Scientific notation and orders of magnitude

Each decimal place changes by a factor of ten

In a decimal number, each place to the left is worth ten times the next place to its right. The digit 3 in 32 000 represents three ten-thousands; the digit 2 in the same number represents two thousands. The place values explain 32 000 = 3 × 10 000 + 2 × 1000, rather than treating the zeros as decoration.

Powers of ten name those scales compactly. For a positive integer n, 10^n is a product of n factors of ten; 10^0 = 1. A negative integer exponent describes a reciprocal: 10^(−n) = 1/10^n. Therefore 10^−3 = 1/1000 = 0.001. Moving between adjacent powers multiplies or divides the scale by ten, on either side of exponent zero.

Normalize the coefficient and keep the integer exponent

For a nonzero number x, normalized scientific notation is x = a × 10^n, where n is an integer and the coefficient a satisfies 1 ≤ |a| < 10. The absolute value |a| is the coefficient’s magnitude without its sign. Thus 3.2 × 10^4 and −6.4 × 10^2 are normalized, while 32 × 10^3 and 0.32 × 10^5 are equal-value forms that are not normalized.

The interval includes 1 and excludes 10: 1 × 10^4 is normalized, but 10 × 10^3 must become 1 × 10^4. Requiring both the interval and an integer exponent fixes one normalized coefficient and exponent for each nonzero number. In a quantity expression such as 3.2 × 10^4 m, this notation describes the numerical value in the named unit; the metre remains part of the quantity.

Change the coefficient and exponent together to preserve the value

To rewrite 32 000, divide its coefficient by 10 000 and supply that scale as a factor: 32 000 = 3.2 × 10^4. For the small number 0.00032, use 0.00032 = 3.2/10 000 = 3.2 × 10^−4. Reading back checks each result: 3.2 × 10 000 = 32 000, while 3.2 ÷ 10 000 = 0.00032. The decimal shift is a consequence of multiplication or division by a power of ten.

Increasing the exponent by one multiplies its power of ten by ten, so divide the coefficient by ten to compensate. Decreasing the exponent by one requires multiplying the coefficient by ten. In general, a × 10^n = (a/10) × 10^(n+1) = (10a) × 10^(n−1). For example, 32 × 10^3 = 3.2 × 10^4 = 0.32 × 10^5. These are exact rewrites of the same number; normalization selects the middle form.

The coefficient’s sign and the exponent have different jobs

Every power 10^n is positive, even when n is negative. Therefore 3.2 × 10^−4 is positive and small, while −3.2 × 10^−4 is negative and has the same magnitude. A minus sign in the exponent means a reciprocal scale; a minus sign in the coefficient determines the number’s sign. Likewise, −640 = −6.4 × 10^2 has a positive exponent and a negative value.

A negative metre value can describe a signed displacement along an axis whose positive direction has been chosen. It does not describe a segment’s length: a segment length is nonnegative. When using a negative example, name the signed quantity and its reference direction, rather than attaching a negative sign to an ordinary length.

Zero is a separate case. For every integer n, 0 × 10^n = 0, but coefficient zero fails 1 ≤ |a| < 10. Zero therefore has no normalized scientific-notation exponent. You may display a zero quantity as 0 m or, for a common-scale calculation, as 0 × 10^n m; that does not assign it a normalized exponent.

Use the full ratio to compare quantities in the same unit

First express quantities of the same kind in the same unit. For positive lengths L = a × 10^n m and S = b × 10^k m, with S nonzero, their ratio is L/S = (a/b) × 10^(n−k). The metres cancel. The exponent difference supplies a power-of-ten factor, but the coefficient ratio also matters. If the quantities are signed and the question concerns their magnitudes, compare |L|/|S| instead.

For stipulated lengths 3.2 × 10^4 m and 8 × 10^2 m, the ratio is (3.2/8) × 10^(4−2) = 0.4 × 10^2 = 40. Their exponents differ by two, yet the first length is exactly forty times the second, rather than one hundred times. Equal coefficients would make the ratio exactly 10^(n−k). A rough scale comparison and an exact ratio answer different levels of detail.

State which order-of-magnitude convention you are using

This lesson uses decade intervals: for a nonzero numerical value x in a fixed unit, its order bin is the integer n satisfying 10^n ≤ |x| < 10^(n+1). This n is the exponent in its normalized scientific notation. In metres, 32 000 lies in bin 4 because 10^4 ≤ 32 000 < 10^5; −640 lies in bin 2 because 10^2 ≤ 640 < 10^3. Zero belongs to no such interval, so it has no order bin.

The lower endpoint is included and the upper endpoint is excluded. Consequently 9999 is in bin 3 and 10 000 is in bin 4. The named unit matters: 32 000 m = 32 km, but their numerical values lie in bins 4 and 1 respectively. These bins describe the numbers in those fixed units, while the quantity itself stays the same. Compare bins only after agreeing on a common unit.

Some sources use the nearest power of ten on a logarithmic scale instead. Within 10^n to 10^(n+1), that convention switches at √10 × 10^n, where √10 is about 3.162; a tie needs an explicit rule. This is the multiplicative halfway point, so a coefficient such as 6 would select the next power under that convention even though it remains in decade bin n here. You do not need logarithms to use this lesson’s interval test; read another source’s definition before comparing its label with ours.

Multiply or divide coefficients and powers, then normalize

Multiplication uses (a × 10^n)(b × 10^k) = (ab) × 10^(n+k). Multiply the coefficients and add the integer exponents. With stipulated exact lengths, (2.4 × 10^5 m)(3 × 10^−3 m) = 7.2 × 10^2 m². The resulting unit is m², an area unit, because the two metre factors multiply. If the coefficient is outside the normalized interval, rewrite it with a compensating exponent change.

For division, require a nonzero divisor and use (a × 10^n)/(b × 10^k) = (a/b) × 10^(n−k). Divide the coefficients and subtract the divisor’s exponent, keeping its sign inside the subtraction. For example, (8 × 10^−4)/(2 × 10^−6) = 4 × 10^(−4−(−6)) = 4 × 10^2. Units divide too: an area in m² divided by a nonzero length in m leaves m. Keep exact fractions if a quotient does not terminate as a decimal, and round only for the final requested report.

Put added or subtracted terms on one scale

Addition and subtraction require compatible quantities expressed in the same unit. They also need a common power of ten before the coefficients can be combined. For example, 3.2 × 10^4 m + 6.4 × 10^2 m = 3.2 × 10^4 m + 0.064 × 10^4 m = 3.264 × 10^4 m. Changing the second coefficient from 6.4 to 0.064 exactly compensates for increasing its exponent from 2 to 4.

At a common scale, (a × 10^n) ± (b × 10^n) = (a ± b) × 10^n. Do not add or subtract exponents for these operations. A coefficient sum can cross a decade boundary: 8 × 10^3 m + 4 × 10^3 m = 12 × 10^3 m = 1.2 × 10^4 m. A difference can be negative, so retain its sign. Exact cancellation gives zero, which is reported separately rather than forced into the normalized interval.

Keep units and distinguish a notation change from a unit change

Rewriting 32 000 m as 3.2 × 10^4 m changes only the number’s form. Converting it to 32 km also changes the unit, using the exact relation 1 km = 10^3 m. Both preserve the same quantity. SI prefixes express exact powers of ten, so they can often make an expression convenient. The prefix symbol’s case matters: m means milli when used as a prefix, while M means mega; the final m in mm or Mm is the metre unit symbol.

Apply prefix factors to the whole unit structure. Since 1 mm = 10^−3 m, squaring gives 1 mm² = 10^−6 m², rather than 10^−3 m². For mass, prefixes attach to gram: 1 mg = 10^−3 g, while 1 kg = 10^3 g. Keep unit cancellation and the dimension check from the preceding lesson. Scientific notation organizes a numerical scale; it cannot make unlike quantities addable or supply a missing physical relationship.

Choose a power and watch the coefficient preserve the quantity

Begin with the explorer’s stipulated exact displacement 32 000 m along an axis with an agreed positive direction. Its default coefficient is 3.2 and exponent 4, giving 3.2 × 10^4 m. Select exponent 3 or 5 and compare the nearby forms in the table. The coefficient changes to preserve the same quantity, and the explorer identifies which form meets 1 ≤ |a| < 10. Only the form with exponent 4 is normalized for this preset.

Choose one of five exact displacement presets: 32 000 m, 0.00032 m, −640 m, 10 000 m or 0 m. They use the same axis and positive direction. The negative preset represents displacement opposite to that direction; it does not represent a segment length. For each preset, the selected exponent can be any integer from −5 through 5. The explorer computes the coefficient as the preset’s numerical value divided by 10 raised to that selected exponent. Every one of the 5 × 11 = 55 states rewrites the same selected quantity exactly; changing the exponent does not resize it.

Test the small positive preset: its normalized exponent is −4, even though the quantity is positive. Test −640 m: its normalized coefficient is −6.4 and exponent 2. At 10 000 m, coefficient 1 and exponent 4 show the included lower boundary. For 0 m, every selected exponent gives coefficient zero; all forms preserve zero, but none is normalized and zero has no order bin. The presets are stipulated values, so the explorer does not assign measurement uncertainty or significant figures to their digits.

Separate exact rewrites from the information in a report

An exact rewrite does not round a number. The stipulated number 32 000 is exactly 3.2 × 10^4, and its exactness comes from the statement that stipulates it. A measurement report is different: significant figures express a chosen reporting precision. For example, 3.20 × 10^4 has three significant figures, including the trailing zero in its coefficient, while 3.2 × 10^4 has two. They have the same numerical value but can carry different reporting information. A bare whole-number report such as 32 000 does not by itself establish how its trailing zeros were intended.

Retain the measurement information explicitly given with a value. A measured length reported as 12 340 m with stated uncertainty 20 m can be rewritten as (1.234 ± 0.002) × 10^4 m: both the reported value and the uncertainty magnitude use the same scale factor. In kilometres it is 12.34 km with stated uncertainty 0.02 km. These are changes of expression, not improvements to the measurement. The digits alone do not determine an uncertainty bound or the meaning of that uncertainty; preserve any stated method or convention too.

Keep exact arithmetic or unrounded information through intermediate steps, then apply the requested final rounding. For a stipulated exact value 9.996 × 10^4 m, a report to three significant figures is approximately 1.00 × 10^5 m. Rounding first produces coefficient 10.0 at exponent 4, so renormalize it as 1.00 at exponent 5 while retaining three significant figures. Use ≈ for the rounded result and = for an exact rewrite. A final reporting choice does not supply an uncertainty that was never stated.

Check the value, the unit and the reporting choice

Name the quantity and its unit, identify whether the values are stipulated or measured, and choose the operation that answers the question. Use powers of ten to keep the scale visible. Multiply and divide powers by exponent rules; for addition and subtraction, first align both the units and the powers. Preserve signs and require a nonzero divisor. After calculating, normalize each nonzero result and handle zero separately.

Read the result back in ordinary decimal form or reverse an exact operation to check the value. Verify the remaining unit, the coefficient interval and the size of the result. For a comparison, retain the coefficient ratio and specify the unit and order-of-magnitude convention. Preserve stated measurement information, round only when reporting, and normalize again if rounding carries the coefficient to ten. The next lesson, Scale, similarity and maps, uses ratios to connect corresponding lengths while distinguishing changes in length, area and volume.

Worked examples

Try it yourself

1. Write 64 000, 0.00032 and −640 in normalized scientific notation. Explain the negative exponent in the small positive number, and decide whether zero has a normalized form.

   Show the answer and reasoning

   64 000 = 6.4 × 10^4; 0.00032 = 3.2 × 10^−4; −640 = −6.4 × 10^2. The negative exponent specifies a reciprocal scale, and zero has no normalized form under 1 ≤ |a| < 10.

   Each normalized coefficient has magnitude at least one and less than ten, with an integer exponent. Powers of ten are positive; the coefficient supplies the sign. Zero can equal 0 × 10^n at any exponent, but its coefficient never satisfies the normalized interval.

2. Rewrite the explorer’s exact 32 000 m preset using selected exponents 3, 4 and 5. Which form is normalized? Explain what must happen to the coefficient when the exponent increases by one.

   Show the answer and reasoning

   32 000 m = 32 × 10^3 m = 3.2 × 10^4 m = 0.32 × 10^5 m. Only 3.2 × 10^4 m is normalized. Increasing the exponent by one requires dividing the coefficient by ten.

   The power-of-ten factor becomes ten times as large, and the compensating coefficient change preserves the product. The unit and the quantity remain the same in all three forms. Selecting a new exponent is an exact rewrite that preserves the displacement.

3. For stipulated exact values, normalize (4 × 10^−3 m)(5 × 10^2 m) and (6 × 10^6 g)/(2 × 10^3 g). Show the exponent operation and the remaining unit in each case.

   Show the answer and reasoning

   The product is 20 × 10^−1 m² = 2 × 10^0 m². The quotient is 3 × 10^3 with dimension one: its gram units cancel.

   For the product, multiply 4 × 5 and add −3 + 2 = −1, then divide coefficient 20 by ten and increase the exponent to zero. For the quotient, divide 6 by 2 and subtract 6 − 3 = 3. Both divisors and starting values here are stipulated, so no measurement rounding is required.

4. Find and normalize these exact results: 8 × 10^3 m + 4 × 10^3 m; 3 × 10^−4 m − 5 × 10^−4 m; 2.5 × 10^2 m − 2.5 × 10^2 m. Interpret the second as a difference of signed displacements along one axis.

   Show the answer and reasoning

   The results are 1.2 × 10^4 m, −2 × 10^−4 m and 0 m, respectively.

   The common powers allow coefficient operations 8 + 4 = 12, 3 − 5 = −2 and 2.5 − 2.5 = 0. Renormalize the first result, retain the negative sign in the displacement difference, and treat zero separately. Adding or subtracting the exponents would not represent these operations.

5. Compare 3.2 × 10^4 m with 8 × 10^2 m using their exact ratio. Under this lesson’s decade-interval convention, which metre-value bins contain −640 m, 9999 m and 10 000 m? Does 0 m have a bin?

   Show the answer and reasoning

   The ratio is 40. The numerical metre-value bins are 2 for −640 m, 3 for 9999 m and 4 for 10 000 m. Zero has no bin.

   Retain (3.2/8) × 10^(4−2) = 40 when calculating the ratio. Test magnitudes with 10^n ≤ |x| < 10^(n+1): the lower endpoint is included, the upper endpoint excluded. The sign of −640 does not change its magnitude bin, and no finite decade interval contains zero.

6. Express 0.00032 m in millimetres and micrometres. Express 32 000 m in kilometres and normalize its kilometre value. Why does its decade-bin exponent change when the numerical value is expressed in kilometres?

   Show the answer and reasoning

   0.00032 m = 0.32 mm = 320 µm. Also 32 000 m = 32 km = 3.2 × 10^1 km. Its numerical value is in bin 4 for metres and bin 1 for kilometres.

   Use the exact relations 1 mm = 10^−3 m, 1 µm = 10^−6 m and 1 km = 10^3 m. A unit change alters the number that counts the chosen units, while preserving the quantity. An order bin is defined for that numerical value in a fixed unit; compare bins only after choosing a common unit.

7. Rewrite a measured length reported as 12 340 m with stated uncertainty 20 m using one common power of ten. Separately, report the stipulated exact value 9.996 × 10^4 m to three significant figures. Explain the trailing zeros in the rounded coefficient and whether digits alone supply uncertainty.

   Show the answer and reasoning

   The measurement can be expressed as (1.234 ± 0.002) × 10^4 m. The rounded stipulated value is approximately 1.00 × 10^5 m, with three significant figures. Neither expression licenses inferring an unstated uncertainty from its digits.

   Scale the stated value and uncertainty magnitude together without changing the measurement information. Three-significant-figure rounding gives 10.0 × 10^4 m, then normalization gives 1.00 × 10^5 m. Its two coefficient zeros record the requested significant figures. Preserve any explicitly stated uncertainty and its interpretation, and use an approximation sign for a rounded value.

What to carry forward

• Every nonzero number has a normalized form a × 10^n with integer n and 1 ≤ |a| < 10; zero is handled separately.
• The coefficient determines the sign, the exponent determines the power-of-ten scale, and compensating changes preserve the exact value.
• Multiply and divide coefficients and powers; add and subtract only after aligning units and powers, then normalize the result.
• For an exact same-unit comparison, retain both the coefficient ratio and exponent difference; state the fixed unit and convention for an order-of-magnitude label.
• Keep units and stated measurement information, distinguish exact rewrites from rounded reports, and renormalize after a rounding carry.