Binary operations and their propertiesAQA A-Level Further Maths: Revision notes

Section 1

Binary operations and closure

A binary operation $\ast$ on a set $S$ combines any two elements $a$ and $b$ of $S$ to give a result $a\ast b$ . It is closed on $S$ if $a\ast b$ is always in $S$ .

Examples: addition on the integers, multiplication of real numbers, $a\ast b=2a+3b$ on the reals, and multiplication of $2\times2$ matrices.
To show a set is closed, show for general $a,b\in S$ that $a\ast b\in S$ . To show it is not closed, give one pair whose result is outside $S$ (for example, subtraction on the natural numbers: $2-5=-3$ ).
Order matters: $a\ast b$ means $a$ first, then $b$ .

Example: for $a\ast b=a+b-ab$ , $3\ast(-1)=3+(-1)-(3)(-1)=2+3=5$ .

Key termsbinary operationclosed

Common mistake

Substituting into the wrong places. Write $a\ast b$ with $a$ and $b$ clearly replaced before you simplify.

Exam tip

Bracket negative numbers when substituting, for example $a+b-ab$ with $b=-1$ .

Section 2

Modular arithmetic and matrices as operations

Modular arithmetic. On $\{0,1,\ldots,n-1\}$ , addition modulo $n$ ( $\oplus$ ) and multiplication modulo $n$ ( $\otimes$ ) are binary operations: add or multiply as usual, then take the remainder on division by $n$ . For example, $4\otimes5=20\equiv2\pmod6$ and $4\oplus5=9\equiv3\pmod6$ .

Matrix multiplication. Multiplication of $2\times2$ matrices is a binary operation on the set of all $2\times2$ matrices. For a smaller set, check closure by multiplying general members. For matrices of the form $\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}$ , the product of two of them is $\begin{pmatrix} 1 & a+b \\ 0 & 1 \end{pmatrix}$ , which is of the same form, so the set is closed.

Key termsmodular arithmeticremainder

Common mistake

Forgetting to reduce modulo $n$ , giving an answer outside the set.

Exam tip

Remember $x\otimes y$ is the remainder of $xy$ , so an answer such as $12$ in the set modulo 6 becomes $0$ .

Section 3

Commutativity

An operation $\ast$ is commutative if $a\ast b=b\ast a$ for all $a,b$ in the set.

To prove it, take general $a$ and $b$ and show the two expressions are equal, using known laws. Example: $a+b-ab=b+a-ba$ because addition and multiplication of reals are commutative.
To disprove it, one counter-example is enough. For $a\ast b=2a+3b$ : $1\ast2=8$ but $2\ast1=7$ .
Matrix multiplication is not commutative in general: $AB\neq BA$ for most pairs, so a counter-example with two chosen matrices disproves it. It may still be commutative on a special set, such as matrices $\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}$ with product $\begin{pmatrix} 1 & a+b \\ 0 & 1 \end{pmatrix}$ .

Key termscommutativecounter-example

Common mistake

Testing a few numbers and concluding the operation is commutative. A proof needs general $a$ and $b$ .

Exam tip

Use unequal values, such as 1 and 2, for a counter-example, since equal values never show non-commutativity.

Section 4

Associativity

An operation $\ast$ is associative if $(a\ast b)\ast c=a\ast(b\ast c)$ for all $a,b,c$ .

To prove it, expand both sides separately and show they simplify to the same expression. For $a\ast b=a+b-ab$ : $(a\ast b)\ast c=a+b+c-ab-ac-bc+abc=a\ast(b\ast c)$ .
To disprove it, give one counter-example. For $a\ast b=2a+3b$ : $(1\ast1)\ast1=13$ but $1\ast(1\ast1)=17$ .
Addition and multiplication of integers, modular addition and multiplication, and matrix multiplication are all associative.

Associativity means that a chain such as $a\ast b\ast c$ is unambiguous without brackets.

Key termsassociative

Common mistake

Working out only one of $(a\ast b)\ast c$ and $a\ast(b\ast c)$ . You must expand both.

Exam tip

Expand each side to the same standard form, such as a sum of terms in alphabetical order, so they are easy to compare.

Section 5

Cayley tables

A Cayley table lists every product $a\circ b$ of a finite set: the element on the left (row) is the first element and the element along the top (column) is the second.

For $a\circ b=ab\bmod5$ on $\{1,2,3,4\}$ : $\begin{array}{c|cccc} \circ & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 4 & 1 & 3 \\ 3 & 3 & 1 & 4 & 2 \\ 4 & 4 & 3 & 2 & 1 \end{array}$

Closed: every entry in the table is an element of the set.
Commutative: the table is symmetric about the leading diagonal (top left to bottom right).
Associativity cannot be read from the table easily, so it is proved separately, or known from the operation.

To construct a table, work out each entry carefully using the rule, row by row, and check that each entry is in the set.

Key termsCayley tableleading diagonal

Common mistake

Reading $a\circ b$ with the column first. The row element is the first element, which matters for operations that are not commutative.

Exam tip

Check symmetry in pairs: compare each entry with its mirror image across the leading diagonal.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Binary operations and their properties

The binary operation $\ast$ is defined on the set of real numbers by $a\ast b=2a+3b$ .
Show that $\ast$ is not associative.2 marks
The binary operation $\otimes$ is defined on the set $\{0,1,2,3,4,5\}$ by $x\otimes y=$ the remainder when $xy$ is divided by 6, that is multiplication modulo 6.
Prove that $\otimes$ is commutative.2 marks
The binary operation $\circ$ is defined on the set $S=\{1,2,3,4\}$ by $a\circ b=$ the remainder when $ab$ is divided by 5.
Construct the Cayley table for $\circ$ on $S$ , with the first element of $a\circ b$ given by the row.3 marks

See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).