Number | GCSE - Wyatt's Notes

Number

1. Types of Number

1.1 The Number System

The real numbers can be classified into several nested subsets. Understanding these classifications Is essential for working with the number system fluently.

Definition. The set of natural numbers is $\mathbb{N} = \{1, 2, 3, \ldots\}$ . The set of integers is $\mathbb{Z} = \{\ldots, -2, -1, 0, 1, 2, \ldots\}$ .

A rational number is any number that can be expressed as $\frac{p}{q}$ where $p \in \mathbb{Z}$ $q \in \mathbb{Z} \setminus \{0\}$ And $p$ and $q$ have no common factors other than 1 (i.e. The Fraction is in its lowest terms).

An irrational number is a real number that cannot be expressed as a fraction of two integers. Key examples include $\sqrt{2}$ , $\pi$ And $e$ .

Theorem. $\sqrt{2}$ is irrational.

Proof. Suppose for contradiction that $\sqrt{2} = \frac{p}{q}$ where $p$ and $q$ are coprime Integers with $q \neq 0$ . Then:

$2 = \frac{p^2}{q^2} \implies p^2 = 2q^2$

Since $p^2$ is even, $p$ must be even. Write $p = 2k$ for some integer $k$ . Then:

$4k^2 = 2q^2 \implies q^2 = 2k^2$

So $q^2$ is also even, meaning $q$ is even. But this contradicts the assumption that $p$ and $q$ are Coprime. Therefore $\sqrt{2}$ is irrational. $\blacksquare$

Theorem. $\sqrt{3}$ is irrational.

Proof. Suppose $\sqrt{3} = \frac{p}{q}$ in lowest terms. Then $p^2 = 3q^2$ So $3 \mid p^2$ Hence $3 \mid p$ (since 3 is prime). Write $p = 3k$ : $9k^2 = 3q^2$ So $q^2 = 3k^2$ Giving $3 \mid q$ . This contradicts coprimality. $\blacksquare$

This technique generalises: for any prime $p$ , $\sqrt{p}$ is irrational. The proof structure is Identical in every case.

Proposition. The sum of a rational and an irrational number is irrational.

Proof. Let $r \in \mathbb{Q}$ and $s \notin \mathbb{Q}$ . Suppose $r + s = q \in \mathbb{Q}$ . Then $s = q - r \in \mathbb{Q}$ (since rationals are closed under subtraction), contradicting the Irrationality of $s$ . $\blacksquare$

Proposition. The product of a non-zero rational and an irrational number is irrational.

Proof. Let $r \in \mathbb{Q} \setminus \{0\}$ and $s \notin \mathbb{Q}$ . Suppose $rs = q \in \mathbb{Q}$ . Then $S = \frac{q}{r} \in \mathbb{Q}$ (since $R \neq 0$ ), a contradiction. $\blacksquare$

1.2 Prime Numbers and Factorisation

Definition. A prime number is a natural number greater than 1 that has exactly two factors: 1 and itself.

The prime numbers below 30 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

Theorem (Fundamental Theorem of Arithmetic). Every integer greater than 1 can be written as a Unique product of prime numbers, up to the order of the factors.

This uniqueness is surprisingly powerful. It underpins the entire theory of divisibility and is the Reason prime factorisation is such a central tool.

Example. Write $1260$ as a product of primes.

$1260 = 126 \times 10 = (63 \times 2) \times (2 \times 5) = (7 \times 9) \times 2^2 \times 5 = 2^2 \times 3^2 \times 5 \times 7$

Example. Write $3960$ as a product of primes.

$3960 = 396 \times 10 = (4 \times 99) \times (2 \times 5) = 2^2 \times 9 \times 11 \times 2 \times 5 = 2^3 \times 3^2 \times 5 \times 11$

Theorem (Euclid). There are infinitely many prime numbers.

Proof. Suppose there are finitely many primes $p_1, p_2, \ldots, p_n$ . Consider $N = p_1 p_2 \cdots p_n + 1$ . For each prime $P_i$ , $N$ leaves remainder 1 when divided by $P_i$ , so No $p_i$ divides $N$ . Either $N$ is prime (contradicting that the list was complete) or $N$ has a Prime factor not in the list (also a contradiction). $\blacksquare$

1.3 Highest Common Factor and Lowest Common Multiple

Given two integers $a$ and $b$ Their highest common factor (HCF) is the largest integer that Divides both $a$ and $b$ . Their lowest common multiple (LCM) is the smallest positive integer That is a multiple of both.

If the prime factorisations are $a = p_1^{\alpha_1} p_2^{\alpha_2} \cdots$ and $b = p_1^{\beta_1} p_2^{\beta_2} \cdots$ Then:

$\mathrm{HCF(a, b) = p_1^{\min(\alpha_1, \beta_1)} p_2^{\min(\alpha_2, \beta_2)} \cdots$

$\mathrm{LCM(a, b) = p_1^{\max(\alpha_1, \beta_1)} p_2^{\max(\alpha_2, \beta_2)} \cdots$

Relationship: For any positive integers $a$ and $b$ :

$\mathrm{HCF(a, b) \times \mathrm{LCM(a, b) = a \times b$

Proof of the relationship. Write $a = \prod p_i^{\alpha_i}$ and $b = \prod p_i^{\beta_i}$ . Then:

$\mathrm{HCF \times \mathrm{LCM = \prod p_i^{\min(\alpha_i, \beta_i)} \cdot \prod p_i^{\max(\alpha_i, \beta_i)} = \prod p_i^{\min(\alpha_i, \beta_i) + \max(\alpha_i, \beta_i)} = \prod p_i^{\alpha_i + \beta_i} = ab \quad \blacksquare$

Worked Example. Find the HCF and LCM of $84$ and $210$ .

$84 = 2^2 \times 3 \times 7, \qquad 210 = 2 \times 3 \times 5 \times 7$

$\mathrm{HCF = 2^{\min(2,1)} \times 3^{\min(1,1)} \times 5^{\min(0,1)} \times 7^{\min(1,1)} = 2 \times 3 \times 7 = 42$

$\mathrm{LCM = 2^{\max(2,1)} \times 3^{\max(1,1)} \times 5^{\max(0,1)} \times 7^{\max(1,1)} = 2^2 \times 3 \times 5 \times 7 = 420$

Verification: $42 \times 420 = 17640 = 84 \times 210$ . $\checkmark$

Worked Example (Higher Tier). Find the HCF and LCM of $180$ , $252$ And $396$ .

$180 = 2^2 \times 3^2 \times 5, \qquad 252 = 2^2 \times 3^2 \times 7, \qquad 396 = 2^2 \times 3^2 \times 11$

$\mathrm{HCF = 2^2 \times 3^2 = 36$

$\mathrm{LCM = 2^2 \times 3^2 \times 5 \times 7 \times 11 = 4 \times 9 \times 385 = 13860$

1.4 Divisibility Tests and Prime Testing

To test whether a number $n$ is prime, you only need to check divisibility by primes up to $\sqrt{n}$ . If none divide $n$ Then $n$ is prime.

Worked Example. Is $211$ prime?

$\sqrt{211} \approx 14.5$ So we check primes up to $13$ : 2, 3, 5, 7, 11, 13.

Not divisible by 2 (odd), 3 (digit sum $= 4$ ), or 5 (does not end in 0 or 5).
$211 = 30 \times 7 + 1$ Not divisible by 7.
$211 = 19 \times 11 + 2$ Not divisible by 11.
$211 = 16 \times 13 + 3$ Not divisible by 13.

Therefore $211$ is prime.

2. Fractions, Decimals, and Percentages

2.1 Fraction Arithmetic

Addition and subtraction: $\frac{a}{b} \pm \frac{c}{d} = \frac{ad \pm bc}{bd}$

Multiplication: $\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}$

Division: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc}$

Worked Example. Evaluate $\frac{3}{4} + \frac{2}{5} - \frac{1}{3}$ .

$\frac{3}{4} + \frac{2}{5} - \frac{1}{3} = \frac{45 + 24 - 20}{60} = \frac{49}{60}$

Worked Example (Higher Tier). Simplify $\frac{2\frac{3}{4}}{1\frac{1}{3}}$ .

Convert to improper fractions: $\frac{11}{4} \div \frac{4}{3} = \frac{11}{4} \times \frac{3}{4} = \frac{33}{16}$ .

Worked Example (Higher Tier). Evaluate $\left(\frac{2}{3}\right)^{-2} \times \frac{9}{16}$ .

$\left(\frac{3}{2}\right)^2 \times \frac{9}{16} = \frac{9}{4} \times \frac{9}{16} = \frac{81}{64}$

2.2 Recurring Decimals to Fractions

A recurring decimal has digits that repeat infinitely. We can convert these to exact fractions Using algebra.

Worked Example. Convert $0.\dot{3}\dot{6}$ to a fraction.

Let $x = 0.363636\ldots$

The repeating block has 2 digits, so multiply by 100:

$100x = 36.363636\ldots$

$x = 0.363636\ldots$

Subtracting: $99x = 36$ So $x = \frac{36}{99} = \frac{4}{11}$ .

General rule: If the repeating block has $n$ digits, multiply by $10^n$ Subtract the original, And simplify.

Worked Example (Higher Tier). Convert $0.1\dot{6}\dot{3}$ to a fraction.

Let $x = 0.163163163\ldots$

The repeating block has 3 digits, so multiply by 1000:

$1000x = 163.163163\ldots$

$x = 0.163163\ldots$

$999x = 163 \implies x = \frac{163}{999}$

Check: $\gcd(163, 999)$ . Since $163$ is prime and $999 = 3^3 \times 37$ They are coprime. So $x = \frac{163}{999}$ .

Worked Example. Convert $0.4\dot{7}$ to a fraction.

Here one digit ( $4$ ) does not repeat and two digits ( $7$ ) repeat. Let $x = 0.47777\ldots$

Multiply by 10: $10x = 4.7777\ldots$

Multiply by 100: $100x = 47.7777\ldots$

Subtract: $90x = 43 \implies x = \frac{43}{90}$ .

2.3 Percentages

A percentage represents a fraction out of 100. The key operations are:

Percentage of an amount: $P\%$ of $A = \frac{P}{100} \times A$
Percentage change: $\frac{\mathrm{change}{\mathrm{original} \times 100\%$
Percentage increase/decrease: $\mathrm{new = \mathrm{original \times \left(1 \pm \frac{P}{100}\right)$

Worked Example. A coat costs 120 pounds. It is reduced by 15% in a sale, then the sale price is Increased by 15%. What is the final price?

After the reduction: $120 \times 0.85 = 102$ pounds.

After the increase: $102 \times 1.15 = 117.30$ pounds.

Theorem. A percentage increase of $P\%$ followed by a percentage decrease of $P\%$ (or vice Versa) always results in a net decrease. The net effect is a decrease of $\frac{P^2}{100}\%$ .

Proof. The factor for increase is $\left(1 + \frac{P}{100}\right)$ and for decrease is $\left(1 - \frac{P}{100}\right)$ . The combined factor is:

$\left(1 + \frac{P}{100}\right)\left(1 - \frac{P}{100}\right) = 1 - \frac{P^2}{10000}$

This is always less than 1 for $P \neq 0$ Confirming a net decrease of $\frac{P^2}{100}\%$ . $\blacksquare$

Worked Example (Higher Tier). A quantity increases by 20% one year and decreases by 20% the Next. What is the overall percentage change?

Combined factor: $1.2 \times 0.8 = 0.96$ A net decrease of 4%.

By the theorem: $\frac{20^2}{100} = 4\%$ decrease. $\checkmark$

2.4 Reverse Percentages

Worked Example. After a 20% increase, a price is 336 pounds. Find the original price.

The original price is $100\%$ And after the increase it is $120\%$ . So:

$\mathrm{original = \frac{336}{1.20} = 280 \mathrm{ pounds$

Worked Example (Higher Tier). A shop offers “15% off the sale price.” A customer pays 34 pounds. What was the original price before the sale?

The customer pays 85% of the sale price. Sale price $= \frac{34}{0.85} = 40$ pounds.

If the sale itself was, say, a 20% discount on the original: original $= \frac{40}{0.80} = 50$ Pounds.

2.5 Compound Interest and Depreciation

For compound growth at rate $r\%$ per period over $n$ periods:

$A = P\left(1 + \frac{r}{100}\right)^n$

For depreciation:

$A = P\left(1 - \frac{r}{100}\right)^n$

Worked Example. 2000 pounds is invested at 3.5% compound interest per year. Find the value after 6 years, giving your answer to the nearest penny.

$A = 2000 \times 1.035^6 = 2000 \times 1.22925\ldots = 2458.51 \mathrm{ pounds$

Worked Example (Higher Tier). A car bought for 18000 pounds depreciates at 12% per annum. After How many whole years will its value first fall below 8000 pounds?

We need $18000 \times 0.88^n \lt 8000$ So $0.88^n \lt \frac{8000}{18000} = \frac{4}{9}$ .

Taking logarithms: $n \ln 0.88 \lt \ln\!\left(\frac{4}{9}\right)$ .

Since $\ln 0.88 \lt 0$ The inequality reverses: $n \gt \frac{\ln(4/9)}{\ln 0.88} = \frac{-0.811}{-0.128} \approx 6.33$ .

So after 7 years the value first falls below 8000 pounds.

Worked Example. 5000 pounds is invested at 4% compound interest. Find the total interest earned After 3 years.

$A = 5000 \times 1.04^3 = 5000 \times 1.124864 = 5624.32 \mathrm{ pounds$

Total interest $= 5624.32 - 5000 = 624.32$ pounds.

3. Indices and Surds

3.1 Laws of Indices

For positive integers $m$ and $n$ And non-zero base $a$ :

Law	Expression
Multiplication	$a^m \times a^n = a^{m+n}$
Division	$a^m \div a^n = a^{m-n}$
Power of a power	$(a^m)^n = a^{mn}$
Power of a product	$(ab)^n = a^n b^n$
Negative index	$a^{-n} = \frac{1}{a^n}$
Zero index	$a^0 = 1$
Fractional index	$a^{1/n} = \sqrt[n]{a}$
Mixed fractional	$a^{m/n} = \left(\sqrt[n]{a}\right)^m$

Why $a^0 = 1$ : Consider $a^m \div a^m = a^{m-m} = a^0$ . But $a^m \div a^m = 1$ . Therefore $a^0 = 1$ for all $a \neq 0$ .

Worked Example. Simplify $\frac{8a^3 b^2 \times 3a^{-1} b^4}{6a^2 b^{-3}}$ .

$\frac{8 \times 3}{6} \cdot a^{3 + (-1) - 2} \cdot b^{2 + 4 - (-3)} = 4 \cdot a^0 \cdot b^9 = 4b^9$

Worked Example (Higher Tier). Simplify $\left(\frac{27x^6}{8y^{-3}}\right)^{-2/3}$ .

$= \left(\frac{8y^{-3}}{27x^6}\right)^{2/3} = \frac{8^{2/3} \cdot y^{-3 \times 2/3}}{27^{2/3} \cdot x^{6 \times 2/3}} = \frac{4 \cdot y^{-2}}{9 \cdot x^4} = \frac{4}{9x^4 y^2}$

3.2 Surds

A surd is an irrational number expressed as the root of an integer, for example $\sqrt{3}$ or $\sqrt[3]{5}$ .

Rules of surds:

$\sqrt{a} \times \sqrt{b} = \sqrt{ab}, \qquad \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}, \qquad (\sqrt{a})^2 = a$

Rationalising the denominator: To remove a surd from the denominator, multiply top and bottom by The surd (or its conjugate if the denominator is a binomial).

Worked Example. Simplify $\frac{6}{\sqrt{3}}$ .

$\frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$

Worked Example. Rationalise $\frac{5}{2 - \sqrt{3}}$ .

Multiply by the conjugate $2 + \sqrt{3}$ :

$\frac{5(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{5(2 + \sqrt{3})}{4 - 3} = 5(2 + \sqrt{3}) = 10 + 5\sqrt{3}$

Worked Example (Higher Tier). Simplify $\frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} - \sqrt{3}}$ .

$\frac{(\sqrt{7} + \sqrt{3})^2}{7 - 3} = \frac{7 + 2\sqrt{21} + 3}{4} = \frac{10 + 2\sqrt{21}}{4} = \frac{5 + \sqrt{21}}{2}$

Worked Example (Higher Tier). Expand and simplify $(3 + 2\sqrt{5})(1 - \sqrt{5})$ .

$= 3 - 3\sqrt{5} + 2\sqrt{5} - 2 \times 5 = 3 - \sqrt{5} - 10 = -7 - \sqrt{5}$

Worked Example (Higher Tier). Simplify $\sqrt{50} + 2\sqrt{8} - 3\sqrt{18}$ .

Write each in simplest surd form:

$= 5\sqrt{2} + 2 \times 2\sqrt{2} - 3 \times 3\sqrt{2} = 5\sqrt{2} + 4\sqrt{2} - 9\sqrt{2} = 0$

Theorem. If $a + b\sqrt{c} = d + e\sqrt{c}$ where $a, b, d, e$ are rational and $\sqrt{c}$ is Irrational, then $a = d$ and $b = e$ .

Proof. If $a + b\sqrt{c} = d + e\sqrt{c}$ Then $(a - d) = (e - b)\sqrt{c}$ . If $e \neq b$ Then $\sqrt{c} = \frac{a - d}{e - b} \in \mathbb{Q}$ Contradicting the irrationality of $\sqrt{c}$ . Therefore $e = b$ and hence $a = d$ . $\blacksquare$

This theorem is used frequently in solving equations involving surds.

3.3 Standard Form

A number in standard form is written as $a \times 10^n$ where $1 \leq a \lt 10$ and $n$ is an Integer.

Worked Example. Calculate $\frac{4.5 \times 10^8}{3 \times 10^{-2}}$ Giving your answer in Standard form.

$\frac{4.5 \times 10^8}{3 \times 10^{-2}} = 1.5 \times 10^{8 - (-2)} = 1.5 \times 10^{10}$

Worked Example. The population of a city is $2.4 \times 10^6$ . The average income is $3.1 \times 10^4$ pounds per year. Find the total income, in standard form.

$2.4 \times 10^6 \times 3.1 \times 10^4 = 7.44 \times 10^{10} \mathrm{ pounds$

Worked Example (Higher Tier). The speed of light is approximately $3 \times 10^8$ m/s. The Distance from the Sun to the Earth is approximately $1.5 \times 10^{11}$ m. How many minutes does Light take to travel from the Sun to the Earth?

$\mathrm{Time = \frac{1.5 \times 10^{11}}{3 \times 10^8} = 500 \mathrm{ seconds = \frac{500}{60} \approx 8.33 \mathrm{ minutes$

4. Upper and Lower Bounds

When a measurement is given to a specified degree of accuracy, the true value lies within a range.

Definition. If a quantity $x$ is given as $a$ to the nearest unit, then:

The upper bound of $x$ is $a + 0.5$
The lower bound of $x$ is $a - 0.5$

For $x$ rounded to $d$ decimal places, the bounds are $a \pm 0.5 \times 10^{-d}$ .

For $x$ rounded to $s$ significant figures, the bounds are $a \pm 0.5 \times 10^{\lfloor \log_{10} a \rfloor - s + 1}$ .

Worked Example. A rectangle has length $8.4$ cm and width $5.2$ cm, both measured to 1 decimal Place. Find the upper and lower bounds for the area.

Bounds for length: $8.35 \leq l \lt 8.45$

Bounds for width: $5.15 \leq w \lt 5.20$

Upper bound of area: $8.45 \times 5.20 = 43.94 \mathrm{ cm^2$
Lower bound of area: $8.35 \times 5.15 = 43.0025 \mathrm{ cm^2$

Worked Example (Higher Tier). $x = 6.3$ and $y = 2.7$ Both correct to 1 decimal place. Find the Lower bound of $\frac{x}{y}$ .

Lower bound of $\frac{x}{y} = \frac{\mathrm{lower(x)}{\mathrm{upper(y)} = \frac{6.25}{2.75} = \frac{25}{11} \approx 2.27$ .

Upper bound of $\frac{x}{y} = \frac{\mathrm{upper(x)}{\mathrm{lower(y)} = \frac{6.35}{2.65} = \frac{127}{53} \approx 2.40$ .

General principle for bounds:

Operation	Upper bound	Lower bound
$a + b$	upper $(a)$ + upper $(b)$	lower $(a)$ + lower $(b)$
$a - b$	upper $(a)$ - lower $(b)$	lower $(a)$ - upper $(b)$
$a \times b$	upper $(a)$ $\times$ upper $(b)$	lower $(a)$ $\times$ lower $(b)$
$a \div b$	upper $(a)$ $\div$ lower $(b)$	lower $(a)$ $\div$ upper $(b)$

Worked Example (Higher Tier). $a = 12.4$ cm and $b = 3.7$ cm, both to 1 d.p. Find the upper Bound of $a^2 - b^2$ .

Upper bound of $a^2 - b^2$ : $\mathrm{upper(a)^2 - \mathrm{lower(b)^2 = 12.45^2 - 3.65^2 = 155.0025 - 13.3225 = 141.68 \mathrm{ cm^2$ .

5. Estimation and Approximation

5.1 Rounding

Decimal places: Count digits after the decimal point.
Significant figures: Count from the first non-zero digit.

Worked Example. Round $0.004063$ to 2 significant figures.

The first significant figure is 4, the second is 0. The next digit is 6, so round up: $0.0041$ .

Worked Example. Round $0.004063$ to 3 significant figures.

The third significant figure is 6, and the next digit is 3, so round down: $0.00406$ .

5.2 Estimation

Replace numbers with approximate values ( 1 significant figure) to get a quick estimate.

Worked Example. Estimate $\frac{3.97 \times 18.4}{0.498}$ .

$\approx \frac{4 \times 20}{0.5} = \frac{80}{0.5} = 160$

Worked Example. Estimate $\sqrt{51} + \frac{4.9 \times 7.8}{3.1}$ .

$\approx \sqrt{49} + \frac{5 \times 8}{3} = 7 + \frac{40}{3} \approx 7 + 13.3 = 20.3$

5.3 Error Intervals

An error interval for a rounded value $x$ is the range of possible true values.

Worked Example. $x$ is rounded to the nearest integer as 7. Write down the error interval for $x$ .

$6.5 \leq x \lt 7.5$

Note: the lower bound is inclusive (values of exactly 6.5 round up to 7), but the upper bound is Exclusive (values of exactly 7.5 round up to 8).

Worked Example (Higher Tier). $p = 42.6$ is correct to 3 significant figures. Write down the Error interval for $p$ .

$42.55 \leq p \lt 42.65$

Worked Example (Higher Tier). $x = 0.0304$ is correct to 3 significant figures. Write down the Error interval for $x$ .

$0.03035 \leq x \lt 0.03045$

5.4 Truncation

Truncation is different from rounding. When a number is truncated to a given number of decimal Places, all digits beyond that point are discarded (not rounded).

Example. Truncate $3.749$ to 1 decimal place: $3.7$ (not $3.8$ ).

Example. Truncate $\pi$ to 3 decimal places: $3.141$ (not $3.142$ ).

Cutoff is 5 or greater. Be sure to read the question carefully.

6. Direct and Inverse Proportion

6.1 Direct Proportion

$y$ is directly proportional to $x$ if $y = kx$ for some constant $k$ (the constant of Proportionality).

We write $y \propto x$ .

Worked Example. $y$ is directly proportional to $x$ . When $x = 5$ , $y = 30$ . Find $y$ when $x = 8$ .

$30 = 5k \implies k = 6$

$y = 6 \times 8 = 48$

6.2 Inverse Proportion

$y$ is inversely proportional to $x$ if $y = \frac{k}{x}$ .

We write $y \propto \frac{1}{x}$ .

Worked Example. $y$ is inversely proportional to $x^2$ . When $x = 3$ , $y = 12$ . Find $y$ when $x = 6$ .

$12 = \frac{k}{9} \implies k = 108$

$y = \frac{108}{36} = 3$

Worked Example (Higher Tier). The time $t$ taken to fill a tank is inversely proportional to the Square of the radius $r$ of the pipe. When $r = 2$ cm, $t = 45$ minutes. Find $t$ when $r = 5$ cm.

$t = \frac{k}{r^2} \implies 45 = \frac{k}{4} \implies k = 180$

$t = \frac{180}{25} = 7.2 \mathrm{ minutes$

6.3 Proportionality with Powers and Roots

Worked Example (Higher Tier). $y$ is directly proportional to $\sqrt{x}$ . When $x = 9$ $y = 12$ . Find $y$ when $x = 25$ .

$y = k\sqrt{x} \implies 12 = 3k \implies k = 4$

$y = 4\sqrt{25} = 20$

7. Factors, Multiples, and Primes in Context

7.1 Squares, Cubes, and Roots

Operation	Symbol	Example
Square	$n^2$	$7^2 = 49$
Cube	$n^3$	$4^3 = 64$
Square root	$\sqrt{n}$	$\sqrt{81} = 9$
Cube root	$\sqrt[3]{n}$	$\sqrt[3]{27} = 3$

It is essential to memorise squares up to $15^2 = 225$ and cubes up to $5^3 = 125$ for efficient Exam work.

7.2 Triangular and Other Sequences

Triangular numbers: $T_n = \frac{n(n+1)}{2}$ Giving the sequence $1, 3, 6, 10, 15, 21, \ldots$

These represent the number of dots that can form an equilateral triangle with $n$ dots on each side.

Square numbers: $S_n = n^2$ Giving $1, 4, 9, 16, 25, \ldots$

Cube numbers: $C_n = n^3$ Giving $1, 8, 27, 64, 125, \ldots$

Proposition. Every square number is either a multiple of 4 or one more than a multiple of 4.

Proof. If $n$ is even, $n = 2k$ and $n^2 = 4k^2$ A multiple of 4. If $n$ is odd, $n = 2k + 1$ And $n^2 = 4k^2 + 4k + 1 = 4(k^2 + k) + 1$ One more than a multiple of 4. $\blacksquare$

7.3 Rules of Divisibility

Divisible by	Rule
2	Last digit is even
3	Sum of digits is divisible by 3
4	Last two digits form a number divisible by 4
5	Last digit is 0 or 5
6	Divisible by both 2 and 3
8	Last three digits form a number divisible by 8
9	Sum of digits is divisible by 9
10	Last digit is 0
11	Alternating sum of digits is divisible by 11

Worked Example. Is $734856$ divisible by 11?

Alternating sum: $7 - 3 + 4 - 8 + 5 - 6 = -1$ . Since $-1$ is not divisible by 11, no.

Worked Example. Is $81624$ divisible by 8?

Last three digits: $624$ . Check $624 \div 8 = 78$ . Yes, divisible by 8.

7.4 LCM Applications (Higher Tier)

LCM problems arise frequently in real-world contexts: synchronising events, finding repeating Patterns, and scheduling.

Worked Example. Bus A arrives every 12 minutes and Bus B arrives every 18 minutes. They both Arrive together at 08:00. When will they next arrive together?

We need $\mathrm{LCM(12, 18) = 36$ minutes.

Next simultaneous arrival: 08:36.

Worked Example. Three lights flash every 6, 8, and 12 seconds respectively. They all flash at Time $t = 0$ . When will they next all flash together?

$\mathrm{LCM(6, 8, 12) = 24$ seconds.

8. Number Theory Proofs (Higher Tier)

8.1 Proof by Exhaustion

List all possible cases and verify each one.

Example. Prove that every prime greater than 3 is of the form $6n \pm 1$ for some integer $n$ .

Every integer is of the form $6n$ , $6n+1$ , $6n+2$ , $6n+3$ , $6n+4$ Or $6n+5$ .

$6n$ : divisible by 6 (not prime for $n \geq 1$ )
$6n + 2 = 2(3n + 1)$ : even (not prime)
$6n + 3 = 3(2n + 1)$ : divisible by 3 (not prime)
$6n + 4 = 2(3n + 2)$ : even (not prime)

The only remaining forms are $6n + 1$ and $6n + 5 = 6(n + 1) - 1$ . $\blacksquare$

8.2 Proof by Deduction

Example. Prove that the sum of any three consecutive integers is divisible by 3.

Let the three consecutive integers be $n$ , $n + 1$ , $n + 2$ .

Sum $= n + (n + 1) + (n + 2) = 3n + 3 = 3(n + 1)$ .

Since $n + 1$ is an integer, the sum is a multiple of 3. $\blacksquare$

Example. Prove that the sum of any four consecutive integers is never a prime number.

Let the integers be $n$ , $n + 1$ , $n + 2$ , $n + 3$ .

Sum $= 4n + 6 = 2(2n + 3)$ .

This is always even and always greater than 2 (since the smallest possible sum is $0 + 1 + 2 + 3 = 6$ ). The only even prime is 2, and the sum is always at least 6, so it can never be Prime. $\blacksquare$

Example. Prove that $n^2 + n$ is always even for all integers $n$ .

$n^2 + n = n(n + 1)$ . Among any two consecutive integers, one must be even. Therefore their product Is even. $\blacksquare$

Example. Prove that the difference between the squares of any two consecutive odd numbers is Divisible by 8.

Let the consecutive odd numbers be $2k + 1$ and $2k + 3$ .

$(2k + 3)^2 - (2k + 1)^2 = (4k^2 + 12k + 9) - (4k^2 + 4k + 1) = 8k + 8 = 8(k + 1)$ .

Since $k + 1$ is an integer, this is a multiple of 8. $\blacksquare$

Common Pitfalls

Confusing HCF and LCM. The HCF uses the minimum power of each prime; the LCM uses the maximum.
Incorrect bounds for subtraction and division. For positive quantities, the upper bound of $a - b$ is upper $(a)$ - lower $(b)$ Not upper $(a)$ - upper $(b)$ .
Forgetting that $a^0 = 1$ for any $a \neq 0$ . This is a definition, not a pattern.
Mishandling negative indices. $a^{-n} = \frac{1}{a^n}$ Not $-a^n$ .
Rationalising denominators incorrectly. When the denominator is $a + \sqrt{b}$ Multiply by $a - \sqrt{b}$ Not by $\sqrt{a} - \sqrt{b}$ .
Assuming compound percentage changes cancel. A 20% increase followed by a 20% decrease gives $1.2 \times 0.8 = 0.96$ A net decrease of 4%.
Truncation vs rounding. Truncation discards digits; rounding considers the next digit.
Using the wrong bound for division. To maximise $\frac{a}{b}$ (positive), maximise the numerator and minimise the denominator.
Assuming surds can cancel partially. $\sqrt{2} + \sqrt{3}$ cannot be simplified further.

Practice Questions

Express $540$ as a product of prime factors. Hence find the HCF and LCM of $540$ and $324$ .
Convert $0.2\dot{7}$ to a fraction in its lowest terms.
Simplify $\frac{(2\sqrt{3})^3}{\sqrt{27}}$ .
A car depreciates at 12% per year. If it was bought for 18000 pounds, find its value after 4 years to the nearest pound.
$a = 6.3$ and $b = 2.7$ Both correct to 1 decimal place. Find the lower bound of $\frac{a}{b}$ .
$y$ is directly proportional to the cube of $x$ . When $x = 2$ , $y = 40$ . Find $x$ when $y = 1080$ .
Simplify $\frac{12x^5 y^{-2}}{3x^{-1} y^4} \times (xy^3)^2$ .
Rationalise the denominator of $\frac{\sqrt{5} + 3}{\sqrt{5} - 1}$ .
Calculate $\frac{6.2 \times 10^{-3} \times 4.8 \times 10^7}{1.2 \times 10^2}$ in standard form.
Prove that the sum of any three consecutive integers is divisible by 3.
Simplify $\frac{(3 + \sqrt{2})^2 - (3 - \sqrt{2})^2}{\sqrt{2}}$ .
Prove that the product of $n(n + 1)(n + 2)$ is always divisible by 6 for positive integers $n$ .
A number $x$ is truncated to 2 decimal places as $3.47$ . Write down the error interval for $x$ .
Express $0.1\dot{2}\dot{3}$ as a fraction in its lowest terms.
Two lights flash every 15 seconds and 24 seconds. They flash together at noon. At what times before 1 pm will they flash together?

Intuition

Number theory is the bedrock of mathematics — every other topic builds on it. Prime factorisation is like a molecular formula: it breaks every number into its irreducible elements, revealing structure that is invisible on the surface. The Fundamental Theorem of Arithmetic guarantees that this decomposition is unique, which is why prime factorisation is so powerful for finding HCFs and LCMs. Bounds are about precision: when you round a measurement, you are accepting uncertainty, and bounds quantify that uncertainty. The key insight is that numbers are not just values — they have properties (divisibility, primality, parity) that determine how they behave in calculations.

Worked Examples

Example 1:

A typical exam question on Number requires you to apply your knowledge to an unfamiliar context. Read the question carefully, identify the key concept being tested, and structure your answer using the appropriate terminology.

Example 2:

Multi-step problems in Number often combine two or more concepts. Break the problem down: identify what you need to find, recall the relevant formula or principle, substitute values, and state your answer with correct units or formatting.

flowchart TD
    A[1_Number] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary

This topic covers the mathematical techniques and concepts related to number, including key theorems, methods, and problem-solving approaches.

Key concepts include:

arithmetic and geometric sequences
series and sigma notation
recurrence relations
convergence tests
mathematical induction

Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.