Number | GCSE - Wyatt's Notes

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. $\blacksquare$

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). $\blacksquare$

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 $396$.

$$180 = 2^2 \times 3^2 \times 5, \qquad 252 = 2^2 \times 3^2 \times 7, \qquad 396 = 2^2 \times 3^2 \times 11$$ $$\mathrm{HCF} = 2^2 \times 3^2 = 36$$ $$\mathrm{LCM} = 2^2 \times 3^2 \times 5 \times 7 \times 11 = 4 \times 9 \times 385 = 13860$$

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.

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 $\left(\frac{2}{3}\right)^{-2} \times \frac{9}{16}$.

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

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

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. $\checkmark$

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.

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.

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

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

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

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.

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?

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

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:

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

• 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: $0.00406$.

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

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

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 $x$.

$$0.03035 \leq x \lt 0.03045$$

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 $3.142$).

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