Write numbers in scientific (standard) form as a number between 1 and 10 times a power of 10, convert back to a numeral, round to a given number of significant figures, and operate with numbers in standard form

What this dot point is asking

NESA wants you to do four things. Write very large and very small numbers in standard form (also called scientific notation), $a \times 10^{n}$ with the front number $a$ between $1$ and $10$. Convert a number in standard form back to an ordinary numeral. Count and round to a given number of significant figures. And operate with numbers in standard form by combining the powers of $10$. The arithmetic is light, but the marks turn on three things done precisely. Get the power of $10$ right, including its sign. Count significant figures correctly through the zero cases. And re-normalise the answer, that is, shift the point so the front number stays between $1$ and $10$.

The answer

A measurement like the Earth's distance from the Sun, about $149\,600\,000$ km, or an atom's radius, about $0.00000000025$ m, is clumsy to write out and easy to miscount. Standard form rewrites any number as a single digit before the decimal point, times a power of $10$:

The front number $a$ holds the significant digits, and the power $n$ records the size. The rule for $n$ follows from what a power of $10$ does. Each step in the power multiplies or divides by $10$, which moves the decimal point one place. So count how far the decimal point moves to get one non-zero digit in front, and that count is the power. A large number needs the point moved left, which gives a positive power. A small number (less than $1$) needs it moved right, which gives a negative power.

Writing a large number in standard form

For a number bigger than $10$, put the decimal point after the first non-zero digit and count the places back to where it started (the right-hand end). The Earth-Sun distance is

because the point moves $8$ places left. Notice $a = 1.496$ keeps only the meaningful digits and the trailing zeros are absorbed into the power. A useful sanity check: $10^{8}$ is a hundred million, and $1.496 \times$ a hundred million is about $150$ million, which matches.

Writing a small number in standard form

For a number between $0$ and $1$, the point moves the other way, to the right, until it sits just after the first non-zero digit, and the power is negative. An atom's radius of

has the point moving $10$ places right. The negative power is not a negative number; $2.5 \times 10^{-10}$ is a tiny positive length. Counting the places carefully is the whole job here, because it is easy to be one out.

Converting back to an ordinary numeral

To undo standard form, read the power as the number of places to move the point and fill the gaps with zeros. A positive power moves the point right (the number grows); a negative power moves it left (the number shrinks):

In the first the point moves $5$ places right; in the second it moves $4$ places left, so there are three zeros between the point and the $6$.

Reading standard form on a calculator

A calculator shows standard form to save space. A display like 6.022E23 or 6.022^23 means $6.022 \times 10^{23}$, not $6.022^{23}$. The E (or a small raised number) stands for "times ten to the power". A display of 4.5E-7 means $4.5 \times 10^{-7}$. To enter standard form, use the calculator's EXP or \times 10^{x} key rather than typing 10 and a power separately. That way the machine treats it as one number. Always rewrite the calculator's shorthand into proper standard form, $a \times 10^{n}$, in your written answer.

Significant figures

Significant figures are the digits in a number that carry real information about its size and accuracy, as opposed to zeros that are only holding place value. The rules for which digits count are:

• Every non-zero digit is significant. In $4\,378$ all four digits count.
• Zeros between non-zero digits are significant. In $56\,091$ all five digits count, including the $0$.
• Leading zeros (at the start of a small number) are not significant; they only place the decimal point. $0.00871$ has three significant figures: $8$, $7$, $1$.
• Trailing zeros after a decimal point are significant, because there is no reason to write them unless they were measured. $51.340$ has five significant figures.

The one genuinely ambiguous case is trailing zeros in a whole number with no decimal point. A figure like $8\,000$ could be accurate to one, two, three or four significant figures, and you cannot tell which from the numeral alone. Standard form removes the ambiguity: $8 \times 10^{3}$ shows one significant figure, $8.0 \times 10^{3}$ shows two, $8.00 \times 10^{3}$ shows three. This is one of the quiet reasons standard form is the natural home for significant figures.

Rounding to a given number of significant figures

To round to $n$ significant figures, keep the first $n$ significant digits, then look at the next digit: if it is $5$ or more, round the last kept digit up; otherwise leave it. The place-holding zeros (or the power of $10$) are then set to keep the number the right size. For example,

In the first, the first three significant figures are $4, 7, 8$, the next digit is $2$ so we round down, and the trailing zeros keep the value near $48\,000$. In the second, the leading zeros do not count, so the first three significant figures are $3, 0, 4$, and the next digit $6$ rounds the $4$ up to $5$.

The case that catches people is when rounding up causes a carry. Rounding $0.09962$ to two significant figures: the first two significant figures are $9, 9$, the next digit is $6$, so $99$ rounds up to $100$. That extra digit changes the size, and the tidy way to record it is in standard form, $1.0 \times 10^{-1}$ (which equals $0.10$). The same happens with $9\,970$ to two significant figures: $99$ becomes $100$, giving $10\,000 = 1.0 \times 10^{4}$. Whenever the leading digits are all $9$s, expect the power to step up by one.

Multiplying and dividing in standard form

Standard form makes multiplying and dividing big or small numbers easy, because the front numbers and the powers are handled separately. Group the front numbers together and the powers together; multiplying powers of $10$ adds the indices, and dividing subtracts them:

For example $(3.0 \times 10^{8}) \times (4.2 \times 10^{-3}) = 12.6 \times 10^{5}$. The front number $12.6$ is not between $1$ and $10$, so re-normalise: $12.6 = 1.26 \times 10^{1}$, which lifts the power by one to give $1.26 \times 10^{6}$. Re-normalising is the step most often forgotten; an answer is only in standard form once the front number is a single non-zero digit before the point.

How exam questions ask about scientific notation and significant figures

The wording varies, but each phrasing maps to one task:

• "Write / express in standard form (scientific notation)" means rewrite as $a \times 10^{n}$ with $1 \le a < 10$; decide the sign of the power from whether the number is large or small.
• "Write as an ordinary numeral / basic numeral" means undo standard form by moving the point the number of places given by the power.
• "Correct to $n$ significant figures" means keep $n$ significant digits and round on the next digit; watch the requested number, and watch for a carry.
• "Evaluate and express your answer in standard form" means combine the front numbers and the powers, then re-normalise the front number to lie between $1$ and $10$.
• "How many times larger / heavier / longer" means divide one quantity by the other; in standard form, divide the front numbers and subtract the powers.
• "Express this measurement in ... (a different unit)" combines unit conversion with standard form: convert first, then write the result as $a \times 10^{n}$.