Number System Class 9: notes and important questions

Number System: revision notes

1. From counting numbers to real numbers

Natural numbers \(\mathbb N = \{1, 2, 3, \ldots\}\) \(\subset\) whole numbers (with \(0\)) \(\subset\) integers \(\mathbb Z\) \(\subset\) rational numbers \(\mathbb Q\) \(\subset\) real numbers \(\mathbb R\). The irrational numbers are the real numbers that are not rational.

2. Rational numbers and decimals

A rational number can be written \(\dfrac pq\) with integers \(p, q\) and \(q \neq 0\). Between any two rational numbers there are infinitely many more (for example their average).

• In lowest terms, \(\dfrac pq\) has a terminating decimal exactly when \(q\) has no prime factors other than \(2\) and \(5\) (\(\dfrac78 = 0.875\)); otherwise the decimal is non-terminating recurring (\(\dfrac{5}{11} = 0.\overline{45}\)).
• Recurring to fraction: for \(x = 0.\overline{47}\), \(100x - x = 47\), so \(x = \dfrac{47}{99}\). For \(x = 0.1\overline{6}\): \(100x - 10x = 15\), so \(x = \dfrac{15}{90} = \dfrac16\).

3. Irrational numbers and the number line

An irrational number has a decimal expansion that neither terminates nor repeats: \(\sqrt2, \sqrt3, \sqrt5, \pi\). If \(n\) is a natural number that is not a perfect square, \(\sqrt n\) is irrational. Every real number is a point on the number line and every point is a real number; \(\sqrt n\) can be marked with right triangles (the square-root spiral: legs \(\sqrt n\) and \(1\) give hypotenuse \(\sqrt{n + 1}\)).

• Rational \(\pm\) irrational is irrational; a non-zero rational \(\times\) irrational is irrational.
• Irrational \(+\) irrational can be rational: \(\sqrt2 + (-\sqrt2) = 0\).
• Proof that \(\sqrt2\) is irrational: suppose \(\sqrt2 = \dfrac pq\) in lowest terms; then \(p^2 = 2q^2\), so \(p\) is even, \(p = 2m\), so \(q^2 = 2m^2\) and \(q\) is even too: a contradiction.

4. Surds and rationalising

For positive \(a, b\): \(\sqrt{ab} = \sqrt a\sqrt b\), \(\sqrt{\dfrac ab} = \dfrac{\sqrt a}{\sqrt b}\), \((\sqrt a + \sqrt b)(\sqrt a - \sqrt b) = a - b\), \((\sqrt a + \sqrt b)^2 = a + b + 2\sqrt{ab}\).

To rationalise \(\dfrac{1}{\sqrt a + \sqrt b}\) multiply top and bottom by \(\sqrt a - \sqrt b\): \(\dfrac{\sqrt a - \sqrt b}{a - b}\).

5. Laws of exponents

For \(a, b \gt 0\) and rational \(m, n\): \(a^m a^n = a^{m+n}\), \(\dfrac{a^m}{a^n} = a^{m-n}\), \((a^m)^n = a^{mn}\), \(a^m b^m = (ab)^m\), \(a^0 = 1\), \(a^{-m} = \dfrac{1}{a^m}\), \(a^{\frac pq} = \left(\sqrt[q]{a}\right)^p\).

Worked example 1

Write \(2.\overline{3}\) as a fraction.

\(x = 2.333\ldots\), \(10x - x = 21\), so \(x = \dfrac{21}{9} = \dfrac73\).

Worked example 2

Rationalise \(\dfrac{4}{\sqrt7 - \sqrt3}\).

\(\dfrac{4(\sqrt7 + \sqrt3)}{7 - 3} = \sqrt7 + \sqrt3\).

Worked example 3

Evaluate \(32^{\frac35}\).

\(32 = 2^5\), so \(32^{\frac35} = 2^3 = 8\).

Common errors

• \(\sqrt{a + b} = \sqrt a + \sqrt b\) (false: \(\sqrt{9 + 16} = 5 \neq 7\)).
• Calling a value used for \(\pi\) in a question (such as a fraction) equal to \(\pi\): it is only a rational approximation; \(\pi\) is irrational.
• \(a^{-m} = -a^m\) (it is \(\dfrac1{a^m}\)), and \(a^m \times a^n = a^{mn}\) (it is \(a^{m+n}\)).
• In \(0.1\overline{6}\), treating the whole of \(16\) as repeating.

Exam tips

• Write numbers as powers of a prime (\(27 = 3^3\), \(16 = 2^4\)) before using the laws of exponents.
• In irrationality proofs, state the assumption ("suppose it is rational, in lowest terms") and the contradiction clearly.