N9 - Standard form

N9 - Standard form

Standard form writes a very large or very small number with a manageable coefficient and a power of ten. You will convert, compare and calculate in this form, both by hand and with a calculator. You will also learn to read a calculator display without losing the exponent.

Writing large and small numbers in standard form

A tiny component has mass 3×10−5 g3\times10^{-5}\,\mathrm g. A bag contains 1.2×1041.2\times10^4 of them. Is the total mass less than 1 g1\,\mathrm g? The powers look far apart, but standard form will make the calculation short.

First, consider a single number. 5 820 000=5.82×1 000 000=5.82×1065\,820\,000=5.82\times1\,000\,000=5.82\times10^6. The coefficient 5.825.82 is easy to read, while the power records the scale.

You already know

Multiplying or dividing by powers of ten changes place value. Also, 100=110^0=1, 10−1=11010^{-1}=\frac1{10}, and 10−2=110010^{-2}=\frac1{100}.

For a small number, the power is negative. 0.000582=5.82÷10 000=5.82×10−40.000582=5.82\div10\,000=5.82\times10^{-4}. The negative index makes a small positive value; it does not make the value negative.

Standard form

A positive number in the form A×10nA\times10^n, where 1⩽A<101\leqslant A<10 and nn is an integer. For example, 4.2×1034.2\times10^3 is in standard form. 42×10242\times10^2 has the same value, but its coefficient is too large to be standard form.

A coefficient of 11 is allowed; a coefficient of 1010 is not. The number 6.46.4 itself has standard form 6.4×1006.4\times10^0. Zero has no representation satisfying the definition above.

Standard form and place value

A×10n1⩽A<10,n an integer10−k=110kk a positive integer\begin{aligned} A\times10^n& && 1\leqslant A<10,\quad n\text{ an integer} \\ 10^{-k}&=\frac1{10^k} && k\text{ a positive integer} \end{aligned}

Choose the power that restores the original value after writing the coefficient between 11 and 1010.

Not on the formulae sheet: learn it.

Count place-value steps, then check their direction. Large numbers need a positive index; positive numbers below 11 need a negative one.

Worked example 1: a large number

Question. Write 5 820 0005\,820\,000 in standard form. Non-calculator.

Plan. Use coefficient 5.825.82, then count how many powers of ten restore the original number.

5 820 000=5.82×1 000 000coefficient between 1 and 10=5.82×106six place-value steps\begin{aligned} 5\,820\,000&=5.82\times1\,000\,000 && \text{coefficient between }1\text{ and }10 \\ &=5.82\times10^6 && \text{six place-value steps} \end{aligned}

Why is the index positive? We must enlarge 5.825.82 to get the original large number.

Answer. 5.82×1065.82\times10^6.

Check. Multiplying 5.825.82 by 10610^6 gives 5 820 0005\,820\,000.

Change only the scale: keep coefficient 5.825.82, but place the original number below 11.

Worked example 2: a small number

Question. Write 0.0005820.000582 in standard form. Non-calculator.

Plan. Find the power that makes 5.825.82 ten thousand times smaller.

0.000582=5.82÷10 000four place-value steps=5.82×10−4division gives negative index\begin{aligned} 0.000582&=5.82\div10\,000 && \text{four place-value steps} \\ &=5.82\times10^{-4} && \text{division gives negative index} \end{aligned}

Why count four steps rather than the zeros? The first non-zero digit moves from the ten-thousandths column to the ones column. Counting only visible zeros can miss a step.

Answer. 5.82×10−45.82\times10^{-4}.

Check. 10−4=0.000110^{-4}=0.0001, and 5.82×0.0001=0.0005825.82\times0.0001=0.000582.

The same scale pattern continues through index zero.

[DIAGRAM: asset_name: Standard form power-of-ten ladder; asset_slug: aqa8300_n12_power_ladder; caption: Reducing the exponent by one divides the value by ten, even across zero; description: Six boxes keep coefficient 3.2 while the exponent falls from 3 to minus 2. Their ordinary values are 3200, 320, 32, 3.2, 0.32 and 0.032. Arrows between boxes show division by ten.]
Reducing the exponent by one divides the value by ten, even across zero

Watch out

Wrong: 0.0042=4.2×1030.0042=4.2\times10^3. This makes 42004200, a large number. Right: 0.0042=4.2×10−30.0042=4.2\times10^{-3}. Check the size after choosing the sign.

Converting back is the same place-value operation in reverse.

Comparing standard-form numbers

For positive numbers already in standard form, compare the powers of ten first, then the coefficients if the powers match.

For example, 9.9×103=99009.9\times10^3=9900 is less than 1.1×104=11 0001.1\times10^4=11\,000. The larger coefficient does not win against a whole extra power of ten. This works because every coefficient is at least 11 and less than 1010.

With the same exponent, coefficients decide: 2.4×10−5<7.1×10−52.4\times10^{-5}<7.1\times10^{-5}. With different negative exponents, remember their order: 10−6<10−510^{-6}<10^{-5}. A more negative exponent means a smaller positive value.

Coefficient

The number multiplying the power of ten. In 7.1×10−57.1\times10^{-5}, the coefficient is 7.17.1, and the exponent is −5-5. The exponent is not part of the coefficient.

Check that each number is normalised before using the exponent shortcut. 32×10232\times10^2 is not in standard form: rewrite it as 3.2×1033.2\times10^3.

Key idea

For positive standard-form numbers, a larger exponent means a larger number; only equal exponents require a coefficient comparison.

Standard form also makes enormous exact numbers easy to name. A googol is 11 followed by one hundred zeros. Writing its standard form, 1×101001\times10^{100}, avoids writing all 101101 digits. The exponent describes scale, not the number of digits needed to write the coefficient.

Multiplying and dividing in standard form

Separate the ordinary coefficients from the powers of ten, then put the final answer back into standard form.

For multiplication, multiply the coefficients and add the indices. For division, divide the coefficients and subtract the denominator's index. These are the same index laws you already know, applied to base 1010.

Products and quotients in standard form

(A×10m)(B×10n)=(AB)×10m+nA×10mB×10n=AB×10m−n\begin{aligned} (A\times10^m)(B\times10^n)&=(AB)\times10^{m+n} \\ \frac{A\times10^m}{B\times10^n}&=\frac AB\times10^{m-n} \end{aligned}

Here AA, BB are positive, and mm, nn are integers. Adjust the result so its coefficient is at least 11 and less than 1010.

Not on the formulae sheet: learn it.

That final adjustment preserves value. For example, 24×104=2.4×10524\times10^4=2.4\times10^5: dividing the coefficient by 1010 means multiplying the power of ten by 1010.

Worked example 3: multiplying and normalising

Question. Work out (6×107)(4×10−3)(6\times10^7)(4\times10^{-3}). Give your answer in standard form. Non-calculator.

Plan. Multiply the coefficients, combine the powers and adjust the coefficient.

(6×107)(4×10−3)=(6×4)×107+(−3)group matching factors=24×104evaluate=2.4×105normalise coefficient\begin{aligned} (6\times10^7)(4\times10^{-3}) &=(6\times4)\times10^{7+(-3)} && \text{group matching factors} \\ &=24\times10^4 && \text{evaluate} \\ &=2.4\times10^5 && \text{normalise coefficient} \end{aligned}

Why increase the exponent? Changing 2424 to 2.42.4 divides by 1010, so the power must supply an extra factor of 1010.

Answer. 2.4×1052.4\times10^5.

Check. 60 000 000×0.004=240 00060\,000\,000\times0.004=240\,000.

Change only multiplication to division. The same coefficients now divide, and the indices subtract.

Worked example 4: dividing by a small number

Question. Work out 6×1074×10−3\frac{6\times10^7}{4\times10^{-3}}. Give your answer in standard form. Non-calculator.

Plan. Divide the coefficients and subtract the complete denominator index.

6×1074×10−3=64×107−(−3)divide equal bases=1.5×1010subtracting −3 adds 3\begin{aligned} \frac{6\times10^7}{4\times10^{-3}} &=\frac64\times10^{7-(-3)} && \text{divide equal bases} \\ &=1.5\times10^{10} && \text{subtracting }-3\text{ adds }3 \end{aligned}

Why is the result so large? Dividing by 0.0040.004, a number below 11, makes the original value larger.

Answer. 1.5×10101.5\times10^{10}.

Check. 1.5×1010×4×10−3=6×1071.5\times10^{10}\times4\times10^{-3}=6\times10^7.

A division may instead produce a coefficient below 11. For example, 0.4×106=4×1050.4\times10^6=4\times10^5: multiplying the coefficient by 1010 requires reducing the exponent by 11.

Watch out

Wrong: 107÷10−3=10410^7\div10^{-3}=10^4. Right: subtract the denominator's whole index, 7−(−3)=107-(-3)=10, so the power is 101010^{10}. Brackets around a negative index prevent this slip.

When a question asks for standard form, an intermediate form such as 24×10324\times10^3 has the right value but is not yet the requested answer.

Adding and subtracting in standard form

Addition needs matching powers of ten before you can collect coefficients.

Think of money: you would not add a number of pounds to a number of pence without first matching the units. Likewise, 2.8×104=0.28×1052.8\times10^4=0.28\times10^5. Rewriting one term puts both terms on the same scale.

Adding and subtracting with matching powers

For numbers AA, BB and an integer nn,

A×10n±B×10n=(A±B)×10nA\times10^n\pm B\times10^n=(A\pm B)\times10^n

First rewrite the numbers with the same power of ten. If the result is positive, normalise it afterwards. If the result is zero, write 00.

Not on the formulae sheet: learn it.

On Paper 1, matching a nearby power usually takes less writing than converting everything to ordinary numbers.

Worked example 5: adding different scales

Question. Work out 7.5×105+2.8×1047.5\times10^5+2.8\times10^4. Give your answer in standard form. Non-calculator.

Plan. Rewrite the smaller term using 10510^5, then add coefficients.

2.8×104=0.28×105match scales7.5×105+2.8×104=(7.5+0.28)×105collect coefficients=7.78×105add\begin{aligned} 2.8\times10^4&=0.28\times10^5 && \text{match scales} \\ 7.5\times10^5+2.8\times10^4 &=(7.5+0.28)\times10^5 && \text{collect coefficients} \\ &=7.78\times10^5 && \text{add} \end{aligned}

Why use 0.280.28? Each unit of 10510^5 is ten times as large as a unit of 10410^4, so we need one tenth as many.

Answer. 7.78×1057.78\times10^5.

Check. 750 000+28 000=778 000750\,000+28\,000=778\,000.

Change only the plus sign to minus. The scale conversion stays exactly the same.

Worked example 6: subtracting different scales

Question. Work out 7.5×105−2.8×1047.5\times10^5-2.8\times10^4. Give your answer in standard form. Non-calculator.

Plan. Use the same matching powers, then subtract coefficients.

7.5×105−2.8×104=(7.5−0.28)×105match 105=7.22×105subtract\begin{aligned} 7.5\times10^5-2.8\times10^4 &=(7.5-0.28)\times10^5 && \text{match }10^5 \\ &=7.22\times10^5 && \text{subtract} \end{aligned}

Why not subtract the indices? Subtracting indices is a division rule. Here we are subtracting two values.

Answer. 7.22×1057.22\times10^5.

Check. 722 000+28 000=750 000722\,000+28\,000=750\,000.

Even when the original powers match, check the final coefficient. 4.2×106−3.9×106=0.3×106=3×1054.2\times10^6-3.9\times10^6=0.3\times10^6=3\times10^5.

Watch out

Wrong: 105+104=10910^5+10^4=10^9. That confuses addition with multiplication. Right: 100 000+10 000=110 000=1.1×105100\,000+10\,000=110\,000=1.1\times10^5.

The same method works for small decimals. Matching exponents keeps the zeros out of the working.

Calculator entry and displays

A calculator's short scientific display is another way of showing a coefficient times a power of ten.

A display such as 3.6E93.6\mathrm{E}9 means 3.6×1093.6\times10^9, not 3.6×93.6\times9. Similarly, 3.6E−13.6\mathrm{E}{-1} means 3.6×10−1=0.363.6\times10^{-1}=0.36. Some calculators show a small raised exponent instead of the letter E\mathrm E.

You already know

Calculator papers allow powers and brackets. Enter the whole numerator and denominator separately when dividing expressions; a negative exponent belongs inside the exponent field.

Many calculators have a key labelled ×10x\times10^x, EXP or EE for entering the power of ten. That key already supplies “times ten to the power”; do not type an extra factor of 1010. It is different from the exponential function key exe^x.

For a check that works on any scientific calculator, use the ordinary power key and brackets. Keep every digit from the display until the final accuracy is requested. The display conventions and entry rules are not on the formulae sheet: learn them.

Worked example 7: reading a large display

Question. Work out (1.2×104)(3×105)(1.2\times10^4)(3\times10^5). Give your answer in standard form. Calculator allowed.

Plan. Enter both bracketed factors, then interpret the complete scientific display.

enter: (1.2×104)×(3×105)power key with bracketsdisplay: 3.6E9read coefficient and exponentvalue: 3.6×109write standard form\begin{aligned} \text{enter: }&(1.2\times10^4)\times(3\times10^5) && \text{power key with brackets} \\ \text{display: }&3.6\mathrm E9 && \text{read coefficient and exponent} \\ \text{value: }&3.6\times10^9 && \text{write standard form} \end{aligned}

Why copy the exponent too? The displayed 3.63.6 gives only the coefficient; the exponent carries almost all of this number's size.

Answer. 3.6×1093.6\times10^9.

Check. By hand, 1.2×3=3.61.2\times3=3.6 and 4+5=94+5=9.

Change only the sign of the second exponent. This gives the component masses from the opening question.

Worked example 8: interpreting a small total

Question. Each component has mass 3×10−5 g3\times10^{-5}\,\mathrm g. A bag contains 1.2×1041.2\times10^4 components. Work out the total mass and decide whether it is less than 1 g1\,\mathrm g. Calculator allowed.

Plan. Multiply the number of components by the mass of each one, then read the exponent.

enter: (1.2×104)×(3×10−5)negative exponent in second factordisplay: 0.36or3.6E−1same complete valuemass=3.6×10−1=0.36 gconvert for comparison\begin{aligned} \text{enter: }&(1.2\times10^4)\times(3\times10^{-5}) && \text{negative exponent in second factor} \\ \text{display: }&0.36\quad\text{or}\quad3.6\mathrm E{-1} && \text{same complete value} \\ \text{mass}&=3.6\times10^{-1}=0.36\,\mathrm g && \text{convert for comparison} \end{aligned}

Why multiply? Each of the 12 00012\,000 components contributes the same mass.

Answer. 0.36 g0.36\,\mathrm g, which is less than 1 g1\,\mathrm g.

Check. By hand, 4+(−5)=−14+(-5)=-1, so the total is 3.6×10−1 g3.6\times10^{-1}\,\mathrm g.

The opening answer is small even though there are many components. The positive exponent for the count and the negative exponent for the mass combine, rather than compete.

Watch out

Wrong: copying 7.2E−67.2\mathrm E{-6} as 7.2×1067.2\times10^6. Right: keep the minus sign, 7.2×10−67.2\times10^{-6}. A missing sign changes the value by a factor of 101210^{12}.

Read the display as a number, then carry out any operation the question requests.

Quick Recap

  • Standard form is A×10nA\times10^n, with 1⩽A<101\leqslant A<10 and integer nn.
  • Multiply or divide the coefficients and use the index laws for the powers of ten.
  • For addition and subtraction, match the powers first, then collect coefficients.
  • Normalise the final coefficient and read a calculator's exponent, including its sign.

The opening bag has total mass 3.6×10−1 g=0.36 g3.6\times10^{-1}\,\mathrm g=0.36\,\mathrm g, so it is lighter than 1 g1\,\mathrm g.