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Scientific Notation

Scientists write the mass of an electron and the distance to the Sun without drowning in zeros. Standard form is the tidy shorthand that makes very large and very small numbers easy to read, compare and calculate with.

MYP 4Extended MathsNumberCriteria A · C · D~9 min read

Written by Fayek · MYP student

What standard form is

Standard form (also called scientific notation) writes any number as a single digit before the decimal point, multiplied by a power of ten:

\[ a \times 10^{n}, \qquad \text{where } 1 \le a < 10 \text{ and } n \text{ is an integer.} \]

The two conditions do all the work. The number \(a\), called the coefficient, must be at least 1 but under 10, so there is exactly one non-zero digit in front of the point. The power \(n\) records how big or small the number is: a positive \(n\) means a large number, a negative \(n\) means a small one.

Coefficient
The value \(a\) in \(a \times 10^n\), always in the range \(1 \le a < 10\).
Exponent
The integer power \(n\), equal to the number of places the decimal point moves.

Converting to standard form

Slide the decimal point until just one non-zero digit sits in front of it, then count the moves. Moving left gives a positive power (the original was large); moving right gives a negative power (the original was small).

Worked example

Write \(48\,500\) and \(0.0032\) in standard form.

1
For \(48\,500\), place the point after the 4 to get \(4.85\). The point moved 4 places to the left, so \(n = 4\).
2
So \(48\,500 = 4.85 \times 10^{4}\).
3
For \(0.0032\), the point moves right past three zeros to sit after the 3, giving \(3.2\). Moving right means a negative power, \(n = -3\).
\(48\,500 = 4.85 \times 10^{4}\) and \(0.0032 = 3.2 \times 10^{-3}\)

Keep the coefficient in range

\(48.5 \times 10^{3}\) has the right value but is not standard form, because \(48.5\) is bigger than 10. The coefficient must satisfy \(1 \le a < 10\), so it becomes \(4.85 \times 10^{4}\).

Converting from standard form

Reading a number back is the same move in reverse. The power tells you how many places to shift the point: shift right for a positive power, left for a negative one, filling any gaps with zeros.

Standard formMove the pointOrdinary number
\(6.1 \times 10^{5}\)5 places right\(610\,000\)
\(2.9 \times 10^{2}\)2 places right\(290\)
\(7.4 \times 10^{-3}\)3 places left\(0.0074\)

Multiplying and dividing in standard form

This is where standard form pays off. Handle the coefficients and the powers separately: multiply or divide the coefficients as normal, and use the index laws \(10^{m} \times 10^{n} = 10^{m+n}\) and \(10^{m} \div 10^{n} = 10^{m-n}\) on the tens. Then tidy the answer back into range.

Worked example

Work out \((6 \times 10^{4}) \times (5 \times 10^{-2})\), giving your answer in standard form.

1
Multiply the coefficients: \(6 \times 5 = 30\).
2
Add the powers of ten: \(10^{4} \times 10^{-2} = 10^{4 + (-2)} = 10^{2}\).
3
Combine: \(30 \times 10^{2}\). But \(30\) is not in the range \(1 \le a < 10\).
4
Rewrite \(30 = 3 \times 10^{1}\), so \(30 \times 10^{2} = 3 \times 10^{1} \times 10^{2} = 3 \times 10^{3}\).
\(3 \times 10^{3}\) (that is \(3000\))

A quick sanity check: \(6 \times 10^{4} = 60\,000\) and \(5 \times 10^{-2} = 0.05\), and \(60\,000 \times 0.05 = 3000\). The standard-form answer agrees.

Worked example

Work out \((8 \times 10^{7}) \div (2 \times 10^{3})\).

1
Divide the coefficients: \(8 \div 2 = 4\).
2
Subtract the powers: \(10^{7} \div 10^{3} = 10^{7 - 3} = 10^{4}\).
3
The coefficient \(4\) is already in range, so no tidying is needed.
\(4 \times 10^{4}\) (that is \(40\,000\))

Where this is assessed

Choosing and applying the index laws is Criterion A, and writing a properly formatted final answer with \(1 \le a < 10\) is Criterion C. Standard form also turns up in Criterion D real-life contexts such as astronomy, populations and atomic scales.

Check yourself

1. Write \(45\,600\) in standard form. +

Move the point after the 4 to get \(4.56\). It moved 4 places left, so the answer is \(\mathbf{4.56 \times 10^{4}}\).

2. Write \(3.2 \times 10^{-3}\) as an ordinary number. +

A power of \(-3\) shifts the point 3 places left, filling zeros: \(3.2 \to 0.0032\). The answer is \(\mathbf{0.0032}\).

3. Work out \((9 \times 10^{6}) \div (3 \times 10^{2})\) in standard form. +

Coefficients: \(9 \div 3 = 3\). Powers: \(10^{6} \div 10^{2} = 10^{4}\). The coefficient is in range, so the answer is \(\mathbf{3 \times 10^{4}}\) (that is \(30\,000\)).

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