Calculating with standard form — Mathematics

Before this

Standard form
Index notation and the basic laws · coming soon

This leads to

Rounding and significant figures
Index laws · coming soon

Eight minutes in the past

When you look at the Sun, you do not see it as it is now. You see it as it was a few minutes ago, because its light takes time to reach you. How many minutes?

The Earth is about 150 000 000 000 m from the Sun, and light travels at about 300 000 000 m every second. To find the time, divide the distance by the speed. Written out in full, that is a long division with a page of zeros. In standard form, it takes one line:

\frac{1.5 \times 1 0 ^{11}}{3 \times 1 0 ^{8}} = \frac{1.5}{3} \times 1 0^{11 - 8} = 0.5 \times 1 0^{3} = 500 s

About 500 seconds, which is about 8⅓ minutes. (NASA gives the Earth's average distance from the Sun as $149.598 \times 1 0^{6}$ km, and the speed of light is exactly 299 792 458 m/s. Rounded, these are the two numbers used here.)

The last lecture showed how to write very large and very small numbers. It promised that standard form also makes calculations with them surprisingly simple. Here is that promise kept. The secret is in the powers of ten, which follow simple rules.

Why the powers add when you multiply

Remember what a power of ten means. $1 0^{3}$ means $10 \times 10 \times 10$ : three tens multiplied together. $1 0^{4}$ means four tens multiplied together. So:

1 0^{3} \times 1 0^{4} = (10 \times 10 \times 10) \times (10 \times 10 \times 10 \times 10) = 1 0^{7}

There are three tens and then four more, so seven tens in all. That is the whole reason for the rule: when you multiply powers of ten, you add the powers. Check with ordinary numbers: $1000 \times 10000 = 10000000$ , which is $1 0^{7}$ .

Division works the other way. In $1 0^{5} \div 1 0^{2}$ , two of the five tens on top cancel with the two tens underneath, leaving three: $1 0^{5} \div 1 0^{2} = 1 0^{3}$ . When you divide powers of ten, you subtract the powers. Check: $100000 \div 100 = 1000$ .

The same rules work with negative powers, because a negative power means dividing by ten. For example, $1 0^{5} \times 1 0^{- 2}$ means multiplying by ten five times and then dividing by ten twice. That leaves $1 0^{3}$ , and $5 + (- 2) = 3$ . (A later lecture on the index laws proves these rules for any base, not just 10.)

Multiplying and dividing in standard form

$(a \times 1 0^{m}) \times (b \times 1 0^{n}) = (a \times b) \times 1 0^{m + n}$

$(a \times 1 0^{m}) \div (b \times 1 0^{n}) = (a \div b) \times 1 0^{m - n}$

Then check that the number in front is at least 1 and less than 10. If it is not, adjust it.

Why can the numbers in front and the powers be handled separately? Because multiplication can be done in any order. $(3 \times 1 0^{4}) \times (2 \times 1 0^{6})$ is four numbers multiplied together, so you may group them as $(3 \times 2) \times (1 0^{4} \times 1 0^{6}) = 6 \times 1 0^{10}$ .

The step that is easy to forget

Multiply $(4 \times 1 0^{5}) \times (5 \times 1 0^{3})$ . The numbers in front give $4 \times 5 = 20$ , and the powers give $1 0^{5 + 3} = 1 0^{8}$ . So the answer is $20 \times 1 0^{8}$ .

That is the correct value, but it is not in standard form, because 20 is not less than 10. Fix it by writing 20 as $2 \times 1 0^{1}$ :

20 \times 1 0^{8} = 2 \times 1 0^{1} \times 1 0^{8} = 2 \times 1 0^{9}

Division can go wrong in the other direction. In the sunlight calculation, $1.5 \div 3 = 0.5$ , which is less than 1. Writing 0.5 as $5 \times 1 0^{- 1}$ gives $0.5 \times 1 0^{3} = 5 \times 1 0^{- 1} \times 1 0^{3} = 5 \times 1 0^{2}$ . The rule for the fix is always the same:

If the number in front is 10 or more, divide it by 10 and add 1 to the power.
If the number in front is less than 1, multiply it by 10 and subtract 1 from the power.

The value does not change, because you multiply one part by 10 and divide the other by 10. Step through a full calculation below.

Step through — how far is a light-year?

A light-year is the distance light travels in one year. Press Next to add one step. Read the reason before you go on.

Each step gives one line and the reason it is allowed.

Step 1. Distance = speed × time = (3.00 × 10⁸) × (3.16 × 10⁷) m: light travels about 3.00 × 10⁸ m each second, and a year is about 3.16 × 10⁷ seconds.
Step 2. = (3.00 × 3.16) × (10⁸ × 10⁷): multiplication can be done in any order, so group the numbers in front and the powers.
Step 3. 3.00 × 3.16 = 9.48: multiply the numbers in front.
Step 4. 10⁸ × 10⁷ = 10¹⁵: eight tens and seven more tens make fifteen tens, so add the powers.
Step 5. = 9.48 × 10¹⁵ m: 9.48 is at least 1 and less than 10, so this is already in standard form.

Result: one light-year is about 9.5 × 10¹⁵ m. The accepted value is about 9.46 × 10¹⁵ m; the small difference comes from rounding the speed of light and the length of a year.

Stop and think

A student works out $(2 \times 1 0^{3}) \times (3 \times 1 0^{4})$ and writes $6 \times 1 0^{12}$ . What went wrong?

Show the answer

The student multiplied the powers, $3 \times 4 = 12$ , instead of adding them. Count the tens: three tens and four more make seven. The answer is $6 \times 1 0^{7}$ . Check with ordinary numbers: $2000 \times 30000 = 60000000$ , which is $6 \times 1 0^{7}$ . Multiplying the powers is the right rule for a different situation, a power raised to another power, which the lecture on the index laws explains.

Adding and subtracting: line up the places first

Addition is different, and this is where most mistakes happen. Try $(4.5 \times 1 0^{5}) + (3 \times 1 0^{4})$ . It is tempting to add the 4.5 and the 3. But they are not the same kind of thing. The 4.5 counts hundred-thousands, and the 3 counts ten-thousands. Adding them would be like adding 4 kilograms to 3 grams and calling it 7 kilograms.

Write them as ordinary numbers in place-value columns, and the problem disappears:

To add, line up the place values. The 3 of

3 \times 1 0^{4}

sits in the

1 0^{4}

column, under the 5, not under the 4. The total is 480 000, which is

4.8 \times 1 0^{5}

In standard form, the same lining up means making the powers equal before adding. Rewrite $4.5 \times 1 0^{5}$ as $45 \times 1 0^{4}$ . Now both numbers count ten-thousands:

45 \times 1 0^{4} + 3 \times 1 0^{4} = 48 \times 1 0^{4} = 4.8 \times 1 0^{5}

That is it. That is the whole idea of this lecture. Multiplying and dividing only need the powers to be combined, so the powers can be different. Adding and subtracting need the places to match, so the powers must be made the same first. Multiplication changes the size. Addition only combines amounts of the same size.

For a simple addition or subtraction, converting to ordinary numbers is also fine: $450000 + 30000 = 480000$ . On a calculator, you can type the numbers in standard form directly. Many calculators have a key marked ×10ˣ or EXP for this.

The working below adds two numbers in standard form. One line is wrong. Find it.

Spot the error — adding 4 × 10³ and 3 × 10²

One line of this working is wrong. Tap the line you think is wrong. Every line has an explanation.

The working, with the mistake marked and corrected:

Line 1. We want 4 × 10³ + 3 × 10². (Correct.)
Line 2. Both numbers are already in standard form. (Correct.)
Line 3. Add the numbers in front and keep the larger power: (4 + 3) × 10³. The mistake: 4 counts thousands and 3 counts hundreds, so they cannot be added directly.
Line 4. = 7 × 10³. (This follows from the mistake.)
Line 5. 7 is between 1 and 10, so the answer is in standard form. (True of 7 × 10³, but the value is wrong. The correct answer is 40 × 10² + 3 × 10² = 43 × 10² = 4.3 × 10³.)

Worked examples

Worked Example 1 — Multiplying with an adjustment

Work out $(4 \times 1 0^{5}) \times (5 \times 1 0^{3})$ .

Numbers in front: $4 \times 5 = 20$ . Powers: $1 0^{5 + 3} = 1 0^{8}$ . So the product is $20 \times 1 0^{8}$ .

20 is not less than 10, so divide it by 10 and add 1 to the power: $2 \times 1 0^{9}$ .

Check: $400000 \times 5000 = 2000000000$ , which is $2 \times 1 0^{9}$ .

Worked Example 2 — Dividing with an adjustment

Work out $(1.5 \times 1 0^{8}) \div (6 \times 1 0^{3})$ .

Numbers in front: $1.5 \div 6 = 0.25$ . Powers: $1 0^{8 - 3} = 1 0^{5}$ . So the answer is $0.25 \times 1 0^{5}$ .

0.25 is less than 1, so multiply it by 10 and subtract 1 from the power: $2.5 \times 1 0^{4}$ .

Check: $150000000 \div 6000 = 25000$ .

Worked Example 3 — Negative powers

Work out (a) $(2.4 \times 1 0^{- 3}) \times (5 \times 1 0^{- 4})$ and (b) $(8 \times 1 0^{- 6}) \div (4 \times 1 0^{- 9})$ .

(a) $2.4 \times 5 = 12$ , and $1 0^{- 3 + (- 4)} = 1 0^{- 7}$ . So $12 \times 1 0^{- 7} = 1.2 \times 1 0^{- 6}$ .

(b) $8 \div 4 = 2$ , and $1 0^{- 6 - (- 9)} = 1 0^{- 6 + 9} = 1 0^{3}$ . So the answer is $2 \times 1 0^{3}$ .

Part (b) surprises people: two tiny numbers give 2000. But $4 \times 1 0^{- 9}$ is much smaller than $8 \times 1 0^{- 6}$ , and dividing by a much smaller number gives a large answer. Check: $0.000 008 \div 0.000 000004 = 2000$ .

Worked Example 4 — Subtracting

Work out $(6.3 \times 1 0^{8}) - (9 \times 1 0^{6})$ .

Make the powers equal. Use the smaller power, $1 0^{6}$ , so that both numbers in front are whole numbers: $6.3 \times 1 0^{8} = 630 \times 1 0^{6}$ .

630 \times 1 0^{6} - 9 \times 1 0^{6} = 621 \times 1 0^{6} = 6.21 \times 1 0^{8}

Check: $630000000 - 9000000 = 621000000$ .

Three classic mistakes

Trap 1 — Multiplying the powers

$1 0^{3} \times 1 0^{4} = 1 0^{7}$ , not $1 0^{12}$ . When you multiply powers of ten, you add the powers, because you are counting the tens.

Trap 2 — Adding with different powers

$4 \times 1 0^{3} + 3 \times 1 0^{2}$ is not $7 \times 1 0^{3}$ . The places must match before the numbers can be added. Make the powers equal first, or write both numbers out in full.

Trap 3 — Stopping at 20 × 10⁸

A correct value is not finished until the number in front is at least 1 and less than 10. Check it at the end of every calculation, and adjust the power by one if needed.

Try it yourself

Give each answer in standard form. Work on paper first. Open a hint only when you are stuck.

Work out (a) $(3 \times 1 0^{4}) \times (2 \times 1 0^{6})$ and (b) $(9 \times 1 0^{7}) \div (3 \times 1 0^{2})$ .

Hint 1
Handle the numbers in front and the powers separately.

Hint 2
Add the powers for (a). Subtract them for (b).

Full answer
(a) $6 \times 1 0^{10}$ . (b) $3 \times 1 0^{5}$ . Both numbers in front are already between 1 and 10. Check (b): $90000000 \div 300 = 300000$ .
Work out $(7 \times 1 0^{- 3}) \times (6 \times 1 0^{5})$ .

Hint 1
$7 \times 6 = 42$ , and $- 3 + 5 = 2$ .

Hint 2
42 is too big. Divide it by 10 and add 1 to the power.

Full answer
$42 \times 1 0^{2} = 4.2 \times 1 0^{3}$ . Check: $0.007 \times 600000 = 4200$ .
Work out $(3.2 \times 1 0^{6}) + (1.5 \times 1 0^{6})$ and $(2.4 \times 1 0^{5}) + (8 \times 1 0^{3})$ .

Hint 1
In the first sum the powers already match. In the second, they do not.

Hint 2
Write $2.4 \times 1 0^{5}$ as $240 \times 1 0^{3}$ .

Full answer
First: $4.7 \times 1 0^{6}$ . Second: $240 \times 1 0^{3} + 8 \times 1 0^{3} = 248 \times 1 0^{3} = 2.48 \times 1 0^{5}$ . Check: $240000 + 8000 = 248000$ .
A proton has a mass of about $1.67 \times 1 0^{- 27}$ kg. About how many protons would have a total mass of 1 kg? Then estimate how many protons would match the mass of the Earth, $5.97 \times 1 0^{24}$ kg.

Hint 1
Number of protons = total mass ÷ mass of one proton. Write 1 kg as $1 \times 1 0^{0}$ .

Hint 2
$1 \div 1.67 \approx 0.599$ , and $0 - (- 27) = 27$ . For the Earth, multiply your first answer by $5.97 \times 1 0^{24}$ .

Full answer
$1 \div (1.67 \times 1 0^{- 27}) \approx 0.599 \times 1 0^{27} = 5.99 \times 1 0^{26}$ protons in 1 kg. For the Earth: $5.99 \times 1 0^{26} \times 5.97 \times 1 0^{24} \approx 35.8 \times 1 0^{50} = 3.58 \times 1 0^{51}$ . (The next lecture explains how many digits to keep in answers like these.) Check by size: the powers alone give $1 0^{27} \times 1 0^{24} = 1 0^{51}$ , which matches. This is the "51 powers of ten" between a proton and the Earth from the last lecture.

The bigger idea

Standard form splits every number into a size (the power of ten) and some digits (the number in front). Multiplication and division then split in the same way: sizes combine by adding or subtracting powers, and digits combine by ordinary arithmetic. Addition cannot be split like this, because it only works between amounts of the same size.

This is why scientists can work with the distance to a star and the mass of an atom in the same calculation without losing track. The powers keep the sizes honest. In the next lecture, the digits get the same care: how many of them are worth keeping.

Key concepts in this lesson

To multiply numbers in standard form, multiply the numbers in front and add the powers of ten.
To divide numbers in standard form, divide the numbers in front and subtract the powers of ten.
The power rules come from counting the tens: three tens times four more tens make seven tens.
To add or subtract, first make the powers equal, because only amounts of the same place value can be combined.
After every calculation, check that the number in front is at least 1 and less than 10, and adjust the power by one if it is not.