Converting Numbers to and from Standard Form

Hello! This final module brings together place value and indices. You will use powers of ten to write very large and very small numbers compactly, then convert them back to ordinary numbers. The aim is not just to count decimal places: it is to understand why the power of ten preserves the number’s value.

By the end of the lesson, you should be able to convert in both directions, decide whether a power should be positive or negative, and check that an answer is genuinely in standard form.

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What standard form means

Our place-value system is built around ten. Multiplying by $10$ makes a number ten times larger; dividing by $10$ makes it ten times smaller. An index records how many times this happens:

$$10^3=1000,10^0=1,10^{-3}=\frac{1}{1000}=\mathrm{0.001.}$$

Standard form, also called scientific notation, uses this pattern to write a number as

$$a\times 10^n,$$

where $1\le a<10$ and $n$ is an integer (a positive or negative whole number, or zero). The number $a$ is the coefficient; $n$ is the index or exponent. The coefficient may equal $1$, but it may not equal $10$.

For example, $4.5\times 10^3$ is in standard form, while $45\times 10^2$ is not: $45$ is too large for the coefficient. Both expressions have the same value, but only the first follows the standard-form rule.

The two examples in this image show the sign rule at a glance. In each case, check that multiplying the coefficient by the stated power of ten gives the original number.

A useful sense-check follows from $1\le a<10$: a positive index gives a number at least $10$; a negative index gives a number between $0$ and $1$; and an index of zero leaves $a$ unchanged. For instance,

$$7.2=7.2\times 10^0.$$

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Ordinary number to standard form

First, make the coefficient a number from $1$ up to—but not including—$10$. Then count how many places its decimal point is from the original decimal point. Finally, choose the sign that gives the original number its correct size.

Consider $72,000,000$. Its coefficient must be $7.2$. To get back from $7.2$ to $72,000,000$, you multiply by ten seven times, so

$$72,000,000=7.2\times 10^7.$$

For a small number, consider $0.000406$. Keep the zero between the $4$ and $6$: the coefficient is $4.06$, not $4.6$. Dividing $4.06$ by ten four times gives the original number, so

$$0.000406=4.06\times 10^{-4}.$$

The direction you move the decimal point while finding the coefficient can be confusing: you moved it right to turn $0.000406$ into $4.06$, yet the final index is negative. To choose the sign reliably, ask: What operation takes my coefficient back to the original number? Here, you must divide, so the index is negative.

Watch the conversions below before reading another pair of examples.

Sometimes the coefficient is a single digit. The first non-zero digit in $0.00000009$ is in the eighth decimal place. Moving from $9$ to that position requires division by ten eight times:

$$0.00000009=9\times 10^{-8}.$$

Do not find this index merely by counting visible zeros. Counting place-value moves, or checking that $9\times 10^{-8}$ gives the original decimal, is more dependable.

For additional worked examples, read the large- and small-number methods in BBC Bitesize.

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Standard form to ordinary notation

Now reverse the process. A positive index means multiply the coefficient by ten that many times; in the written number, the decimal point ends up further right. A negative index means divide by ten that many times; the decimal point ends up further left. Add placeholder zeros where needed.

For a positive index, start with $3.06\times 10^5$. Multiplying by $100,000$ gives

$$3.06\times 10^5=306,000.$$

The zero in $3.06$ still matters: $306,000$ is not the same number as $36,000$.

For a negative index, start with $5.02\times 10^{-4}$. Divide $5.02$ by ten four times:

$$5.02\times 10^{-4}=\mathrm{0.000502.}$$

You can check the decimal-place count by converting your answer back. Starting at the decimal point in $0.000502$, it takes four places to put the point after the $5$, producing $5.02$. Because the ordinary number is smaller than $1$, the index is negative.

Think “scale,” not “zero count.” The index tells you how many factors of ten change the coefficient’s size. Zeros simply hold empty place-value positions when you write the result.

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Checking answers and spotting near misses

A conversion is complete only when both parts are right: the expression has the correct value and its coefficient satisfies $1\le a<10$.

Suppose you are given $0.52\times 10^{-3}$ and asked to write it in standard form. The coefficient $0.52$ is too small. Multiplying it by $10$ makes $5.2$, so you must divide the power-of-ten part by $10$ to keep the overall value unchanged:

$$0.52\times 10^{-3}=5.2\times 10^{-4}.$$

This also passes a size check: $5.2\times 10^{-4}=0.00052$, a small positive number. Similarly, $10\times 10^4$ has an invalid coefficient of $10$; its standard-form version is $1\times 10^5$.

When checking GCSE-style conversions, use these three questions in order:

1. Coefficient: Is it at least $1$ but less than $10$?
2. Index: Is it an integer, with a sign that matches the number’s size?
3. Value: If you convert back, do you recover the original number—including any zeros between non-zero digits?

These checks catch different errors. For example, $4.6\times 10^{-4}$ has a valid coefficient and a plausible negative index, but it does not equal $0.000406$. Converting back reveals the lost zero.

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Key takeaways

Standard form writes a positive number as $a\times 10^n$, with $1\le a<10$ and an integer $n$. To convert into standard form, find the coefficient, count the place-value moves, and choose the index’s sign by asking what takes the coefficient back to the original number. To convert out, multiply or divide by ten the stated number of times, filling empty places with zeros.

You have reached the final topic in this unit. Work through the lesson checks and use any mistakes to identify whether you need more practice with the coefficient, the number of places, or the sign before attempting a full-unit test.