2.1 - Algebraic symbols and indices

Algebra lets you describe many calculations with one statement. You will read letters as numbers, use the usual rules of arithmetic, and simplify integer powers. The final part extends powers to roots for Higher learners.

Letters and relationships

How can one short expression count the cards in any number of packs? Suppose each pack holds 6 cards and you have 4 loose cards. One pack gives $6+4=10$ ; two packs give $6\times2+4=16$ . We want a statement that works before we choose the number of packs.

You already know

A product counts equal groups. Powers record repeated multiplication, and brackets tell you which operations belong together.

Let $p$ be the number of packs. Then $6p+4$ is the total number of cards. Here $p$ can be a nonnegative integer: a count of whole packs cannot be negative or fractional.

Expression, equation and formula

An expression describes a value without an equals sign, such as $6p+4$ . An equation states that two expressions are equal, such as $6p+4=40$ . A formula is an equation relating quantities, such as $C=6p+4$ , where $C$ is the total number of cards.

The letter's role depends on the question. A variable can take different values. An unknown is a number to be found from a stated relationship. In $6p+4$ , we can vary $p$ . In $6p+4=40$ , the total fixes the unknown $p$ . A formula is also an equation; these names describe how we use it.

An identity is an equality true for every allowed value. For example, $ab=ba$ is an identity for real numbers. In contrast, $p+1=5$ holds only when $p=4$ . Letters do not have permanent jobs: $x$ can be a variable or an unknown.

Key idea

A letter stands for a number. Read its meaning and allowed values from the situation.

The same expression can describe a small or large collection. This is why algebra is useful: you write the relationship once, then choose the input later.

Arithmetic with letters

Letters follow the arithmetic rules you use with numbers. You can reorder a product because $a\times b=b\times a$ , and regroup it because $(ab)c=a(bc)$ . Addition has these properties too. Subtraction and division generally do not: $7-2\ne2-7$ .

Coefficient

A coefficient is the numerical multiplier of a term. In $3ab$ , the coefficient is 3. The notation means $3\times a\times b$ , with the number written first. It does not mean $3+a+b$ .

Repeated letters multiply: $aa=a^2$ , whereas $a+a=2a$ . A fraction bar groups its whole numerator: $\frac{a+b}{3}$ divides the sum by 3. It differs from $a+\frac b3$ . These conventions are not supplied in either tier's formulae sheet: learn them.

Worked example 1: arranging a product

Question. Simplify $3ab\times2a$ .

Plan. Group the numbers, then group matching letters.

\begin{aligned} 3ab\times2a&=(3\times2)(a\times a)b &&\text{reorder factors}\\ &=6a^2b &&\text{multiply and use a power} \end{aligned}

Why this step? Reordering makes the repeated factor $a$ visible.

Answer. $6a^2b$ .

Check. At $a=2$ , $b=3$ , the original gives $18\times4=72$ , and $6(2)^2(3)=72$ .

Now change only the final letter. Watch which variable repeats; the numerical calculation stays the same.

Worked example 2: changing the repeated letter

Question. Simplify $3ab\times2b$ .

Plan. Group the numerical factors and the two factors of $b$ .

\begin{aligned} 3ab\times2b&=(3\times2)a(b\times b) &&\text{reorder factors}\\ &=6ab^2 &&\text{write the repeated factor} \end{aligned}

Why this step? The power belongs to $b$ , because $b$ is the letter used twice.

Answer. $6ab^2$ .

Check. At $a=2$ , $b=3$ , both expressions give $108$ .

Division acts on the coefficient too: $\frac{15a}{3}=5a$ . Operation order is unchanged: at $x=2$ , $4+3x^2=4+3(4)=16$ . Square before multiplying, then add.

Watch out

The line $a\times a=2a$ is wrong: it confuses multiplication with addition. The correct line is $a\times a=a^2$ . At $a=3$ , the two proposed answers are 9 and 6, so they cannot be identical.

Try the product with a different numerical coefficient. A numerical check can expose an error, but one matching value does not prove an identity.

Zero and integer powers

Powers count factors, not additions: $x^3=xxx$ . Multiplying $x^3$ by $x^2$ joins three factors and two factors, so the result is $x^5$ . Dividing cancels factors; raising a power to a power repeats the whole group.

Integer index laws

x^m x^n=x^{m+n},\qquad \frac{x^m}{x^n}=x^{m-n},\qquad (x^m)^n=x^{mn}

For integer indices, use these wherever the original expressions are defined. For $x\ne0$ ,

x^0=1,\qquad x^{-n}=\frac1{x^n}\quad(n\text{ a positive integer})

Not in either tier's formulae sheet: learn them.

Dividing $x^3$ by itself gives 1, while the division law gives $x^0$ . This explains the zero-power rule for nonzero $x$ . Continuing backwards through the powers gives reciprocals. A negative index does not make a number negative: $2^{-3}=\frac18$ . We do not use these rules to assign a value to $0^0$ .

Worked example 3: a positive denominator index

Question. For $x\ne0$ , simplify $\frac{x^3x^{-5}}{x}$ .

Plan. Add the numerator indices, then subtract the denominator index.

\begin{aligned} \frac{x^3x^{-5}}{x}&=\frac{x^{3+(-5)}}{x^1} &&\text{same-base product}\\ &=x^{-2-1}=x^{-3} &&\text{subtract indices}\\ &=\frac1{x^3} &&\text{use the reciprocal} \end{aligned}

Why this step? The denominator has index 1, even though that index is usually unwritten.

Answer. $\frac1{x^3}$ , for $x\ne0$ .

Check. At $x=2$ , $\frac{8\times(1/32)}2=\frac18$ .

Change only the denominator's index from positive to negative. Subtracting a negative is the new decision.

Worked example 4: a negative denominator index

Question. For $x\ne0$ , simplify $\frac{x^3x^{-5}}{x^{-1}}$ .

Plan. Combine the numerator, then subtract the whole denominator index.

\begin{aligned} \frac{x^3x^{-5}}{x^{-1}}&=\frac{x^{-2}}{x^{-1}} &&\text{add numerator indices}\\ &=x^{-2-(-1)}=x^{-1} &&\text{subtract the negative}\\ &=\frac1x &&\text{write a positive index} \end{aligned}

Why this step? Division subtracts the index $-1$ ; it does not subtract 1.

Answer. $\frac1x$ , for $x\ne0$ .

Check. At $x=2$ , $\frac{8\times(1/32)}{1/2}=\frac12$ .

For a power of a product, every factor is raised: $(3x)^2=9x^2$ . Brackets also identify a negative base: $(-2)^4=16$ , while $-2^4=-16$ . As a surprising scale check, just ten factors of 2 already give $2^{10}=1024$ , more than a thousand.

Watch out

The line $x^2+x^3=x^5$ is wrong. The add-indices law applies to a product, not a sum. At $x=2$ , the sum is $4+8=12$ , while $x^5=32$ . Leave unlike powers as a sum.

In a bracketed power, Pearson expects the coefficient to be raised as well as each letter. Write that step so it is easy to check.

Fractional powers (Higher)

A fractional index extends the same laws to roots. Since $(\sqrt{x})^2=x$ , the index on a square root must be $\frac12$ : multiplying it by 2 gives 1.

Fractional index laws

For $x>0$ and positive integers $m$ and $n$ ,

x^{1/n}=\sqrt[n]{x},\qquad x^{m/n}=\left(\sqrt[n]{x}\right)^m,\qquad x^{-m/n}=\frac1{x^{m/n}}

Not in the Higher formulae sheet: learn them. Fractional indices are not required at Foundation.

The denominator names the root; the numerator names the power. An even root denotes the nonnegative principal root, so $16^{1/2}=4$ , not $\pm4$ . Solving an equation is a different task. Zero is allowed for positive fractional powers but not negative ones. Odd roots can be real for negative numbers, such as $\sqrt[3]{-8}=-2$ ; even roots of negative numbers are not real. Positive bases keep the general laws here unambiguous.

Worked example 5: root before power

Question. Evaluate $16^{3/4}$ .

Plan. Take the fourth root first, then cube it.

\begin{aligned} 16^{3/4}&=(\sqrt[4]{16})^3 &&\text{denominator means root}\\ &=2^3 &&\text{since }2^4=16\\ &=8 &&\text{cube the root} \end{aligned}

Why this step? Root first keeps the arithmetic small and exact.

Answer. 8.

Check. $8^4=4096=16^3$ , as the index law predicts.

Now change only the sign of the index. The root and cube stay the same; the negative sign adds a reciprocal.

Worked example 6: a negative fractional index

Question. Evaluate $16^{-3/4}$ .

Plan. Find the positive fractional power, then take its reciprocal.

\begin{aligned} 16^{-3/4}&=\frac1{16^{3/4}} &&\text{negative means reciprocal}\\ &=\frac1{(\sqrt[4]{16})^3} &&\text{root then power}\\ &=\frac18 &&\text{use }2^3=8 \end{aligned}

Why this step? A negative index changes the operation to a reciprocal, not the sign of the answer.

Answer. $\frac18$ .

Check. $16^{3/4}\times16^{-3/4}=8\times\frac18=1$ .

The same reasoning gives $27^{-2/3}=\frac1{3^2}=\frac19$ . For $x>0$ , $(x^{1/2})^3=x^{3/2}$ ; multiplication and division still add and subtract indices.

Watch out

The line $16^{3/4}=\frac{16^3}{4}$ is wrong. The fraction is an index, not a division outside the power. The correct meaning is $(\sqrt[4]{16})^3$ .

The opening card question needs no chosen input: $6p+4$ counts every allowed number of packs. Powers give the same economy when factors repeat.

Quick Recap

Letters represent numbers; the situation decides whether they vary or are unknown.
Expressions follow ordinary arithmetic order and conventions.
Same-base products add indices; quotients subtract them; a power of a power multiplies them.
Zero and negative integer indices belong to both tiers; fractional indices add roots at Higher.

The opening answer is $6p+4$ : six cards per pack, plus four loose cards, for any allowed $p$ .

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2.1 - Algebraic symbols and indices

Letters and relationships

Arithmetic with letters

Zero and integer powers

Fractional powers (Higher)

Quick Recap

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