The algebra rules, order rules, density, absolute value and index laws that every later unit relies on, with proofs of basic inequalities.
Every identity in algebra comes from a short list of rules. For all real $a, b, c$:
| Rule | Addition | Multiplication |
|---|---|---|
| Closure | $a + b \in \mathbb{R}$ | $ab \in \mathbb{R}$ |
| Commutative | $a + b = b + a$ | $ab = ba$ |
| Associative | $(a + b) + c = a + (b + c)$ | $(ab)c = a(bc)$ |
| Identity | $a + 0 = a$ | $a \cdot 1 = a$ |
| Inverse | $a + (-a) = 0$ | $a \cdot \dfrac{1}{a} = 1$ for $a \neq 0$ |
| Distributive | $a(b + c) = ab + ac$ |
Subtraction and division are not separate operations: $a - b = a + (-b)$ and $a \div b = a \cdot \dfrac{1}{b}$. That is why they lack the commutative and associative rules: $8 - 3 \neq 3 - 8$ and $8 - (5 - 2) = 5$ but $(8 - 5) - 2 = 1$.
Zero product rule. If $ab = 0$ then $a = 0$ or $b = 0$. This is why we can solve $(x - 2)(x + 3) = 0$ by setting each bracket to zero.
For real $a, b$ exactly one of $a < b$, $a = b$, $a > b$ is true. The order obeys:
From these: $x^2 \geq 0$ for every real $x$, and $x^2 = 0$ only for $x = 0$. This one fact proves most inequalities.
Squares and reciprocals. Squaring does not always keep the order: $-3 < 2$ but $(-3)^2 = 9 > 4 = 2^2$. But if $0 < a < b$ then $b^2 - a^2 = (b - a)(b + a) > 0$, so $a^2 < b^2$. Likewise, for $0 < a < b$: $\dfrac{1}{a} - \dfrac{1}{b} = \dfrac{b - a}{ab} > 0$, so $\dfrac{1}{a} > \dfrac{1}{b}$.
Example: Prove that $a^2 + b^2 \geq 2ab$ for all real $a, b$, and deduce that $\dfrac{a + b}{2} \geq \sqrt{ab}$ for $a, b \geq 0$. $a^2 + b^2 - 2ab = (a - b)^2 \geq 0$, so $a^2 + b^2 \geq 2ab$, with equality exactly when $a = b$. For $a, b \geq 0$ put $a = x^2$, $b = y^2$ with $x = \sqrt{a}$, $y = \sqrt{b}$. Then $x^2 + y^2 \geq 2xy$ says $a + b \geq 2\sqrt{ab}$, that is $\dfrac{a + b}{2} \geq \sqrt{ab}$. This is the arithmetic mean $\geq$ geometric mean inequality for two numbers.
Between any two real numbers $a < b$ there is a rational number, and there is an irrational number. So there are infinitely many of each. For instance, the midpoint $\dfrac{a + b}{2}$ is between them. When $a$ and $b$ are both rational, the midpoint $\dfrac{a + b}{2}$ is rational and $a + \dfrac{b - a}{\sqrt{2}}$ is irrational, and both lie between $a$ and $b$. Real numbers have no "next" number: there is no smallest real number greater than $0$.
The absolute value (modulus) of $x$ is $$\lvert x \rvert = \begin{cases} x & x \geq 0 \ -x & x < 0. \end{cases}$$ It is the distance from $x$ to $0$ on the number line, so $\lvert x \rvert \geq 0$, and $\lvert a - b \rvert$ is the distance between $a$ and $b$. Key facts:
For $a > 0$ and any rational $m, n$ (with $n \neq 0$ where it is a divisor): $$a^m a^n = a^{m+n}, \quad \dfrac{a^m}{a^n} = a^{m-n}, \quad (a^m)^n = a^{mn}, \quad a^{-n} = \dfrac{1}{a^n}, \quad a^0 = 1,$$ and $a^{1/n} = \sqrt[n]{a}$, so $a^{m/n} = \left(\sqrt[n]{a}\right)^m$.
Example: Evaluate $\left(\dfrac{27}{8}\right)^{-2/3}$. A negative power means the reciprocal: $\left(\dfrac{27}{8}\right)^{-2/3} = \left(\dfrac{8}{27}\right)^{2/3}$. Take the cube root first, which is easier: $\sqrt[3]{\dfrac{8}{27}} = \dfrac{2}{3}$. Then square: $\left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9}$.
Exam tip: To prove an inequality in Paper I, write "LHS $-$ RHS" and simplify until you see a square such as $(a - b)^2$, then say it is $\geq 0$. Never multiply or divide an inequality by an expression whose sign you have not checked, and say where equality holds if the question asks for it.