Rational and irrational numbers

Every number you have ever written down — a page number, a cricket score, $\pi$ in a circle formula, the diagonal of a unit square — belongs somewhere inside one giant nested system of number sets. Class 9 arithmetic draws the first sharp line inside that system: numbers that can be written exactly as a fraction of integers, called rational, versus numbers whose decimal expansion runs forever without ever settling into a repeating block, called irrational. The split sounds abstract, but it is entirely decidable from a number's decimal expansion, and it decides real exam questions: is $\sqrt{16}$ rational, is $2.\overline{7}$ rational, can you always squeeze a rational and an irrational number between $2$ and $3$ ? This section builds the classification rule, proves that recurring decimals are always fractions in disguise, and drills the traps examiners love — perfect-square roots, "non-terminating means irrational," and the surprising fact that two irrationals can add up to something perfectly rational.

The core insight

A number is rational exactly when it can be written as $\dfrac{p}{q}$ with integers $p, q$ and $q \neq 0$ — equivalently, when its decimal expansion terminates or repeats — and irrational otherwise.

The core idea

The nested number system. Every number set a Class 9 student meets sits inside the next:

N \subset W \subset Z \subset Q \subset R

where: $N = 1, 2, 3, \dots$ are the Natural numbers, $W = 0, 1, 2, \dots$ are the Whole numbers, $Z = \dots, -2, -1, 0, 1, 2, \dots$ are the Integers, $Q$ is the set of Rational numbers, and $R$ is the set of Real numbers. Irrational numbers are exactly the leftovers, $R \setminus Q$ — real numbers that are not rational.

The rational test. A number is rational if it can be written as $\dfrac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$ . Every integer is automatically rational, since $n = \dfrac{n}{1}$ . Equivalently — and this is the test you actually use on a decimal — a number is rational exactly when its decimal expansion terminates (like $0.75$ ) or repeats in a fixed block forever (like $0.333\ldots$ or $0.1\overline{45}$ ).

The irrational test. A number is irrational if it cannot be written as a ratio of two integers at all. Its decimal expansion goes on forever without ever locking into a repeating pattern — non-terminating and non-recurring. Famous examples are $\sqrt2$ , $\sqrt3$ , $\sqrt5$ , and $\pi$ .

Density and arithmetic behaviour. Between any two rational numbers there lie infinitely many rational numbers, and infinitely many irrational numbers too — both types pack the real line densely, with no "next" number after any given one. For arithmetic: rational $\pm$ , $\times$ , $\div$ (non-zero divisor) rational stays rational; rational $\pm$ non-zero irrational is always irrational; but irrational $+$ irrational can land back on a rational number, e.g. $\sqrt2 + (-\sqrt2) = 0$ .

Figure

A branching diagram showing natural numbers inside whole numbers inside integers inside rational numbers inside real numbers with irrational numbers as the leftover region of real numbers

Derivation AConverting a recurring decimal to p/q

The main grounded technique of this topic is showing that every recurring decimal genuinely is a rational number, by producing its exact $\dfrac{p}{q}$ form through a subtraction trick that cancels the infinite repeating tail; the same three-step method works whether the repeating block starts right after the decimal point or after some non-repeating digits, and mastering it is what lets you settle any "is this decimal rational" question with certainty rather than a guess.

Step 1. Let $x$ equal the recurring decimal, and identify the length of the repeating block. For $x = 0.\overline{6} = 0.6666\ldots$ , the repeating block is a single digit, "6".

Step 2. Multiply $x$ by $10^k$ , where $k$ is the block length, so the decimal point shifts exactly one block to the right and the repeating tails of $10^k x$ and $x$ line up perfectly. Here $k = 1$ , so $10x = 6.6666\ldots$

Step 3. Subtract $x$ from $10^k x$ . The infinite repeating tails cancel exactly, leaving a finite equation: $10x - x = 6.6666\ldots - 0.6666\ldots$ , so $9x = 6$ .

Step 4. Solve for $x$ as a ratio of integers and simplify. $x = \dfrac{6}{9} = \dfrac{2}{3}$ . Since $2$ and $3$ are integers with $3 \neq 0$ , this confirms $x$ is rational by definition — no recurring decimal can escape this reduction, which is why "repeats forever" is just as rational as "terminates." Verified: $10x - x = 9x = 6 \Rightarrow x = 6/9 = 2/3$ ; checking, $2 \div 3 = 0.6666\ldots$ matches the original decimal.

Worked example

Classify each number as rational or irrational, with a reason, and if rational express it as $\dfrac{p}{q}$ : (a) $\sqrt{16}$ (b) $2.\overline{7}$ (c) $0.101001000100001\ldots$

Step 1. Evaluate (a). $\sqrt{16} = 4$ because $4^2 = 16$ ; $16$ is a perfect square, so its square root is a whole number, $4 = \dfrac{4}{1}$ . This is rational.

Step 2. Convert (b) using the subtraction method. Let $x = 2.\overline{7} = 2.7777\ldots$ ; the repeating block has length $k = 1$ , so $10x = 27.7777\ldots$ . Subtracting, $10x - x = 27.7777\ldots - 2.7777\ldots$ , giving $9x = 25$ , so $x = \dfrac{25}{9}$ . This is rational.

Step 3. Examine (c). The digits follow a pattern of "one more zero each time between the 1s," but the block of zeros keeps growing — it never locks into a fixed repeating group. This decimal is non-terminating and non-recurring, so it is irrational.

Final answer: (a) rational, $4 = \dfrac{4}{1}$ ; (b) rational, $\dfrac{25}{9}$ ; (c) irrational.

Competitive stretch

Find a rational and an irrational number strictly between $2$ and $3$ . A rational choice is the average, $\dfrac{2+3}{2} = \dfrac{5}{2} = 2.5$ , a clean ratio of integers with $2 < 2.5 < 3$ . An irrational choice can be built directly as $2.020020002\ldots$ , a decimal between $2$ and $3$ whose growing block of zeros never repeats; alternatively $\sqrt5 \approx 2.236$ works too, since $4 < 5 < 9$ gives $2 < \sqrt5 < 3$ .

Common traps

Assuming $\sqrt{n}$ is always irrational — it is irrational only when $n$ is not a perfect square; $\sqrt{25} = 5$ is rational.
Believing every non-terminating decimal is irrational — a decimal that repeats forever, even in a long block, is still rational, as the subtraction method proves.
Thinking "irrational" means "cannot be written down or does not really exist" — irrational numbers are exact, well-defined real numbers; they simply have no exact fraction form.
Forgetting $q \neq 0$ in $\dfrac{p}{q}$ — division by zero is undefined and plays no role in deciding rationality.
Assuming that because two numbers both involve a square root they must both be irrational — always check whether the number under the root is a perfect square first.

Ask the tutor: Want to practise classifying trickier decimals or converting longer repeating blocks to $\dfrac{p}{q}$ — ask your AI tutor for a fresh set of examples.

Why does the subtraction trick always work, even for very long repeating blocks

Multiplying $x$ by $10^k$ , where $k$ is the length of the repeating block, shifts the decimal point exactly one full block to the right. Because the block repeats forever, the digits after the new decimal point in $10^k x$ are identical to the digits after the decimal point in $x$ . Subtracting $x$ from $10^k x$ cancels that infinite tail completely, leaving a finite whole number on the right and $(10^k - 1)x$ on the left — always solvable for a ratio of integers.

Is 0.999999 repeating forever actually equal to 1

Yes, exactly, not approximately. Let $x = 0.\overline{9}$ . Then $10x = 9.\overline{9}$ , so $10x - x = 9.\overline{9} - 0.\overline{9} = 9$ , giving $9x = 9$ , so $x = 1$ . The subtraction method that classifies recurring decimals as rational also proves this identity — it is not a rounding convention but an equality.

How can pi be irrational when we use 22/7 or 3.14 for it in problems

Both $\dfrac{22}{7}$ and $3.14$ are only close rational approximations of $\pi$ , used because $\pi$ 's true decimal expansion never terminates and never repeats, so it cannot be written exactly as any ratio of integers. Check: $\dfrac{22}{7} = 3.142857\ldots$ , which already differs from $\pi = 3.14159265\ldots$ after two decimal places. For everyday classroom accuracy the approximation is fine, but $\pi$ itself stays irrational.

Rational and irrational numbers

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads