Standard Form (AI SL)
Standard form is a compact way to write very large or very small numbers as \(a\times10^k\), where \(1\le a<10\) and \(k\) is an integer. This topic covers writing numbers in and out of standard form, and the rules for multiplying, dividing, adding and subtracting numbers once they're written this way - all common territory for real-world contexts like astronomy, finance and data storage.
What the syllabus says
This topic maps onto one point in the official IB Applications & Interpretation syllabus.
| Code | Syllabus content |
|---|---|
| SL1.1 | Operations with numbers in the form \(a\times10^k\) where \(1\le a<10\) and \(k\) is an integer. Calculator or computer notation is not acceptable - for example \(5.2\text{E}30\) is not acceptable and should be written as \(5.2\times10^{30}\). |
This is a shared SL point common to both Applications & Interpretation and Analysis & Approaches.
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is standard form?
Standard form (also called scientific notation) writes a number as \(a\times10^k\), where \(a\) is between 1 and 10 (never reaching 10) and \(k\) is an integer. It makes very large or very small numbers compact and easy to compare.
e.g. \(45\,000 = 4.5\times10^{4}\).
How do you multiply numbers in standard form?
Multiply the two coefficients together, and add the two powers of 10. If the resulting coefficient is 10 or more, renormalise by shifting the decimal point and adjusting the power.
e.g. \((6.3\times10^{7})\times(4.0\times10^{-3}) = 25.2\times10^{4} = 2.52\times10^{5}\).
How do you divide numbers in standard form?
Divide the two coefficients, and subtract the power of 10 in the denominator from the power in the numerator.
e.g. \(\dfrac{3.6\times10^{12}}{6\times10^{7}} = 0.6\times10^{5} = 6\times10^{4}\).
How do you add or subtract numbers in standard form?
You can't just add the coefficients unless the powers of 10 already match. Rewrite one number so both share the same power of 10, then add or subtract the coefficients.
e.g. \(4.5\times10^{6}+8\times10^{5} = 4.5\times10^{6}+0.8\times10^{6} = 5.3\times10^{6}\).
How do you convert to an ordinary number?
Move the decimal point \(k\) places to the right if \(k\) is positive (a large number), or \(k\) places to the left if \(k\) is negative (a small number), filling any gaps with zeros.
e.g. \(1.5\times10^{8} = 150\,000\,000\).
Key formulas
There's really one definition and three operation rules to remember on this topic. The two tables below summarise them at a glance - the explanations underneath go into more depth on each one.
Formula reference
None of these are printed as formulas in the official booklet - standard form is a notation convention and calculation technique, treated as prior/common knowledge you're expected to apply directly.
| Rule | Used for | Booklet? |
|---|---|---|
| \(a\times10^k,\ 1\le a<10,\ k\in\mathbb{Z}\) | Definition of standard form | Not in the formula booklet - notation convention |
| \((a\times10^m)\times(b\times10^n)=ab\times10^{m+n}\) | Multiplying in standard form | Not in the formula booklet - prior knowledge |
| \(\dfrac{a\times10^m}{b\times10^n}=\dfrac{a}{b}\times10^{m-n}\) | Dividing in standard form | Not in the formula booklet - prior knowledge |
| Align powers of 10, then add/subtract coefficients | Adding or subtracting in standard form | Not in the formula booklet - prior knowledge |
Multiplying/dividing vs adding/subtracting
The two families of operations follow genuinely different rules - mixing them up is the single biggest source of errors on this topic.
| Feature | Multiply / divide | Add / subtract |
|---|---|---|
| What happens to the powers | Add (multiply) or subtract (divide) them directly | Must be made equal first - no shortcut |
| What happens to the coefficients | Multiply or divide them directly | Add or subtract them, once powers match |
| Extra step needed? | Only if the result isn't in the \(1\le a<10\) range | Always - align the powers before combining |
| Example | \((6.3\times10^{7})\times(4.0\times10^{-3})=2.52\times10^{5}\) | \(4.5\times10^{6}+8\times10^{5}=5.3\times10^{6}\) |
Converting to and from standard form
Every question on this topic starts by recognising which direction you're converting.
Ordinary number → standard form
Count how many places the decimal point moves to leave exactly one non-zero digit before it - that count (with its sign) is \(k\).
Standard form → ordinary number
Move the decimal point \(k\) places right (large numbers, \(k>0\)) or left (small numbers, \(k<0\)), padding with zeros as needed.
Calculator notation isn't acceptable
Your GDC may display \(5.2\text{E}30\), but that's calculator shorthand, not a valid final answer - always write it as \(5.2\times10^{30}\).
Operations in standard form
Once numbers are in \(a\times10^k\) form, these rules let you combine them without ever expanding to the full ordinary number.
Multiplying
\[ab\times10^{m+n}\]
Multiply the coefficients, add the powers.
Dividing
\[\dfrac{a}{b}\times10^{m-n}\]
Divide the coefficients, subtract the powers.
Adding / subtracting
Align the powers of 10 first, then combine the coefficients.
There's no shortcut that skips this alignment step.
Renormalising
If the coefficient lands outside \(1\le a<10\), shift the decimal point and adjust the power to compensate.
e.g. \(25.2\times10^4 \to 2.52\times10^5\).
Worked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
\((6.3\times10^{7})\times(4.0\times10^{-3}).\)
(a) Evaluate it, in standard form to 3 significant figures.
(b) Find \((6.3\times10^{7})\div(4.0\times10^{-3}),\) in standard form.
(c) Find \((6.3\times10^{7})+(4.0\times10^{-3}),\) in standard form to 3 significant figures.
Worked solution
(a) \(6.3\times4.0=25.2\) and \(10^{7+(-3)}=10^{4}.\) M1
\(25.2\times10^{4}=2.52\times10^{5}\) (3 s.f.). A1
(b) \(6.3\div4.0=1.575\) and \(10^{7-(-3)}=10^{10}.\) M1
\(1.575\times10^{10}.\) A1
(c) \(4.0\times10^{-3}\) is negligible next to \(6.3\times10^{7}\), so the sum is dominated by the larger term. M1
\(6.3\times10^{7}+4.0\times10^{-3}\approx6.30\times10^{7}\) (3 s.f.). A1
Two masses are \(4.5\times10^{6}\) kg and \(8\times10^{5}\) kg.
(a) Find their total, in standard form.
(b) Find the difference between the two masses, in standard form.
(c) Express the smaller mass as a percentage of the larger mass.
Worked solution
(a) \(4.5\times10^{6}+0.8\times10^{6}.\) M1
\(4.5\times10^{6}+0.8\times10^{6}=5.3\times10^{6}\) kg. A1
(b) \(4.5\times10^{6}-0.8\times10^{6}.\) M1
\(4.5\times10^{6}-0.8\times10^{6}=3.7\times10^{6}\) kg. A1
(c) \(\dfrac{8\times10^{5}}{4.5\times10^{6}}\times100.\) M1
\(\dfrac{8\times10^{5}}{4.5\times10^{6}}\times100\approx17.8\%.\) A1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Writing calculator notation as the final answer. \(5.2\text{E}30\) is your calculator's shorthand, not acceptable IB notation - always write it out as \(5.2\times10^{30}\).
- Leaving the coefficient outside \(1\le a<10\). A result like \(25.2\times10^{4}\) is correct arithmetic but not standard form - it must be renormalised to \(2.52\times10^{5}\).
- Adding or subtracting before aligning the powers of 10. You can't add \(4.5\times10^{6}\) and \(8\times10^{5}\) by just adding \(4.5\) and \(8\) - rewrite one term so both share the same power first.
- Sign errors when subtracting a negative index. Dividing \(10^{7}\) by \(10^{-3}\) means subtracting \(-3\) from \(7\), giving \(10^{10}\) - it's easy to miscount the sign and get \(10^{4}\) instead.
Using your GDC
Every step below is a real button sequence, not a vague "use your calculator" hint - covering the TI-84 Plus, TI-Nspire, and Casio fx-9860/fx-CG50. Pick your model to filter down to just the steps that apply to you.
For very large or very small numbers - avoids typing long strings of zeros and prevents rounding errors.
- Scientific notation means \(a\times10^n\), e.g. \(3.2\times10^{8}\) or \(4.5\times10^{-3}\).
- Use 2nd → , (EE) to enter the ×10 part: type 3.2 2nd , 8 to enter \(3.2\times10^{8}\). Do NOT type ×10^ separately.TI-84
- Use the EE key (or type ×10^ from the keyboard template) to enter scientific notation. Or just type 3.2×10^8 using the ^ key.Nspire
- Use the ×10ˣ key (EXP key) - type 3.2 then EXP then 8. Do NOT type ×10^ manually.Casio
- To display answers in scientific notation: on TI-84 press MODE and choose SCI; on Casio set the display mode in SET UP.
Tip: A common mistake is typing ×10^ instead of using the EE/EXP key - this gives ×10×... (multiplication, then a power) rather than proper scientific notation.
See the full GDC guide for more calculator models and topics.
Ready to practise properly?
Standard form questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
Why can't I write my calculator's E notation as my final answer?
The IB syllabus explicitly says calculator or computer notation like 5.2E30 is not acceptable as a final answer - it must be written as 5.2 times 10 to the power 30, with the multiplication sign and the power shown properly.
How do I add two numbers in standard form?
First rewrite both numbers with the same power of 10 (often the larger one), then add the coefficients and keep that power. Only after adding do you renormalise the result back into the 1 to less than 10 range if needed.
What does 1 is less than or equal to a, which is less than 10, mean?
It's the rule that defines standard form: the coefficient a must be at least 1 and strictly less than 10, so there's exactly one non-zero digit before the decimal point. 25.2 times 10 to the 4 is not standard form; 2.52 times 10 to the 5 is.
Will I need standard form on the non-calculator paper?
Yes - standard form is common content and can appear on any paper. The rules for multiplying, dividing, adding and subtracting powers of 10 work identically whether or not you have a calculator.
Sub-topics
Standard Form broken down into its individual skills, each with its own focused page.
Related topics
More Number & Algebra topics from the same AI SL syllabus unit, in case you want to keep going.