and advised him to study at Cambridge. Boole declined because his school’s income kept his parents. The symbolical method shaped everything he did later: he learned to ask which laws a system of symbols obeys, and to compute with those laws alone.

Two results of this period matter most. In 1841 his “Exposition of a general theory of linear transformations” showed that certain polynomial expressions in the coefficients of a form, such as the discriminant, change only by a power of the determinant under a linear change of variables. Arthur Cayley read it and in 1845 began the systematic theory of invariants, crediting Boole’s paper as his starting point (Lesson 10). In 1843 Boole wrote “On a general method in analysis”, on operator methods for linear differential equations (Lesson 09). The Royal Society published it in the Philosophical Transactions in 1844 and awarded him a Royal Medal in November 1844, its first gold medal for a mathematical paper.

From 1842 he corresponded with Augustus De Morgan in London, a friendship that lasted the rest of his life.

Lesson 03

Logic, and the call to Cork, 1847–1849

Early in 1847 a public dispute broke out between De Morgan and the Edinburgh philosopher Sir William Hamilton over who deserved credit for the idea of “quantifying the predicate” in syllogisms. Boole had long wondered whether logic could be treated by algebra. The quarrel prompted him to write out his ideas, and in 1847 he published The Mathematical Analysis of Logic, a pamphlet of 82 pages. De Morgan’s Formal Logic appeared at nearly the same time.

The pamphlet’s central claim was that the laws of reasoning about classes are the laws of an algebra whose symbols obey all the usual rules except one: for a class symbol \(x\), the product \(xx\) equals \(x\). Selecting the sheep from a flock and then selecting the sheep again gives just the sheep. From this one law, Boole derived the classical syllogisms and the principle of contradiction (Lesson 04).

A chair without a degree

In 1845 the British government founded three Queen’s Colleges in Ireland, at Belfast, Cork and Galway, open to students of every religion. Boole applied for a chair in 1846, with testimonials from De Morgan, Philip Kelland, Cayley and William Thomson (later Lord Kelvin). The appointments were delayed. His father died in December 1848. In August 1849 Boole was announced as the first Professor of Mathematics at Queen’s College, Cork, and he took up the post in November. He was 33, had no university degree, and now had a salary and time for research.

Part II · The mathematics
Lesson 04

The algebra of classes

For the general reader. Boole’s idea was to write “sheep” as a letter \(x\), “black things” as \(y\), and “black sheep” as the product \(xy\). With 1 for “everything” and 0 for “nothing”, “not sheep” becomes \(1-x\). Most rules of school algebra still hold, with one new law: \(x^2=x\). Remarkably, that law is enough to do logic by calculation.

Goals
  • State Boole’s laws for elective symbols and the role of \(x^2=x\).
  • Prove the expansion theorem and interpret its constituents.
  • Understand the modern model that makes Boole’s partial operations rigorous.

Elective symbols and the laws

In The Laws of Thought (1854), a letter \(x\) stands for a class, and \(xy\) for the class of things that are both \(x\) and \(y\). The symbol 1 is the universe of discourse and 0 the empty class. Boole wrote \(x+y\) for the union only when \(x\) and \(y\) have no members in common, and \(x-y\) only when \(y\) is contained in \(x\). The laws he used are

\[ xy=yx,\qquad x(y+z)=xy+xz,\qquad x(y-z)=xy-xz,\qquad x^2=x. \]

From \(x^2=x\) he derived \(x(1-x)=0\): nothing is both \(x\) and not \(x\). Boole presented this as an algebraic proof of Aristotle’s principle of contradiction. He noted that the equation \(x^2=x\) has only the roots 0 and 1, and concluded that his logic is the algebra of quantities restricted to these two values.

Making the system rigorous

Boole’s partial addition troubled his readers, since a calculation might pass through terms like \(x+x\) that mean nothing as classes. Theodore Hailperin and later Stanley Burris gave a clean model. Fix a set \(U\), and let \(R=\mathbb Z^U\) be the ring of integer-valued functions on \(U\) with pointwise operations. A class \(X\subseteq U\) is its indicator function. The idempotents of \(R\) (elements with \(e^2=e\)) are exactly the indicator functions. Then \(x+y\) is idempotent if and only if \(xy=0\), and \(x-y\) is idempotent if and only if \(xy=y\), which is precisely Boole’s restriction.

Theorem (Rule of 0 and 1)

Let \(p,q\in\mathbb Z[x_1,\dots,x_n]\). The equation \(p(x_1,\dots,x_n)=q(x_1,\dots,x_n)\) holds for all classes \(x_1,\dots,x_n\) in every universe \(U\) (interpreted in \(\mathbb Z^U\)) if and only if it holds whenever each \(x_i\in\{0,1\}\subset\mathbb Z\).

Proof. Operations in \(\mathbb Z^U\) are pointwise. At a point \(u\in U\), each indicator takes a value in \(\{0,1\}\), so \(p(x)(u)=p(x_1(u),\dots,x_n(u))\) is \(p\) evaluated at a 0–1 vector. Conversely, a one-point universe realizes every 0–1 vector. ∎

Boole stated a version of this in Chapter V of The Laws of Thought, and justified intermediate uninterpretable expressions by analogy with \(\sqrt{-1}\) in trigonometry. The theorem shows his instinct was sound for equations.

The expansion theorem

Theorem (Boole’s expansion, 1847/1854)

For any polynomial \(f\) with integer coefficients and class symbols \(x,y,\dots\),

\[ f(x)=f(1)\,x+f(0)\,(1-x), \] \[ f(x,y)=f(1,1)\,xy+f(1,0)\,x(1-y)+f(0,1)\,(1-x)y+f(0,0)\,(1-x)(1-y), \]

and in general \(f(x_1,\dots,x_n)=\sum_{\varepsilon\in\{0,1\}^n} f(\varepsilon)\,C_\varepsilon\), where the constituent \(C_\varepsilon=\prod_{i}x_i^{\varepsilon_i}(1-x_i)^{1-\varepsilon_i}\).

Proof. By the Rule of 0 and 1 it suffices to check 0–1 values. If \(x=\varepsilon\in\{0,1\}^n\), then \(C_{\varepsilon}=1\) and every other constituent is 0, so the right side equals \(f(\varepsilon)\). ∎

The constituents are idempotent, pairwise disjoint (\(C_\varepsilon C_{\varepsilon'}=0\) for \(\varepsilon\ne\varepsilon'\)), and sum to 1: they are the \(2^n\) cells of a Venn diagram. Boole called the numbers \(f(\varepsilon)\) the moduli of \(f\). Two polynomials agree on all classes exactly when their moduli agree.

Corollary (interpreting an equation)

For classes \(x_1,\dots,x_n\), the equation \(f(x_1,\dots,x_n)=0\) holds if and only if \(C_\varepsilon=0\) for every \(\varepsilon\) with \(f(\varepsilon)\ne 0\).

So an equation in Boole’s algebra always says that certain cells of the Venn diagram are empty. In particular “All X is Y” becomes \(x(1-y)=0\), and “No X is Y” becomes \(xy=0\).

Exercise 4.1Expand \(f(x,y)=x+y-2xy\) and say which class it represents. Do the same for \(g(x,y)=x+y-xy\).

Show solution
Moduli of \(f\): \(f(1,1)=0,\ f(1,0)=1,\ f(0,1)=1,\ f(0,0)=0\). So \(f=x(1-y)+(1-x)y\): things that are \(x\) or \(y\) but not both (symmetric difference). For \(g\): moduli \(1,1,1,0\), so \(g=xy+x(1-y)+(1-x)y\), the union \(x\cup y\). Both are idempotent, as their moduli are all 0 or 1.

Exercise 4.2Show that a polynomial \(f\) represents a class (that is, \(f^2=f\) for all classes) if and only if all its moduli lie in \(\{0,1\}\). How many distinct classes can be built from \(n\) given classes?

Show solution
By expansion, \(f^2=\sum_\varepsilon f(\varepsilon)^2C_\varepsilon\), because the constituents are disjoint idempotents. So \(f^2=f\) iff \(f(\varepsilon)^2=f(\varepsilon)\) for all \(\varepsilon\), i.e. \(f(\varepsilon)\in\{0,1\}\). Such an \(f\) is a choice of a subset of the \(2^n\) constituents, so there are \(2^{2^n}\) classes (some may coincide for particular \(x_i\)).

Exercise 4.3Translate “No X is Y” and “All Y is Z” into equations, and decide which cells of the three-circle Venn diagram they empty.

Show solution
\(xy=0\) empties \(xyz\) and \(xy(1-z)\). \(y(1-z)=0\) empties \(xy(1-z)\) and \((1-x)y(1-z)\). Together they say that \(y\) lies inside \((1-x)z\).
Lesson 05

Elimination, solution, propositions

For the general reader. A syllogism such as “all men are mortal, all Greeks are men, therefore all Greeks are mortal” removes the middle term “men”. Boole saw this as elimination of a variable, like eliminating \(y\) from two equations in school algebra. He gave general rules for eliminating any class from any set of premises, and for solving for any class in terms of the others.

Goals
  • Reduce several premises to one equation.
  • Prove and apply Boole’s elimination theorem.
  • Use Boole’s solution rule, including the coefficient 0/0.
  • See how propositional logic fits as the algebra of “times”.

Reduction

In the integer-valued model of Lesson 04, a sum of squares vanishes at a point only when each term does. So several premises \(r_1=0,\dots,r_k=0\) are equivalent to the single equation \(r_1^2+\cdots+r_k^2=0\). When each \(r_i\) is itself idempotent, as in \(x(1-y)\), the squares can be dropped.

Theorem (Boole’s elimination, 1854)

Let \(f(x,y_1,\dots,y_m)\) be a polynomial with integer coefficients. For given classes \(y_1,\dots,y_m\), there exists a class \(x\) with \(f(x,y)=0\) if and only if

\[ f(1,y)\,f(0,y)=0. \]

Proof. Work at each point \(u\) of the universe. A suitable \(x(u)\in\{0,1\}\) exists exactly when \(f(1,y(u))=0\) or \(f(0,y(u))=0\), which is the condition that the product vanish at \(u\). Choosing \(x(u)\) pointwise defines the class \(x\). ∎

The syllogism Barbara. “All X is Y” and “All Y is Z” give \(f=x(1-y)+y(1-z)=0\). Eliminating \(y\): \(f|_{y=1}=1-z\) and \(f|_{y=0}=x\), so \(x(1-z)=0\): All X is Z.

Solving for a class

To solve \(f(x,y)=0\) for \(x\), Boole wrote \(x=\dfrac{f(0,y)}{f(0,y)-f(1,y)}\), a purely formal quotient, and expanded the right-hand side in the remaining variables. Each constituent gets a coefficient \(N/D\), read as follows:

  • \(1\) (or \(m/m\), \(m\ne0\)): the constituent is included in \(x\);
  • \(0\) (or \(0/m\)): it is excluded;
  • \(0/0\): an indefinite part \(v\) of it is included;
  • anything else, such as \(1/0\): the constituent must be empty, a side condition.

The division is a mnemonic. The rule itself can be proved pointwise: on a constituent where \(f(0,y)=f(1,y)=0\), \(x\) is free; where only \(f(1,y)=0\), \(x=1\); where only \(f(0,y)=0\), \(x=0\); where neither vanishes, no \(x\) works.

Secondary propositions

For statements such as “if it rains, the ground is wet”, Boole let \(x\) be the portion of time during which proposition \(X\) is true. Then \(x=1\) says \(X\) is true, \(xy=1\) is “\(X\) and \(Y\)”, \(x+y-xy=1\) is inclusive “or”, and “if \(X\) then \(Y\)” is \(x(1-y)=0\). Checking that a formula is a tautology means checking that its translation equals 1 at every 0–1 assignment. Boole did not draw truth tables, but this is the same computation.

Exercise 5.1Boole’s style of example: “Clean beasts are those which both divide the hoof and chew the cud.” Write \(x\) for clean beasts, \(y\) for those dividing the hoof, \(z\) for those chewing the cud, so \(x=yz\). Solve for \(y\).

Show solution
\(f=x-yz\). Then \(f(y{=}0)=x\), \(f(y{=}1)=x-z\), so \(y=\dfrac{x}{x-(x-z)}=\dfrac{x}{z}\). Expand in \(x,z\): at \((1,1)\) the coefficient is \(1\); at \((1,0)\) it is \(1/0\); at \((0,1)\) it is \(0\); at \((0,0)\) it is \(0/0\). So \(y=xz+v(1-x)(1-z)\) with the condition \(x(1-z)=0\). Hoof-dividers are the clean beasts together with an indefinite part of the unclean beasts that do not chew the cud; and every clean beast chews the cud.

Exercise 5.2From “No X is Y” and “All Z is Y”, eliminate \(y\).

Show solution
\(f=xy+z(1-y)\). \(f(y{=}1)=x\), \(f(y{=}0)=z\). So \(xz=0\): No X is Z (equivalently No Z is X).

Exercise 5.3Using the algebra of times, show that modus ponens is valid: from \(x=1\) and \(x(1-y)=0\) deduce \(y=1\).

Show solution
Substitute \(x=1\) into \(x(1-y)=0\) to get \(1-y=0\), so \(y=1\). In Boole’s reading: at all times when the premises hold, \(Y\) holds.
Lesson 06

From Boole’s algebra to Boolean algebras

For the general reader. Boole’s successors kept his symbols but changed his addition. William Stanley Jevons made \(x+y\) mean “\(x\) or \(y\), or both”, so that \(x+x=x\). Charles Sanders Peirce and Ernst Schröder built the algebra of logic that Peirce called “Boolian”. In the twentieth century it became a standard algebraic structure with a beautiful representation theorem.

Goals
  • State the axioms of a Boolean algebra and of a Boolean ring, and pass between them.
  • Prove that Boolean rings are commutative of characteristic 2.
  • State Stone’s representation theorem and classify finite Boolean algebras.

The modern structure

In 1863–64 Jevons objected to Boole’s refusal of \(x+x=x\). Boole replied that \(x+x=0\) is a meaningful equation (equivalent to \(x=0\)) while \(x+x\) is not a class. Jevons’s Pure Logic (1864) took inclusive union as the basic operation, and that choice prevailed. Peirce in the 1870s and 1880s and Schröder in his Vorlesungen über die Algebra der Logik (1890–1905) developed the algebra further. Edward Huntington gave independent axioms in 1904.

Definition

A Boolean algebra is a set \(B\) with binary operations \(\vee,\wedge\), a unary operation \('\), and elements \(0,1\), such that \(\vee\) and \(\wedge\) are commutative and associative, each distributes over the other, \(x\vee0=x\), \(x\wedge1=x\), \(x\vee x'=1\) and \(x\wedge x'=0\).

A Boolean ring is a ring with identity in which \(x^2=x\) for every \(x\).

Theorem (Stone, 1936)

(a) Every Boolean ring is commutative and satisfies \(x+x=0\).

(b) Boolean rings and Boolean algebras are the same structures: given a Boolean algebra, set \(x+y=(x\wedge y')\vee(x'\wedge y)\) and \(xy=x\wedge y\); given a Boolean ring, set \(x\vee y=x+y+xy\), \(x\wedge y=xy\), \(x'=1+x\).

(c) Every Boolean algebra is isomorphic to a field of subsets of some set, with union, intersection and complement.

Proof of (a). Since \(2x\) is idempotent, \(2x=(2x)^2=4x^2=4x\), so \(2x=0\). Next \(x+y=(x+y)^2=x+xy+yx+y\), so \(xy+yx=0\), hence \(xy=-yx=yx\). ∎

Part (b) is a direct verification. Part (c) uses the Stone space: the set of ring homomorphisms \(B\to\mathbb F_2\) (equivalently, ultrafilters), with \(b\mapsto\{\varphi:\varphi(b)=1\}\). Injectivity needs the Boolean prime ideal theorem, a weak form of the axiom of choice. Note how close Boole’s ring of integer functions is to the Boolean ring \(\mathbb F_2^U\): Boole worked with \(\mathbb Z\) where the modern theory uses \(\mathbb F_2\), which is why his addition was partial.

Corollary (finite Boolean algebras)

Every finite Boolean algebra is isomorphic to the power set of its set of atoms. In particular it has \(2^n\) elements for some \(n\).

Exercise 6.1Show that \(\mathbb Z/6\) is not a Boolean ring, but that its idempotents \(\{0,1,3,4\}\) form a Boolean algebra under \(x\wedge y=xy\), \(x\vee y=x+y-xy\), \(x'=1-x\). Identify the atoms.

Show solution
\(2^2=4\ne2\). The idempotents: \(3^2=9=3\), \(4^2=16=4\). Check closure: \(3\cdot4=0\); \(3\vee4=3+4-0=7=1\); \(3'=-2=4\). So the algebra is \(\{0,3,4,1\}\cong\mathcal P(\{a,b\})\) with atoms 3 and 4. This mirrors \(\mathbb Z/6\cong\mathbb Z/2\times\mathbb Z/3\): the idempotents are \((1,0)\) and \((0,1)\).

Exercise 6.2Prove the corollary: if \(B\) is a finite Boolean algebra with atoms \(a_1,\dots,a_n\), then \(b\mapsto\{i: a_i\le b\}\) is an isomorphism \(B\to\mathcal P(\{1,\dots,n\})\).

Show solution
For \(b\ne0\), finiteness gives an atom below \(b\) (descend through nonzero elements below \(b\)). Let \(c=\bigvee_{a_i\le b}a_i\le b\). If \(b\wedge c'\ne0\) it contains an atom \(a_j\), which lies below \(b\), hence below \(c\), contradicting \(a_j\le c'\). So \(b=c\), and the map is injective. It is surjective since every join of atoms is in \(B\), and distinct atoms have meet 0, so the atoms below \(\bigvee_{i\in S}a_i\) are exactly \(a_i, i\in S\). It preserves \(\wedge,\vee,'\) because atoms are join-prime: \(a\le x\vee y\) implies \(a\le x\) or \(a\le y\).
Lesson 07

Expansion, normal forms and switching circuits

For the general reader. In 1937 Claude Shannon, a master’s student at MIT, noticed that relay circuits obey Boole’s algebra: switches in series behave like a product, switches in parallel like a sum. Designing a circuit became a matter of simplifying an algebraic expression. Every digital computer is built from this observation.

Goals
  • Prove that every Boolean function has a disjunctive normal form, using Boole’s expansion.
  • Count Boolean functions and show functional completeness of NAND.
  • Translate between formulas and series–parallel circuits.

Boolean functions

A Boolean function of \(n\) variables is a map \(\{0,1\}^n\to\{0,1\}\). There are \(2^{2^n}\) of them. Boole’s expansion (Lesson 04) applied to such a function, now in the two-element algebra, is called the Shannon expansion in engineering:

\[ f(x_1,\dots,x_n)=\big(x_1\wedge f(1,x_2,\dots,x_n)\big)\vee\big(x_1'\wedge f(0,x_2,\dots,x_n)\big). \]
Theorem (disjunctive normal form)

Every Boolean function \(f\) equals the join of the minterms \(C_\varepsilon=\bigwedge_i x_i^{\varepsilon_i}\) over all \(\varepsilon\) with \(f(\varepsilon)=1\), where \(x^1=x\) and \(x^0=x'\). Hence \(\{\wedge,\vee,'\}\) is functionally complete.

Proof. Iterate the Shannon expansion over all variables, or note that both sides agree at each \(\varepsilon\), where exactly one minterm equals 1. ∎

Proposition

The single operation \(x\uparrow y=(x\wedge y)'\) (NAND) is functionally complete.

Proof. \(x'=x\uparrow x\), \(x\wedge y=(x\uparrow y)\uparrow(x\uparrow y)\), \(x\vee y=(x\uparrow x)\uparrow(y\uparrow y)\). ∎ Henry Sheffer published this in 1913; Peirce had noticed it earlier in unpublished work.

Shannon’s circuits

Shannon’s thesis, A Symbolic Analysis of Relay and Switching Circuits (1937, published 1938), assigned to each contact a variable that is 1 when the contact conducts. Two contacts in series conduct when both do, a product; in parallel, when either does, a join. A network’s conduction is therefore a Boolean function, and algebraic identities give equivalent, often cheaper, networks. Victor Shestakov in Moscow developed a similar theory in the same years, publishing in 1941.

Exercise 7.1Write the majority function \(m(x,y,z)\) (1 when at least two inputs are 1) in disjunctive normal form, then simplify it to three terms.

Show solution
Minterms at \(110,101,011,111\): \(m=xyz'\vee xy'z\vee x'yz\vee xyz\). Since \(xyz\) can be joined to each of the others, \(m=xy\vee xz\vee yz\).

Exercise 7.2A light is controlled by two switches at the ends of a staircase; flipping either switch changes the state. Find the Boolean function and express it in Boole’s own notation.

Show solution
The function is exclusive or: \(f=xy'\vee x'y\). In Boole’s integer notation this is \(x(1-y)+(1-x)y=x+y-2xy\) (Exercise 4.1). In a Boolean ring it is simply \(x+y\).

Exercise 7.3Show that \(\{\wedge,\vee\}\) alone is not functionally complete.

Show solution
Every function built from \(\wedge,\vee\) and variables is monotone: if \(x\le y\) componentwise then \(f(x)\le f(y)\). Negation is not monotone (\(0\le1\) but \(0'=1>1'=0\)).
Lesson 08

Logic and probability

For the general reader. Most of The Laws of Thought is not about syllogisms. Boole wanted to compute the probability of any event that can be described in words, given the probabilities of other events. His logic told him how events are built from simpler ones; probability theory then gave the numbers. When the data do not fix the answer, the right output is a range of values.

Goals
  • Express the probability of a compound event through its expansion into constituents.
  • Prove the union bound now called Boole’s inequality.
  • Derive Boole’s “conditions of possible experience” and see the link to linear programming.

Events as Boolean functions

If \(A_1,\dots,A_n\) are events and \(f\) is a Boolean function, the event \(f(A_1,\dots,A_n)\) is the disjoint union of the constituents on which \(f=1\). So

\[ P\big(f(A)\big)=\sum_{\varepsilon:\,f(\varepsilon)=1}P(C_\varepsilon), \]

and the problem reduces to the \(2^n\) unknown constituent probabilities \(q_\varepsilon\ge0\) with \(\sum q_\varepsilon=1\). Boole’s own method added an assumption of independence among the “simple” events, which drew criticism from Wilbraham and later from John Venn. Theodore Hailperin showed in 1965 that the assumption-free version of Boole’s problem is a linear programme.

Theorem (Boole’s inequality)

For events \(A_1,\dots,A_n\), \(P\big(\bigcup_i A_i\big)\le\sum_i P(A_i)\), with equality when the events are pairwise disjoint.

Proof. Let \(B_1=A_1\) and \(B_k=A_k\setminus(A_1\cup\cdots\cup A_{k-1})\). The \(B_k\) are disjoint with the same union, and \(P(B_k)\le P(A_k)\). ∎ The attribution of this elementary bound to Boole is traditional.

Theorem (conditions of possible experience, two events)

Numbers \(p_1,p_2,p_{12}\) are the probabilities of \(A,B,A\cap B\) for some events in some probability space if and only if

\[ \max(0,\;p_1+p_2-1)\;\le\;p_{12}\;\le\;\min(p_1,p_2). \]

Proof. The constituent probabilities are \(q_{11}=p_{12}\), \(q_{10}=p_1-p_{12}\), \(q_{01}=p_2-p_{12}\), \(q_{00}=1-p_1-p_2+p_{12}\). They are the probabilities of some events exactly when all four are non-negative, which is the stated inequality. ∎ Boole derived such conditions in general by elimination. Itamar Pitowsky (1989, 1994) showed that Bell’s inequalities in quantum mechanics are conditions of this type.

Exercise 8.1Two witnesses are each right with probability 0.9. Without assuming independence, what are the best bounds on the probability that both are right?

Show solution
\(\max(0,0.9+0.9-1)=0.8\le p_{12}\le0.9\). Independence would give 0.81, inside the range.

Exercise 8.2Find the sharp bounds on \(P(A\cup B\cup C)\) given only \(P(A)=P(B)=P(C)=0.3\).

Show solution
Upper: Boole’s inequality gives 0.9, attained by three disjoint events. Lower: the union contains \(A\), so it is at least 0.3, attained when \(A=B=C\). The answer is \([0.3,\,0.9]\).

Exercise 8.3Show that \(p_1=p_2=p_3=\tfrac12\) and \(P(A_1\cap A_2)=P(A_2\cap A_3)=P(A_1\cap A_3)=0\) is impossible.

Show solution
The events would be pairwise disjoint, so \(P(A_1\cup A_2\cup A_3)=\tfrac32>1\). This is an inequality of Boole’s kind: \(p_1+p_2+p_3-p_{12}-p_{13}-p_{23}\le1\).
Lesson 09

Operators and differential equations

For the general reader. Boole’s logic grew out of a habit he learned in his twenties: treating the instruction “differentiate” as a symbol \(D\) that can be added, multiplied and factored like a number. His prize-winning 1844 memoir extended this to operators that do not commute, and his 1859 Treatise on Differential Equations was a standard text for decades.

Goals
  • Use the symbolic method for linear equations with constant coefficients.
  • Prove the exponential shift rule and apply it to resonance.
  • Handle the Euler–Cauchy equation with the operator \(\theta=x\,d/dx\) and its commutation law.

Constant coefficients

Let \(D=d/dx\) and \(p(D)=\sum_k a_kD^k\) with constant \(a_k\). Since \(D^j D^k=D^{j+k}\), polynomials in \(D\) multiply like ordinary polynomials, so \(p(D)\) factors as \(a_n\prod_i(D-r_i)\).

Proposition (exponential shift)

For any smooth \(u\) and constant \(a\), \(p(D)\big(e^{ax}u\big)=e^{ax}\,p(D+a)\,u\). In particular \(p(D)e^{ax}=p(a)e^{ax}\).

Proof. \(D(e^{ax}u)=e^{ax}(D+a)u\). Apply it \(k\) times and take linear combinations. ∎

Consequently, if \(p(a)\ne0\), a particular solution of \(p(D)y=e^{ax}\) is \(y=e^{ax}/p(a)\). If \(a\) is a root of multiplicity \(m\), write \(p(D)=(D-a)^m q(D)\) with \(q(a)\ne0\); then \(y=\dfrac{x^m e^{ax}}{m!\,q(a)}\), since \((D-a)^m(x^m e^{ax})=e^{ax}D^m x^m=m!\,e^{ax}\).

Non-commuting operators

The 1844 memoir “On a general method in analysis” treated equations with variable coefficients by finding operators that obey simple exchange laws. The basic example is \(\theta=x\,\dfrac{d}{dx}\) together with multiplication by \(x^m\).

Proposition

\(\theta\,x^m=m\,x^m\), and for any polynomial \(f\), \(f(\theta)\,(x^m u)=x^m f(\theta+m)\,u\). Moreover \(x^nD^n=\theta(\theta-1)\cdots(\theta-n+1)\).

Proof. \(\theta(x^mu)=mx^mu+x^m\theta u=x^m(\theta+m)u\); iterate. For the last identity, induct on \(n\). The exchange law gives \(x^n\theta=(\theta-n)x^n\), so \(x^{n+1}D^{n+1}=x^n(xD)D^n=x^n\theta D^n=(\theta-n)\,x^nD^n\), and the inductive hypothesis finishes the proof. ∎

So the Euler–Cauchy equation \(\sum_k a_k x^kD^ky=0\) becomes a constant-coefficient equation in \(t=\log x\).

Exercise 9.1Solve \(y''-3y'+2y=e^{x}\).

Show solution
\(p(D)=(D-1)(D-2)\). Homogeneous: \(c_1e^x+c_2e^{2x}\). Here \(a=1\) is a simple root with \(q(D)=D-2\), \(q(1)=-1\), so \(y_p=xe^x/(1\cdot(-1))=-xe^x\). Check: \(y_p'=-e^x-xe^x\), \(y_p''=-2e^x-xe^x\); then \(y''-3y'+2y=(-2+3)e^x+(-1+3-2)xe^x=e^x\). General solution \(y=c_1e^x+c_2e^{2x}-xe^x\).

Exercise 9.2Solve \(x^2y''-2xy'+2y=0\) for \(x>0\).

Show solution
\(x^2D^2=\theta(\theta-1)\) and \(xD=\theta\), so the equation is \((\theta^2-3\theta+2)y=(\theta-1)(\theta-2)y=0\). Then \(y=c_1x+c_2x^2\).
Lesson 10

Differences, quadrature, invariants

For the general reader. Boole’s 1860 treatise applied the same symbolic thinking to sums and differences, the discrete analogues of integrals and derivatives. A five-point integration formula from that book still carries his name. Twenty years earlier, his observation that some expressions survive a change of coordinates almost untouched had launched invariant theory.

Goals
  • Use the shift operator \(E=e^{hD}\) and the difference operator \(\Delta=E-1\).
  • Verify Boole’s quadrature rule and its degree of exactness.
  • Prove Boole’s 1857 integral identity and his 1841 invariance of the discriminant.

The calculus of finite differences

For step \(h\), set \(Ef(x)=f(x+h)\) and \(\Delta=E-1\). Taylor’s theorem, written symbolically, is \(E=e^{hD}\). Newton’s forward-difference formula is then the binomial expansion \(f(x+kh)=E^kf(x)=\sum_j\binom kj\Delta^jf(x)\), exact for polynomials.

Boole’s rule (1860)

With \(x_k=x_0+kh\), \(f_k=f(x_k)\), and \(f\) six times continuously differentiable,

\[ \int_{x_0}^{x_4}f(x)\,dx=\frac{2h}{45}\big(7f_0+32f_1+12f_2+32f_3+7f_4\big)-\frac{8}{945}h^7f^{(6)}(\xi) \]

for some \(\xi\in(x_0,x_4)\). The rule is exact for polynomials of degree at most 5.

It is the closed Newton–Cotes rule on five points: integrate the degree-4 interpolating polynomial. A misprint in a widely used handbook led to the name “Bode’s rule”, which still circulates.

Theorem (Boole’s identity, 1857)

For every Lebesgue-integrable \(f:\mathbb R\to\mathbb R\),

\[ \int_{-\infty}^{\infty}f\!\left(x-\frac1x\right)dx=\int_{-\infty}^{\infty}f(x)\,dx. \]

Proof. The map \(x\mapsto u=x-1/x\) is increasing on \((-\infty,0)\) and on \((0,\infty)\), each branch a bijection onto \(\mathbb R\). On the two branches the inverses are \(x_\pm(u)=\tfrac12\big(u\pm\sqrt{u^2+4}\big)\), and \(x_+'(u)+x_-'(u)=1\). Changing variables on each branch and adding gives \(\int f(u)\,(x_+'+x_-')\,du=\int f(u)\,du\). ∎

Invariants (1841)

Take a binary quadratic form \(Q=ax^2+2bxy+cy^2\) and a linear substitution \(x=pX+qY,\ y=rX+sY\) with determinant \(\delta=ps-qr\). Boole showed that the discriminant transforms by a power of \(\delta\), and did the same for binary cubics. He also found, independently of Cayley, an invariant of the binary quartic \(ax^4+4bx^3y+6cx^2y^2+4dxy^3+ey^4\), namely \(ace+2bcd-ad^2-eb^2-c^3\), the determinant of the matrix \(\begin{pmatrix}a&b&c\\ b&c&d\\ c&d&e\end{pmatrix}\).

Proposition (Boole, 1841)

If \(Q\) becomes \(a'X^2+2b'XY+c'Y^2\), then \(b'^2-a'c'=\delta^2\,(b^2-ac)\).

Proof. Write \(Q=v^{\mathsf T}Mv\) with \(M=\begin{pmatrix}a&b\\ b&c\end{pmatrix}\) and \(v=Pw\), \(P=\begin{pmatrix}p&q\\ r&s\end{pmatrix}\). The new matrix is \(P^{\mathsf T}MP\), whose determinant is \(\delta^2\det M\), and \(\det M=ac-b^2\). ∎

Exercise 10.1Check that Boole’s rule is exact for \(f(x)=x^4\) on \([0,4]\) with \(h=1\), and show that it is also exact for \(x^5\) on \([-2,2]\).

Show solution
\(\int_0^4x^4dx=1024/5=204.8\). The rule gives \(\tfrac{2}{45}(0+32+12\cdot16+32\cdot81+7\cdot256)=\tfrac{2}{45}\cdot4608=204.8\). On \([-2,2]\) the integrand \(x^5\) is odd, and the rule’s weights are symmetric about 0 while the nodes are \(-2,-1,0,1,2\), so both sides are 0. By symmetry the rule is exact for all odd functions about the midpoint, hence for degree 5.

Exercise 10.2Use Boole’s identity to compute \(\displaystyle\int_{-\infty}^{\infty}\exp\!\Big(-\big(x-\tfrac1x\big)^2\Big)dx\) and \(\displaystyle\int_{-\infty}^{\infty}\frac{dx}{1+(x-1/x)^2}\).

Show solution
They equal \(\int e^{-x^2}dx=\sqrt\pi\) and \(\int\frac{dx}{1+x^2}=\pi\). Multiplying out, the first is \(e^{2}\int e^{-x^2-1/x^2}dx\), so \(\int_{-\infty}^\infty e^{-x^2-1/x^2}dx=\sqrt\pi\,e^{-2}\).

Exercise 10.3Show that \(\Delta^n x^n=n!\,h^n\) and \(\Delta^{n+1}x^n=0\).

Show solution
\(\Delta\) lowers the degree of a polynomial by one and multiplies the leading coefficient by \(nh\): \((x+h)^n-x^n=nhx^{n-1}+\cdots\). Applying it \(n\) times to \(x^n\) gives the constant \(n!\,h^n\); once more gives 0.
Part III · Life and legacy
Lesson 11

Cork, marriage and death, 1849–1864

Boole arrived in Cork in November 1849 as one of the founding professors of a new and contested institution. The Queen’s Colleges were non-denominational, and the Catholic hierarchy condemned them. Boole, a deeply religious man of undogmatic views, wrote to De Morgan in October 1850 of the religious bigotry around him and asked about positions in England. He stayed. In May 1851 he was elected Dean of Science.

In Cork he met Mary Everest, niece of Sir George Everest, the surveyor whose name was given to the mountain. She visited her uncle John Ryall, the college’s professor of Greek, in 1850. They met again at Wickwar in 1852, when she was 20 and he 37, and they corresponded about mathematics. They married at Wickwar on 11 September 1855. Mary Everest Boole later wrote books on the psychology of learning mathematics.

The books

In 1854 he published An Investigation of the Laws of Thought, which he regarded as his most valuable contribution. De Morgan praised it warmly; most philosophers ignored it. He followed it with two textbooks, A Treatise on Differential Equations (1859) and A Treatise on the Calculus of Finite Differences (1860), clear and influential works that went through several editions. His honours included the Keith Medal of the Royal Society of Edinburgh, election as a Fellow of the Royal Society in 1857, and honorary degrees from the universities of Dublin and Oxford.

Death

The Booles had five daughters: Mary Ellen (1856), Margaret (1858), Alicia (1860), Lucy Everest (1862) and Ethel Lilian (1864). In late November 1864 Boole walked from his home at Ballintemple to the college in heavy rain and lectured in his wet clothes. He developed a feverish cold that settled on his lungs and died on 8 December 1864, aged 49. He is buried in the churchyard of St Michael’s Church of Ireland, Blackrock, Cork.

DiscussionBoole’s career was built on correspondence and a journal open to provincial contributors. What does his path suggest about how mathematical talent was found and recognized in early Victorian Britain, and who was still excluded?

Lesson 12

Legacy: from logic to logic gates

Boole’s logic had few readers in his lifetime. Its growth came after his death, through Jevons in England, Peirce in the United States and Schröder in Germany. Peirce stripped away the arithmetic scaffolding and named the result after Boole. Gottlob Frege and Bertrand Russell took logic in another direction, toward quantifiers and the foundations of mathematics, but the propositional core of every modern logic is Boolean. Marshall Stone’s representation theorem of 1936 made Boolean algebras central objects in algebra and topology, and Boolean-valued models became a tool in set theory in the 1960s.

The engineering legacy began with Claude Shannon’s 1937 thesis and Shestakov’s parallel work. Every logic gate, every search query with AND and OR, and the bool type of most programming languages descend from it.

The Boole family

  • Alicia Boole Stott (1860–1940) developed a remarkable intuition for four-dimensional geometry, computed the three-dimensional sections of the regular 4-polytopes, and collaborated with Pieter Hendrik Schoute. She introduced the word “polytope” into English.
  • Lucy Everest Boole (1862–1904) became a pharmaceutical chemist and taught chemistry at the London School of Medicine for Women.
  • Ethel Lilian Voynich (1864–1960) wrote the novel The Gadfly.
  • Mary Ellen married the mathematician Charles Howard Hinton, who wrote on the fourth dimension. Margaret was the mother of the physicist Geoffrey Ingram Taylor.

Honours

A lunar crater bears Boole’s name. University College Cork, the successor of Queen’s College, holds his papers and celebrated his bicentenary in 2015 with the George Boole 200 programme.

Final projectTake one argument from The Laws of Thought (or Exercise 5.1), solve it with Boole’s method, then solve it again in a modern Boolean ring over \(\mathbb F_2\). Write two pages comparing the two computations and explaining where Boole’s partial addition helps or hinders.

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, George Boole, MacTutor History of Mathematics.
  2. George Boole, Wikipedia.
  3. D. MacHale, The Life and Work of George Boole: A Prelude to the Digital Age, Cork University Press, 2014.
  4. University College Cork, George Boole 200.

Mathematics

  1. G. Boole, Exposition of a general theory of linear transformations, Cambridge Mathematical Journal 3 (1841–42).
  2. G. Boole, On a general method in analysis, Phil. Trans. R. Soc. 134 (1844).
  3. G. Boole, The Mathematical Analysis of Logic, Cambridge, 1847 (Project Gutenberg).
  4. G. Boole, An Investigation of the Laws of Thought, London, 1854 (Project Gutenberg).
  5. G. Boole, A Treatise on Differential Equations, 1859; A Treatise on the Calculus of Finite Differences, 1860.
  6. S. Burris, George Boole, Stanford Encyclopedia of Philosophy.
  7. T. Hailperin, Boole’s Logic and Probability, North-Holland, 2nd ed. 1986.
  8. M. H. Stone, The theory of representations for Boolean algebras, Trans. AMS 40 (1936).
  9. C. E. Shannon, A symbolic analysis of relay and switching circuits, Trans. AIEE 57 (1938).
  10. I. Pitowsky, George Boole’s “conditions of possible experience” and the quantum puzzle, Brit. J. Phil. Sci. 45 (1994).
  11. L. Corry, The origins of invariant theory (Encyclopaedia Italiana, Storia).
  12. Boole’s rule, Wikipedia.