Standard Form

Standard form is a compact way of writing very large or very small numbers using powers of ten. It is widely used in science and appears across the Number strand of OCR GCSE Mathematics (J560). This lesson covers writing numbers in standard form and back again, the four operations in standard form (multiplying, dividing, adding and subtracting), and how to enter and read standard form on a calculator for Papers 2 and 3.

This lesson mainly builds AO1 fluency in converting and calculating, with AO2 reasoning when you interpret what a power of ten tells you about size, and AO3 problem-solving in the contextual examples (populations, distances, particle sizes).

Key Vocabulary

Term	Meaning
Standard form	A number written as $A \times 10^{n}$A×10n with $1 \le A < 10$1≤A<10 and $n$n an integer
Power of ten	$10^{n}$10n, where $n$n is positive for large numbers and negative for small
Ordinary number	A number written in full, not in standard form
Coefficient ($A$A)	The number part, always at least $1$1 and less than $10$10
Mantissa	Another name for the coefficient $A$A

Why Standard Form Is Useful

Imagine writing the distance to the Sun (about $150{,}000{,}000$150,000,000 km) or the diameter of a red blood cell (about $0.000\,008$0.000008 m) in full every time. The numbers are awkward, easy to mis-copy, and hard to compare at a glance. Standard form solves this by separating a number into a tidy coefficient and a power of ten that records its size. The Sun's distance becomes $1.5 \times 10^{8}$1.5×108 km and the cell's diameter $8 \times 10^{-6}$8×10−6 m — compact, and the powers $8$8 and $-6$−6 tell you instantly which is large and which is tiny. This is why scientists and your calculator both use it.

What Standard Form Is

A number is in standard form when it is written as

$A \times 10^{n}, \qquad 1 \le A < 10, \quad n \in \mathbb{Z}.$A×10n,1≤A<10,n∈Z.

A positive power means a large number (move the point right); a negative power means a small number between $0$0 and $1$1 (move the point left). The power $n$n counts how many places the decimal point moves. The two conditions are equally important: $A$A must be at least $1$1 and less than $10$10 (so exactly one non-zero digit sits before the decimal point), and $n$n must be a whole number. If either fails, the number is not yet in correct standard form and you must re-normalise it.

Worked Example 1

Write $4{,}730{,}000$4,730,000 in standard form.

Place the decimal point after the first non-zero digit: $4.73$4.73. Count how many places it moved: from $4{,}730{,}000.$4,730,000. to $4.73$4.73 is $6$6 places left, so $n = 6$n=6.

$4{,}730{,}000 = 4.73 \times 10^{6}.$4,730,000=4.73×106.

Answer: $4.73 \times 10^{6}$4.73×106.

Worked Example 2

Write $0.000\,82$0.00082 in standard form.

The first non-zero digit is $8$8, giving $A = 8.2$A=8.2. The point moves $4$4 places right to get there, so the power is $-4$−4.

$0.000\,82 = 8.2 \times 10^{-4}.$0.00082=8.2×10−4.

Answer: $8.2 \times 10^{-4}$8.2×10−4.

Common error: Writing $82 \times 10^{-5}$82×10−5 or $0.82 \times 10^{-3}$0.82×10−3. The coefficient must satisfy $1 \le A < 10$1≤A<10, so $82$82 and $0.82$0.82 are both invalid forms.

Worked Example 3

Write $3.6 \times 10^{5}$3.6×105 and $9.1 \times 10^{-3}$9.1×10−3 as ordinary numbers.

$3.6 \times 10^{5}$3.6×105: move the point $5$5 places right → $360{,}000$360,000. $9.1 \times 10^{-3}$9.1×10−3: move the point $3$3 places left → $0.0091$0.0091.

Answers: $360{,}000$360,000 and $0.0091$0.0091.

Multiplying and Dividing in Standard Form

Deal with the coefficients and the powers separately: multiply/divide the coefficients, and add/subtract the indices (using the index laws). Then adjust so the coefficient is back in range $1 \le A < 10$1≤A<10.

Worked Example 4

Work out $(3 \times 10^{4}) \times (2 \times 10^{6})$(3×104)×(2×106). Give your answer in standard form.

Coefficients: $3 \times 2 = 6$3×2=6. Powers: $10^{4} \times 10^{6} = 10^{10}$104×106=1010.

$(3 \times 10^{4}) \times (2 \times 10^{6}) = 6 \times 10^{10}.$(3×104)×(2×106)=6×1010.

Answer: $6 \times 10^{10}$6×1010.

Worked Example 5

Work out $(8 \times 10^{5}) \div (2 \times 10^{9})$(8×105)÷(2×109).

Coefficients: $8 \div 2 = 4$8÷2=4. Powers: $10^{5} \div 10^{9} = 10^{-4}$105÷109=10−4.

$= 4 \times 10^{-4}.$=4×10−4.

Answer: $4 \times 10^{-4}$4×10−4.

Worked Example 6 (adjusting the coefficient)

Work out $(6 \times 10^{7}) \times (5 \times 10^{3})$(6×107)×(5×103).

Coefficients: $6 \times 5 = 30$6×5=30. Powers: $10^{7} \times 10^{3} = 10^{10}$107×103=1010. This gives $30 \times 10^{10}$30×1010, but $30$30 is not in range. Rewrite $30 = 3 \times 10^{1}$30=3×101:

$30 \times 10^{10} = 3 \times 10^{1} \times 10^{10} = 3 \times 10^{11}.$30×1010=3×101×1010=3×1011.

Answer: $3 \times 10^{11}$3×1011.

Common error: Leaving the answer as $30 \times 10^{10}$30×1010. A standard-form answer must have a coefficient between $1$1 and $10$10.

Adding and Subtracting in Standard Form

For $+$+ and $-$− the powers must match first. Either convert to ordinary numbers, or rewrite one number so both share the same power of ten, then add/subtract the coefficients.

Worked Example 7

Work out $(4.2 \times 10^{6}) + (3 \times 10^{5})$(4.2×106)+(3×105). Give your answer in standard form.

Make the powers match. Rewrite $3 \times 10^{5} = 0.3 \times 10^{6}$3×105=0.3×106. Now add the coefficients:

$4.2 \times 10^{6} + 0.3 \times 10^{6} = 4.5 \times 10^{6}.$4.2×106+0.3×106=4.5×106.

Answer: $4.5 \times 10^{6}$4.5×106.

OCR Exam Tip: For $+$+ and $-$−, you cannot just add the coefficients unless the powers are equal. If unsure, convert both to ordinary numbers, add, then convert back.

Standard Form on a Calculator

On the calculator papers, use the $\boxed{\times 10^{x}}$×10x​ or $\boxed{\text{EXP}}$EXP​ key to enter the power — not the multiplication and "$10$10" keys, which is slower and error-prone. To enter $5.2 \times 10^{-3}$5.2×10−3, type $5.2$5.2, then the power key, then $-3$−3. A display such as $4.6^{08}$4.608 means $4.6 \times 10^{8}$4.6×108; rewrite it properly in your answer.

Worked Example 8

Using a calculator, work out $\dfrac{(7.2 \times 10^{8})}{(1.5 \times 10^{-2})}$(1.5×10−2)(7.2×108)​. Give your answer in standard form.

Enter each value with the power key. Coefficients: $7.2 \div 1.5 = 4.8$7.2÷1.5=4.8. Powers: $10^{8} \div 10^{-2} = 10^{10}$108÷10−2=1010.

$= 4.8 \times 10^{10}.$=4.8×1010.

Answer: $4.8 \times 10^{10}$4.8×1010.

Comparing and Ordering Numbers in Standard Form

To compare numbers in standard form, look at the power of ten first — the larger the power, the larger the number (for positive numbers). Only if two numbers share the same power do you compare their coefficients.

Worked Example 8b

Write these in ascending order: $3.1 \times 10^{5}$3.1×105, $9.8 \times 10^{4}$9.8×104, $1.2 \times 10^{5}$1.2×105, $4.5 \times 10^{6}$4.5×106.

Compare powers: the $10^{4}$104 number is smallest, the $10^{6}$106 number is largest. The two $10^{5}$105 numbers are compared by coefficient: $1.2 < 3.1$1.2<3.1.

Answer: $9.8 \times 10^{4}, \; 1.2 \times 10^{5}, \; 3.1 \times 10^{5}, \; 4.5 \times 10^{6}$9.8×104,1.2×105,3.1×105,4.5×106.

Common error: Comparing the coefficients first and concluding $9.8 \times 10^{4}$9.8×104 is the largest "because $9.8$9.8 is biggest". The power of ten dominates — $9.8 \times 10^{4} = 98{,}000$9.8×104=98,000 is far smaller than $1.2 \times 10^{5} = 120{,}000$1.2×105=120,000.

OCR Exam Tip: With negative powers, a more negative power means a smaller number: $2 \times 10^{-7}$2×10−7 is smaller than $9 \times 10^{-5}$9×10−5, because $10^{-7}$10−7 is much tinier than $10^{-5}$10−5.

Extended Worked Examples

Worked Example 9: Subtraction needing a coefficient adjustment

Work out $(5 \times 10^{4}) - (8 \times 10^{3})$(5×104)−(8×103). Give your answer in standard form.

Match powers: $8 \times 10^{3} = 0.8 \times 10^{4}$8×103=0.8×104. Subtract: $5 \times 10^{4} - 0.8 \times 10^{4} = 4.2 \times 10^{4}$5×104−0.8×104=4.2×104.

Answer: $4.2 \times 10^{4}$4.2×104.

Worked Example 10: A real-world large-number calculation

Light travels about $3 \times 10^{8}$3×108 metres per second. Work out how far it travels in $1$1 hour ($3600$3600 s). Give your answer in standard form.

Distance $=$= speed $\times$× time $= (3 \times 10^{8}) \times 3600$=(3×108)×3600. Write $3600 = 3.6 \times 10^{3}$3600=3.6×103:

$(3 \times 10^{8}) \times (3.6 \times 10^{3}) = 10.8 \times 10^{11} = 1.08 \times 10^{12} \text{ m}.$(3×108)×(3.6×103)=10.8×1011=1.08×1012 m.

Answer: $1.08 \times 10^{12}$1.08×1012 m (note the adjustment from $10.8$10.8 to $1.08 \times 10^{1}$1.08×101).

Worked Example 11: A small-number context

A grain of fine sand has a mass of about $6.5 \times 10^{-5}$6.5×10−5 g. Work out the mass of $20{,}000$20,000 such grains. Give your answer in standard form.

$20{,}000 = 2 \times 10^{4}$20,000=2×104. Mass $= (6.5 \times 10^{-5}) \times (2 \times 10^{4}) = 13 \times 10^{-1} = 1.3 \times 10^{0} = 1.3$=(6.5×10−5)×(2×104)=13×10−1=1.3×100=1.3 g.

Answer: $1.3$1.3 g (or $1.3 \times 10^{0}$1.3×100 g).