Decimal expansions and chapter summary

We close the chapter by looking at rational numbers from one more angle , their decimal expansions , and then we tie everything together with a one-page revision.

Decimal expansions of rationals

Every rational number $\dfrac{p}{q}$ (in lowest terms) has a decimal expansion that is either terminating (it ends, like $0.25$) or non-terminating but recurring (a block of digits repeats forever, like $0.\overline{3}$).

A neat criterion using prime factorisation: a rational $\dfrac{p}{q}$ in lowest terms has a terminating decimal expansion if and only if the prime factorisation of $q$ is of the form $2^m \cdot 5^n$, where $m, n$ are non-negative integers.

Why? A decimal terminates after $k$ places exactly when we can multiply $p/q$ by $10^k$ to get an integer. So we need $q \mid 10^k = 2^k \cdot 5^k$, which means $q$ has no prime factor other than $2$ and $5$.

A few quick illustrations.

• $\dfrac{3}{8}$: $8 = 2^3$. Terminating. Indeed $3/8 = 0.375$.
• $\dfrac{7}{20}$: $20 = 2^2 \times 5$. Terminating. $7/20 = 0.35$.
• $\dfrac{1}{3}$: $3$ is neither $2$ nor $5$. Non-terminating recurring. $1/3 = 0.\overline{3}$.
• $\dfrac{2}{15}$: $15 = 3 \times 5$ has a $3$. Non-terminating recurring. $2/15 = 0.1\overline{3}$.

In short: look at the denominator after reducing to lowest terms. Only $2$s and $5$s in its factorisation? Terminates. Anything else? Recurs.

Theorem: classification

Theorem. Let $x = p/q$ be a rational number in lowest terms ($\gcd(p, q) = 1$). Then $x$ has a terminating decimal expansion if and only if $q = 2^m 5^n$ for some non-negative integers $m, n$.

Equivalently:

• $x$ has a terminating decimal iff the only prime factors of the denominator are $2$ and $5$.
• $x$ has a non-terminating recurring decimal iff the denominator has at least one prime factor other than $2$ or $5$.

Combining with what we already know about irrational numbers: an irrational number has a non-terminating, non-recurring decimal expansion (like $\sqrt{2} = 1.4142135\ldots$ or $\pi = 3.14159265\ldots$). The three types , terminating, non-terminating recurring, non-terminating non-recurring , partition the real numbers.

Chapter summary

The whole chapter sits on a tripod:

1. Fundamental Theorem of Arithmetic , every $n \ge 2$ factorises into primes uniquely.
2. HCF / LCM via prime factorisation , compare exponents prime by prime; min for HCF, max for LCM; and $\operatorname{HCF}(a,b) \cdot \operatorname{LCM}(a,b) = a \cdot b$.
3. Proof by contradiction , assume the opposite, derive a contradiction, conclude.

From this small core we extracted:

• Counting divisors of $n$.
• Solving HCF/LCM word problems (signals, bells, tracks, packing).
• Proving $\sqrt{p}$ irrational for any prime $p$, hence many derived surds irrational.
• Classifying decimal expansions by the prime factorisation of the denominator.

Worked examples

Example 1. Without doing the long division, decide whether $\dfrac{17}{8}$ has a terminating decimal.

$8 = 2^3$. Only the prime $2$ in the denominator. Terminating. (In fact $17/8 = 2.125$.)

Example 2. Decide whether $\dfrac{77}{210}$ has a terminating decimal.

First reduce: $\gcd(77, 210) = 7$, so $77/210 = 11/30$. Now $30 = 2 \times 3 \times 5$ has a $3$. Non-terminating recurring.

Example 3. Find the smallest natural number $k$ such that $\dfrac{k}{225}$ has a terminating decimal expansion.

$225 = 3^2 \times 5^2$. For the fraction to terminate, the $3^2$ in the denominator must cancel. So $k$ must be a multiple of $9$. Smallest positive such $k = 9$. Indeed $9/225 = 1/25 = 0.04$.

Example 4. Determine whether $\dfrac{13}{3125}$ is terminating; if it is, find the decimal expansion.

$3125 = 5^5$. So terminating. Multiply numerator and denominator by $2^5 = 32$: $\frac{13}{3125} = \frac{13 \times 32}{3125 \times 32} = \frac{416}{100000} = 0.00416.$

Example 5. Classify $\dfrac{29}{343}$.

$343 = 7^3$. The denominator has $7$, not just $2$s and $5$s. Non-terminating recurring.

Try it yourself

1. Without dividing, decide whether each terminates: $\dfrac{13}{3125}$, $\dfrac{17}{8}$, $\dfrac{64}{455}$, $\dfrac{15}{1600}$, $\dfrac{29}{343}$, $\dfrac{23}{2^3 \cdot 5^2}$.
2. Find the smallest positive integer $k$ such that $\dfrac{k}{50}$ has a terminating decimal , and the smallest $k$ such that $\dfrac{k}{42}$ does.
3. Express $\dfrac{15}{1600}$ as a decimal.
4. Show that $\dfrac{6}{15}$ has a terminating decimal even though $15$ has a factor of $3$.
5. Without dividing, find the decimal expansion of $\dfrac{43}{2^4 \cdot 5^3}$.
6. Two numbers have HCF $1$ and LCM $90$. Find all possible such pairs (positive integers).
7. State whether the following are rational or irrational, with reasons: $\sqrt{49}$, $\sqrt{50}$, $0.3030030003\ldots$ (a non-recurring pattern), $0.\overline{142857}$.
8. Find HCF$(96, 404)$ using prime factorisation, then state the LCM.
9. Show that $\sqrt{2}$ and $1 + \sqrt{2}$ cannot both be roots of $\dfrac{p}{q}$ form simultaneously.
10. The HCF of $65$ and $117$ is expressible in the form $65m - 117$. Find $m$. (Bonus: any linear combination representation.)

Pitfalls / Insight

• Reduce to lowest terms first before checking the denominator. $\dfrac{6}{15}$ looks non-terminating, but $\dfrac{6}{15} = \dfrac{2}{5}$ , terminating.
• A non-recurring decimal that goes on forever is irrational. Patterns like $0.1010010001\ldots$ are not recurring.
• Don't confuse recurring with terminating. $0.\overline{3}$ goes forever; $0.333$ is just an approximation.

Insight. Chapter 1 rewards a small set of habits: prime-factorise first, write exponents in columns, label each problem as HCF or LCM, and reach for proof by contradiction when "irrational" appears in the statement. Get those habits and the chapter is essentially free marks.