Indices & Surds
Every rule for powers comes from one idea — a power counts how many copies you multiply. Master that, and indices, roots and surds all become the same skill.
An index (or power) is nothing more than a shorthand for repeated multiplication. Writing $a^4$ saves us from writing $a \times a \times a \times a$. That single observation — that a power counts how many copies of the base are multiplied together — is the seed from which every index law grows. If you ever forget a rule, you can rebuild it by going back to this meaning.
Watch how the first law falls straight out of the definition. What is $a^2 \times a^3$? It's $(a \times a) \times (a \times a \times a)$ — five copies of $a$ multiplied together, so $a^5$. We didn't memorise anything; we just counted. That's the whole game.
The core index laws
From the counting idea, three laws follow immediately, for any base $a$ and integers $m, n$:
Multiplication: $a^m \times a^n = a^{m+n}$ — combine the copies, so add the powers.
Division: $a^m \div a^n = a^{m-n}$ — cancel matching copies, so subtract.
Power of a power: $(a^m)^n = a^{mn}$ — $n$ groups of $m$ copies, so multiply.
Three more extend the idea beyond whole-number powers, and this is where the real understanding is tested:
Zero: $a^0 = 1$. Why? Because $a^n \div a^n = a^{n-n} = a^0$, but anything divided by itself is $1$. The two must agree.
Negative: $a^{-n} = \dfrac{1}{a^n}$. A negative power is a reciprocal, forced by the division law running past zero.
Fractional: $a^{1/n} = \sqrt[n]{a}$, and $a^{m/n} = \left(\sqrt[n]{a}\right)^m$. Since $a^{1/2}\times a^{1/2}=a^1=a$, the number $a^{1/2}$ must be the thing that squares to give $a$ — the square root.
For $a^{m/n}$: the denominator is the root, the numerator is the power. Always take the root first — it keeps the numbers small. For $8^{2/3}$, do the cube root ($\sqrt[3]{8}=2$) before squaring ($2^2=4$), rather than cubing $8$ to $512$ and then hunting for its cube root.
Worked example 1
Handle the numbers and the powers as two separate streams — coefficients with coefficients, $x$'s with $x$'s. Trying to do both at once is where slips happen. Notice we never expanded anything; the laws did all the work.
Worked example 2
We did three things in the cleanest order: reciprocal (from the minus), root (from the denominator 4), power (from the numerator 3). Doing them in any order gives the same answer mathematically, but this order keeps every number tiny — you never handle anything bigger than $81$. That's the difference between a two-line answer and a page of arithmetic.
Worked example 3
A surd is a root that can't be written exactly as a fraction — an irrational number like $\sqrt{3}$. We simplify surds by pulling out the largest square factor hiding inside.
Surds behave exactly like algebra: $\sqrt3$ acts as a single 'unit', so $5\sqrt3-2\sqrt3=3\sqrt3$ works just like $5x-2x=3x$. And rationalising is just multiplying by a cleverly-disguised $1$ (here $\tfrac{\sqrt3}{\sqrt3}$) — it changes the appearance without changing the value. Spotting the largest square factor in step 1 is the skill examiners reward: use $25$, not $5$, or you'll be simplifying twice.
$\sqrt{a+b}$ is not $\sqrt a + \sqrt b$. Test it: $\sqrt{9+16}=\sqrt{25}=5$, but $\sqrt9+\sqrt{16}=3+4=7$. The root of a sum is not the sum of the roots — roots only split over multiplication and division, never over addition. This single error costs more marks than any other in this topic.
Check yourself
Given that $\dfrac{2\sqrt{5}+\sqrt{20}}{\sqrt{5}}$ can be written as an integer, find its value.
(Hint: simplify $\sqrt{20}$ first, then split the fraction.)
The trick is recognising that both terms on top are multiples of $\sqrt5$, so the denominator cancels cleanly — a small preview of how surds simplify inside fractions.