Properties of rational numbers

A set of numbers is closed under an operation if combining any two members always produces another member of the same set. Rational numbers are closed under addition, subtraction and multiplication — and under division too, provided the divisor is not $0$.

Order and grouping. For addition and multiplication, order does not matter (commutativity) and grouping does not matter (associativity):

where: $a$, $b$, $c$ are any rational numbers. Subtraction and division satisfy neither: $a-b\neq b-a$ in general, and $(a-b)-c\neq a-(b-c)$ in general.

Identities and inverses. Zero and one are the "do-nothing" numbers for addition and multiplication respectively, and every rational number has a partner that undoes the operation:

where: $-a$ is the additive inverse of $a$, and $\dfrac1a$ is the multiplicative inverse (reciprocal) of $a$, defined only when $a\neq0$, since $\dfrac10$ is undefined. There is no equivalent "identity" for subtraction or division as separate operations — they are built from addition and multiplication in disguise.

Distributive property. Multiplication spreads across a sum: $a\times(b+c)=a\times b+a\times c$. This is the rule that justifies factoring a common term out of two products, and it is the only property that links two different operations rather than describing one operation alone. Read left to right it "expands" a bracket; read right to left it "factors" a common term back out — both directions are the same identity, and exam questions use both, so it pays to recognise the pattern either way round. Closure is worth testing against smaller number systems to see why rational numbers were needed. Natural numbers are not closed under subtraction ($3-5$ is not a natural number), and integers are not closed under division ($3\div2$ is not an integer). Rational numbers repair both gaps — every subtraction of integers and every division by a nonzero rational number stays inside the rational numbers. This is one of the main reasons rational numbers are introduced as their own system rather than treated as a leftover corner of integers: they are the smallest familiar system in which addition, subtraction, multiplication and (nonzero) division can all be carried out without ever leaving the set.

Why subtraction and division break the pattern. A quick numeric check makes the failure concrete rather than abstract. Take $a=5$ and $b=3$: $a-b=2$ but $b-a=-2$, so $a-b\neq b-a$ — order matters for subtraction even though it never mattered for addition. The same happens for grouping: $(8-4)-2=2$ but $8-(4-2)=6$, so subtraction is not associative either. Division behaves the same way — $6\div2=3$ but $2\div6=\dfrac13$, and $(8\div4)\div2=1$ while $8\div(4\div2)=4$. The reason traces back to the identity $a-b=a+(-b)$: subtraction is addition, but with the second number flipped to its negative. Swapping $a$ and $b$ therefore does not just swap two equal-role numbers as it does in $a+b=b+a$ — it changes which one gets flipped, so the results differ.

Use the distributive property to simplify $-\dfrac{3}{5}\times\dfrac{7}{9}+\left(-\dfrac{3}{5}\right)\times\dfrac{2}{9}$, then state the additive and multiplicative inverse of the result.

Step 1. Spot the common factor $-\dfrac35$ in both terms and rewrite using $a\times b+a\times c=a\times(b+c)$: $-\dfrac35\times\dfrac79+\left(-\dfrac35\right)\times\dfrac29=-\dfrac35\times\left(\dfrac79+\dfrac29\right)$.

Step 2. Add inside the bracket: $\dfrac79+\dfrac29=\dfrac99=1$.

Step 3. Multiply: $-\dfrac35\times1=-\dfrac35$. The whole expression simplifies to $-\dfrac35$, matching the verification above.

Step 4. Find the inverses of $-\dfrac35$. The additive inverse is $\dfrac35$, since $-\dfrac35+\dfrac35=0$. The multiplicative inverse is $-\dfrac53$, since $-\dfrac35\times\left(-\dfrac53\right)=1$.

The simplified value is $-\dfrac35$, with additive inverse $\dfrac35$ and multiplicative inverse $-\dfrac53$.

Competitive stretch

Verify associativity of addition for $a=\dfrac12$, $b=-\dfrac23$, $c=\dfrac56$, and then check that the same three numbers do not associate under subtraction, to see both properties side by side.

Addition, left side: $a+b=\dfrac12+\left(-\dfrac23\right)=\dfrac36-\dfrac46=-\dfrac16$, so $(a+b)+c=-\dfrac16+\dfrac56=\dfrac46=\dfrac23$.

Addition, right side: $b+c=-\dfrac23+\dfrac56=-\dfrac46+\dfrac56=\dfrac16$, so $a+(b+c)=\dfrac12+\dfrac16=\dfrac36+\dfrac16=\dfrac46=\dfrac23$. Both sides give $\dfrac23$, so $(a+b)+c=a+(b+c)$ holds, exactly as the property guarantees it must for any three rationals.

Subtraction, for contrast: $(a-b)-c=\left(\dfrac12+\dfrac23\right)-\dfrac56=\dfrac76-\dfrac56=\dfrac26=\dfrac13$, whereas $a-(b-c)=\dfrac12-\left(-\dfrac23-\dfrac56\right)=\dfrac12+\dfrac32=2$. Since $\dfrac13\neq2$, subtraction fails to associate for these same three numbers, confirming with real values what the general argument above predicts.

• Assuming subtraction or division is commutative or associative "because addition and multiplication are" — always test with a quick numeric example such as $5-3\neq3-5$ first.
• Writing $\dfrac10$ as if it were a valid rational number; zero has no multiplicative inverse because nothing times $0$ gives $1$.
• Mixing up the additive identity ($0$, paired with $+$) and the multiplicative identity ($1$, paired with $\times$) in a problem that uses both operations.
• Distributing incorrectly, e.g. $a\times(b+c)=ab+c$ instead of $ab+ac$ — multiplication must reach every term inside the bracket.
• Treating "verify" questions as optional working — a verification answer needs both sides computed independently and compared, not just one side stated as if it were obviously equal to the other.
• Forgetting to state the restriction $b\neq0$ when claiming closure under division, or when writing a reciprocal $\dfrac1b$ — an unqualified claim of closure for division loses marks even if the arithmetic that follows is correct.

Ask the tutor: want to see how these same properties look when you switch from addition to division, or try them with your own numbers?