XS1, XS2, Xa1, Xa2 - Sets, group axioms and examples | OCR A-Level Further Mathematics B (MEI)

A group records a collection of objects together with a way to combine them. XS1 and XS2 make the set and number language precise; Xa1 gives the four group axioms, and Xa2 tests them in arithmetic, complex numbers, matrices and geometric symmetries.

Specify the set before the operation

If multiplication works perfectly for every non-zero rational number, why can adding just one more number destroy the group? The newcomer is zero. To see what goes wrong, we need to specify exactly which objects are allowed.

You already know

Rational numbers are ratios of integers with non-zero denominator. Real numbers include rational and irrational numbers; complex numbers include the reals. Matrix products and compositions act from right to left, and multiplication by a unit complex number rotates the Argand plane.

A set lists objects without counting repetitions or imposing an order: $\{1,2,2\}=\{2,1\}$ . Membership concerns one object; containment concerns a whole set.

Notation	Meaning
$x\in A$ , $x\notin A$	$x$ is, or is not, an element of $A$ .
$A\subseteq B$	Every element of $A$ is in $B$ ; equality is allowed.
$A\subset B$	$A$ is a proper subset: $A\subseteq B$ and $A\ne B$ .
$A\nsubseteq B$	Some element of $A$ is not in $B$ .
$\varnothing$	The empty set, containing no elements.
$A\cup B$ , $A\cap B$	Elements in at least one set, or in both sets.
$A'=U\setminus A$	The complement of $A$ in a stated universal set $U$ .
$A\setminus B$	Elements in $A$ but not in $B$ .
$n(A)=	A

Every set satisfies $A\subseteq A$ and $\varnothing\subseteq A$ . If $A\ne\varnothing$ , then $\varnothing\subset A$ . In particular, $\varnothing$ and $\{\varnothing\}$ differ: the latter contains one object, the empty set.

Worked example 1: overlapping finite sets

Question. Let $U=\{0,1,2,3,4,5,6\}$ , $A=\{0,2,4,6\}$ and $B=\{2,3,5\}$ . Find $A\cap B,A\cup B,A\setminus B,A'$ , and decide whether $B\subseteq A$ .

Plan. Test membership against the two lists, using $U$ only for the complement.

\begin{aligned} A\cap B&=\{2\} &&\text{in both}\\ A\cup B&=\{0,2,3,4,5,6\} &&\text{in either; no repetitions}\\ A\setminus B&=\{0,4,6\} &&\text{in }A\text{ only}\\ A'&=\{1,3,5\} &&\text{in }U\text{ but not }A\\ 3&\in B,\quad3\notin A &&\text{one counterexample to containment} \end{aligned}

Why this step? A complement depends on the universal set; a difference such as $A\setminus B$ depends only on the two named sets.

Answer. The four sets are displayed above; $B\nsubseteq A$ .

Check. $A\cap A'=\varnothing$ and $A\cup A'=U$ ; their sizes are $4+3=7$ .

Change only $B$ to $\{2,4\}$ . Its relationship with $A$ changes from overlapping to contained.

Worked example 2: a proper subset

Question. With the same $U,A$ and the new $B$ , find $A\cap B,A\cup B,A\setminus B$ and $B\setminus A$ . State the subset relationship and $n(B)$ .

Plan. Use the fact that every element of the new $B$ is already in $A$ .

\begin{aligned} A\cap B&=\{2,4\}=B &&\text{both elements are shared}\\ A\cup B&=A &&\text{no new elements}\\ A\setminus B&=\{0,6\},\quad B\setminus A=\varnothing &&\text{difference is directional}\\ B&\subset A,\qquad n(B)=2 &&\text{containment is strict} \end{aligned}

Why this step? $B\subset A$ also requires $B\ne A$ ; here $0\in A\setminus B$ establishes that difference.

Answer. $A\cap B=B$ , $A\cup B=A$ , differences $\{0,6\}$ and $\varnothing$ ; $B$ is a proper subset of $A$ , with two elements.

Check. $n(A\setminus B)+n(B)=2+2=4=n(A)$ .

Set-builder notation gives a rule rather than a finite list. The colon means “such that”. These standard sets will recur throughout the lesson.

Set	Meaning
$\mathbb N=\{1,2,3,\ldots\}$ , $\mathbb N_0=\{0,1,2,\ldots\}$	Positive integers; non-negative integers.
$\mathbb Z=\{\ldots,-2,-1,0,1,2,\ldots\}$	All integers.
$\mathbb Q=\{a/b:a,b\in\mathbb Z,\ b\ne0\}$	Rational numbers.
$\mathbb R$ , $\mathbb C$	Real numbers; complex numbers.
$\mathbb Q^+=\{q\in\mathbb Q:q>0\}$	Positive rational numbers.
$\mathbb R_0^+=\{x\in\mathbb R:x\ge0\}$	Non-negative real numbers.

Thus $\mathbb N\subset\mathbb Z\subset\mathbb Q\subset\mathbb R\subset\mathbb C$ ; irrational numbers form $\mathbb R\setminus\mathbb Q$ . This notation is not printed in the formula booklet.

Watch out

Wrong: $B\in A$ for $B=\{2,4\}$ and $A=\{0,2,4,6\}$ .

Why: the elements of $A$ are numbers; the set $B$ is not one of them.

Right: $B\subset A$ . Every element of $B$ belongs to $A$ , and the sets differ.

Use membership to compare an object with a set, and containment to compare two sets. The next check asks for the resulting sets themselves.

Reason about rational and irrational results

The sum, difference and product of two rational numbers are rational. Their quotient is rational when the divisor is non-zero. This follows by writing both as fractions and combining integer numerators and denominators.

There is no corresponding rule that two irrational inputs must give an irrational output. Cancellation can change the type of number. Use an argument or an exact example rather than a decimal approximation.

Worked example 3: rational plus irrational

Question. Let $r\in\mathbb Q$ and $x\in\mathbb R\setminus\mathbb Q$ . Prove that $r+x$ is irrational. Explain what happens to $rx$ when $r\ne0$ .

Plan. Suppose the result were rational and rearrange to make $x$ rational.

\begin{aligned} r+x=q\in\mathbb Q&\ \Longrightarrow\ x=q-r\in\mathbb Q &&\text{contradiction}\\ rx=q\in\mathbb Q,\quad r\ne0&\ \Longrightarrow\ x=q/r\in\mathbb Q &&\text{contradiction} \end{aligned}

Why this step? Rational numbers stay rational under the subtraction and non-zero division used to isolate $x$ .

Answer. Both $r+x$ and $rx$ , for $r\ne0$ , are irrational. If $r=0$ , the product is the rational number zero.

Check. $3+\sqrt2$ and $3\sqrt2$ fit the result, whereas $0\sqrt2=0$ explains the exception.

Change the rational input to another irrational number. Now neither a rational nor an irrational result is forced.

Worked example 4: two irrational inputs

Question. Give examples showing that the sum of two irrational numbers may be rational or irrational. Do the same for their product.

Plan. Use the known irrational number $\sqrt2$ , arranging cancellation in one case and a non-zero rational multiple in another.

\begin{aligned} \sqrt2+(-\sqrt2)&=0 &&\text{rational sum}\\ \sqrt2+\sqrt2&=2\sqrt2 &&\text{irrational sum by the preceding result}\\ \sqrt2\,\sqrt2&=2 &&\text{rational product}\\ \sqrt2(1+\sqrt2)&=\sqrt2+2 &&\text{irrational product} \end{aligned}

Why this step? $1+\sqrt2$ is irrational by the first example; both inputs in the last product really are irrational.

Answer. Each outcome is possible, with the exact examples shown.

Check. No assertion depends on rounded digits: the conclusions follow from exact identities and the irrationality of $\sqrt2$ .

These number properties are not printed in the formula booklet. They help with closure: one counterexample disproves a universal claim that an operation always stays in a proposed set.

Watch out

Wrong: $\sqrt2+(-\sqrt2)\notin\mathbb Q$ , because both inputs are irrational.

Why: irrational inputs can cancel; their types alone do not determine the sum's type.

Right: $\sqrt2+(-\sqrt2)=0\in\mathbb Q$ . Simplify exactly before classifying the result.

For a closure claim, choose inputs from the proposed set and check whether their exact result stays there.

Separate the four group axioms

A binary operation combines an ordered pair of objects into one result. In a group it maps $G\times G$ into $G$ , so the result stays in the set. The pair $(G,\circ)$ specifies both the objects and the operation; changing either can change the answer.

Group

A non-empty set $G$ with an operation $\circ$ is a group when all four conditions hold:

Axiom	Required statement
Closure	$a\circ b\in G$ for every $a,b\in G$ .
Associativity	$(a\circ b)\circ c=a\circ(b\circ c)$ for every $a,b,c\in G$ .
Identity	There is $e\in G$ with $e\circ a=a\circ e=a$ for every $a\in G$ .
Inverses	For every $a\in G$ , there is $a^{-1}\in G$ with $a\circ a^{-1}=a^{-1}\circ a=e$ .

The group is abelian if it is also commutative: $a\circ b=b\circ a$ for all $a,b\in G$ .

Associativity moves brackets without changing the order of the inputs. Commutativity exchanges inputs. A group need not be abelian. To prove an axiom, use arbitrary elements; one numerical example can disprove an axiom but cannot prove it for every input.

An inverse undoes the stated operation. Under addition the identity is $0$ and the inverse of $a$ is $-a$ ; under ordinary multiplication the identity is $1$ and a non-zero number's inverse is $1/a$ . Do not transfer either rule blindly to a newly defined operation.

For example, $(\mathbb Z,+)$ is an infinite abelian group: sums of integers are integers, addition is associative, zero is the identity, and $-a\in\mathbb Z$ for every integer $a$ . A group can equally have finitely many elements. The axioms and this vocabulary are not supplied in the formula booklet.

Test a newly defined operation

Use the operation's definition every time its symbol appears. For a general associativity proof, expand the two bracketings separately, then state their equality and the conclusion. Determine an identity from its defining equation before solving for inverses.

Worked example 5: an unfamiliar identity

Question. On $G=\mathbb Q\setminus\{0\}$ , define $a\circ b=2ab$ . Show that $(G,\circ)$ is an abelian group.

Plan. Verify closure and associativity, solve for the identity, then find each element's inverse under this operation.

\begin{aligned} a,b\in G&\ \Longrightarrow\ 2ab\in\mathbb Q\setminus\{0\} &&\text{closure}\\ (a\circ b)\circ c&=2(2ab)c=4abc &&\text{first bracketing}\\ a\circ(b\circ c)&=2a(2bc)=4abc &&\text{second bracketing}\\ 2ae=a\quad(a\ne0)&\ \Longrightarrow\ e=\tfrac12 &&\text{identity equation}\\ 2ab=\tfrac12&\ \Longrightarrow\ b=\frac1{4a}\in G &&\text{inverse equation} \end{aligned}

The two bracketings are equal for all $a,b,c\in G$ , so $\circ$ is associative. Also $2ab=2ba$ , so the operation is commutative. The identity and inverse equations hold in both directions.

Why this step? The inverse must give $e=1/2$ , not the identity of ordinary multiplication.

Answer. All four axioms hold, and the group is abelian; $e=1/2$ and $a^{-1}=1/(4a)$ under $\circ$ .

Check. The inverse of $2$ is $1/8$ , since $2\circ(1/8)=2(2)(1/8)=1/2$ .

Change only the set by allowing zero: take $G=\mathbb Q$ . This is the opening newcomer.

Worked example 6: one extra element breaks an axiom

Question. Determine whether $(\mathbb Q,\circ)$ , with the same rule $a\circ b=2ab$ , is a group.

Plan. Retain the valid closure, associativity and identity arguments, then test the new element against the inverse requirement.

\begin{aligned} 2ab&\in\mathbb Q &&\text{closure still holds}\\ (a\circ b)\circ c&=4abc=a\circ(b\circ c) &&\text{associativity still holds}\\ a\circ\tfrac12&=a=\tfrac12\circ a &&\text{identity is still }\tfrac12\\ 0\circ b&=2(0)b=0\ne\tfrac12 &&\text{for every }b\in\mathbb Q \end{aligned}

Why this step? The group axiom requires an inverse for every element; zero is one specific element for which no candidate can work.

Answer. It is not a group: zero has no inverse under $\circ$ .

Check. Non-zero elements still have inverses. That does not repair the failed universal requirement.

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XS1, XS2, Xa1, Xa2 - Sets, group axioms and examples

Specify the set before the operation

Reason about rational and irrational results

Separate the four group axioms

Test a newly defined operation

Continue this lesson.

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