Scientific Notation

Let u denote any agreed-upon unit for a quantity. For example, when measuring distance, u refers to 1 metre, denoted 1 m. When measuring information, u refers to 1 byte, denoted 1 B.

Definition 1. Given a unit u of quantity,

one kilo-unit, denoted ku, is defined to be 1000 units of that quantity,
one mega-unit, denoted Mu, is defined to be 1000 kilo-units of that quantity,
one giga-unit, denoted Gu, is defined to be 1000 mega-units of that quantity,
one tera-unit, denoted Tu, is defined to be 1000 giga-units of that quantity.

Example 1. 1 kB = 1000 B.

Problem 1. Explain why 1 kB = (1 × 10³) B. Deduce the integer α such that

1 TB = (1 × 10^α) B.

(Click for Solution)

Solution. We remark that

$1000 = 10 \times 10 \times 10 = 10^3.$

Therefore,

$\begin{aligned}1\, \mathrm{TB} &= 1000\, \mathrm{GB} = (1 \times 10^3)\, \mathrm{GB}.\end{aligned}$

Repeating the pattern,

$\begin{aligned}1\, \mathrm{TB} &= (1 \times 10^3)\, \mathrm{GB} \\ &= (1 \times 10^3 \cdot 10^3)\, \mathrm{MB} \\ &= (1 \times 10^3 \cdot 10^3 \cdot 10^3)\, \mathrm{kB} \\ &= (1 \times 10^3 \cdot 10^3 \cdot 10^3 \cdot 10^3)\, \mathrm{B} \\ &= (1 \times 10^{12})\, \mathrm B. \end{aligned}$

Therefore, $\alpha = 12$ .

Example 2. Letting the dollar, $1, denote the unit of measurement for money, one million dollars is equal to one mega-dollar. One billion dollars is equal to one giga-dollar. One trillion dollars is equal to one tera-dollar.

Definition 2. Given a unit u of quantity,

one deci-unit, denoted du, is defined to be 0.1 units of that quantity,
one centi-unit, denoted cu, is defined to be 0.01 units of that quantity,
one milli-unit, denoted mu, is defined to be 0.001 units of that quantity,
one micro-unit, denoted μu, is defined to be 0.001 milli-units of that quantity,
one nano-unit, denoted nu, is defined to be 0.001 micro-units of that quantity,
one pico-unit, denoted pu, is defined to be 0.001 nano-units of that quantity.

Example 3. 1 m = 100 cm.

Problem 2. Explain why 1 mm = (1 × 10^–3) m. Deduce the integer α such that

1 pm = (1 × 10^α) m.

(Click for Solution)

Solution. We remark that

$1\, \mathrm{mm} = 0.001\, \mathrm{m}.$

Therefore,

$1\, \mathrm{m} = 1000\, \mathrm{mm} = 10^3\, \mathrm{mm}.$

Dividing both sides by $10^3$ ,

$1\, \mathrm{mm} = (1/10^3)\, \mathrm m = (1 \times 10^{-3})\, \mathrm m.$

Similar as to Problem 1,

$\begin{aligned}1\, \mathrm{pm} &= (1 \times 10^{-3})\, \mathrm{nm} \\ &= (1 \times 10^{-3} \cdot 10^{-3})\, \mathrm{\text{\textmu} m} \\ &= (1 \times 10^{-3} \cdot 10^{-3} \cdot 10^{-3})\, \mathrm{mm} \\ &= (1 \times 10^{-3} \cdot 10^{-3} \cdot 10^{-3} \cdot 10^{-3})\, \mathrm{m} \\ &= (1 \times 10^{-12})\, \mathrm m. \end{aligned}$

Therefore, $\alpha = -12$ .

Theorem 1. For any positive real number N, there exists a unique integer α_N and a unique real number 1 ≤ N₀ < 10 such that

N = N₀ × 10^–α_N.

The right-hand side is called the scientific form of N. We call α_N the order of magnitude of N.

Proof. Using real analysis, there must exist some smallest (and thus, unique) integer α_N such that

1 ≤ N₀ / 10^α_N < 10

Define N₀ := N₀ / 10^α_N, so that N = N₀ × 10^–α_N as required.

Problem 3. Write down the order of magnitude for 1 TB and 1 pm respectively. What do you notice?

(Click for Solution)

Solution. The orders of magnitude are 12 and –12 respectively. The order of magnitude is positive if the quantity is greater than or equal to 1, and negative if the quantity is smaller than 1.

Theorem 2. Suppose α_N ≤ α_M. We have the following properties for orders of magnitude:

α_M + α_N ≤ α_MN ≤ α_M + α_N + 1,
α_M – α_N – 1 ≤ α_M/N ≤ α_M – α_N,
α_M ≤ α_{M + N} ≤ α_M + 1,
α_M – 1 ≤ α_{M – N} ≤ α_M.

Proof. Left as an exercise for the motivated student.

Problem 4. In chemistry, the accepted Avogadro constant N_A, measured in units of mol^–1, is given approximately by the number

N_A ≈ 602 214 076 000 000 000 000 000.

That is, 1 mol is defined to be the number N_A.

The accepted Boltzmann’s constant k_B, measured in units of J K^–1, given approximately by the number

k_B ≈ 0.000 000 000 000 000 000 000 013 806 490.

Express N_A and k_B in scientific form. Hence, evaluate the ideal gas constant R, measured in units of J K^–1 mol^–1, defined by R := N_A · k_B.

(Click for Solution)

Solution. We observe that

$\begin{aligned} N_{\mathrm A} &\approx 602\, 214\, 076\times 10^{15}\\ &= 6.022\, 140\, 760 \times 10^8 \times 10^{15} \\ &= 6.022\, 140\, 760 \times 10^{23} \\ &\approx 6.02 \times 10^{23}. \end{aligned}$

Similarly,

$\begin{aligned} k_{\mathrm B} &= 0.013\, 806\, 490 \times 10^{-21} \\ &= 1.380\, 649 \times 10^{-2} \times 10^{-21} \\ &= 1.380\, 649 \times 10^{-23} \\ &\approx 1.38 \times 10^{-23}. \end{aligned}$

Therefore,

$\begin{aligned}R = N_{\mathrm A} \cdot k_{\mathrm B} &= (6.022\, 140\, 760 \times 10^{23} ) \cdot (1.380\, 649 \times 10^{-23}) \\ &= 6.022\, 140\, 760 \cdot 1.380\, 649 \\ &= 8.314\, 462\, 618\, 200 \\ & \approx 8.31. \end{aligned}$

Problem 5. The ideal gas law states the following, given:

the number N of molecules of a gas,
the temperature T of the gas (measured in Kelvin, denoted K),
the pressure p of the gas (measured in Pascals, denoted Pa),
and the volume V of the gas (measured in m³),

the equation p · V = N · k_B · T holds.

Determine the volume of 0.400 mol of a gas with a pressure of 101 325 Pa and a temperature of 284 K, giving your answer in scientific notation.

(Click for Solution)

Solution. Using Problem 4, $N = 0.400 \cdot N_{\mathrm A}$ .

Substituting the values,

$101\, 325 \cdot V = 0.400 \cdot \underbrace{ N_{\mathrm A} \cdot k_{\mathrm B}}_R \cdot 284.$

Making $V$ the subject,

$\begin{aligned} V &= \frac{0.400 \cdot R \cdot 284}{101\, 325} \\ &= \frac{0.400 \cdot 8.314\, 462\, 618\, 200 \cdot 284}{101\, 325} \\ &= 0.009\, 321\, 716\, 787 \\ &= 9.321\, 716\, 787 \times 10^{-3} \\ &\approx 9.32 \times 10^{-3}. \end{aligned}$

Problem 6. In physics and astronomy, the accepted universal gravitational constant G, measured in units of m³ kg^–1 s^–2, is given approximately by the quantity

G ≈ 0.000 000 000 066 743.

Newton’s law of universal gravitation states that the magnitude F of the gravitational force between two masses M, m (measured in kg) that are separated by a distance of r metres is given by the equation

$\displaystyle F = \frac {GMm}{r^2}.$

Express G in scientific form. Furthermore, given that the Earth has a mass of

5 972 000 000 000 000 000 000 000 kg

and a radius of 6371 km, determine the gravitational acceleration g of an object with 1 kg near the surface of the Earth, measured in units of m s^–2, defined by g := F /m.

(Click for Solution)

Solution. Expressing $G$ in scientific form,

$\begin{aligned} G&\approx 0.000\, 000\, 000\, 066\, 743 \\ &= 0.066\, 743\times 10^{-9} \\ &= 6.674\, 300 \times 10^{-2} \times 10^{-9} \\ &= 6.674\, 300 \times 10^{-11} \\ &\approx 6.67 \times 10^{-11}. \end{aligned}$

Similarly, expressing $M$ in scientific form,

$\begin{aligned} M&= 5\, 972\, 000\, 000\, 000\, 000\, 000\, 000\, 000 \\ &= 5972 \times 10^{21} \\ &= 5.972 \times 10^3 \times 10^{21} \\ &= 5.972 \times 10^{24}. \end{aligned}$

Since the object is near the surface of the Earth,

$\begin{aligned} r &\approx 6371 \times 10^3 \\ &= 6.371 \times 10^3 \times 10^3 \\ &= 6.371 \times 10^6. \end{aligned}$

Substituting the quantities,

$\begin{aligned} g = \frac{F}{m} = \frac{GM}{r^2} &= \frac{(6.674\, 300 \times 10^{-11}) \cdot (5.972 \times 10^{24})}{(6.371 \times 10^6)^2} \\ &= \frac{6.674\, 300 \cdot 5.972}{6.371^2} \times \frac{10^{-11} \cdot {10}^{24}}{(10^6)^2} \\ &= 0.981\, 997\, 342\, 600 \times 10 \\ &= 9.819\, 973\, 426 \\ &\approx 9.82. \end{aligned}$

Problem 7. The Sun has a volume of 1.412 × 10¹⁸ m³, while the Earth has a volume of 1.083 × 10¹² m³. Disregarding issues related to unused empty space (i.e. sphere-packing), how many Earths could fit inside the Sun?

(Click for Solution)

Solution. The total number of Earths that could fit inside the Sun is simply the ratio below:

$\begin{aligned} \frac{1.412 \times 10^{18}}{1.083 \times 10^{12}} &= 1.303\, 785\, 780 \times 10^6 \\ &= 1\, 303\, 785. 780 \\ &\approx 1\, 303\, 785. \end{aligned}$

Therefore, approximately 1 303 785 Earths can fit in the Sun. Surprisingly, after accounting for sphere-packing, at least 982 334 Earths can still fit into the Sun. The Sun is not small.

Problem 8. The Schwarzchild radius is the radius r_S of the largest black hole that could be formed by an object with mass M. It is calculated using the formula

$\displaystyle r_{\mathrm S} = \frac {2GM}{c^2},$

where c ≈ 299 792 458, measured in m s^–1, denotes the speed of light in a vacuum. Determine the Schwarzchild radius for a black hole whose mass is equal to the mass of the Earth.

(Click for Solution)

Solution. Expressing $c$ in scientific notation,

$\begin{aligned}c &\approx 299\, 792\, 458 \\ &= 299.792\, 458 \times 10^6 \\ &= 2.997\, 924\, 580 \times 10^2 \times 10^6 \\ &= 2.997\, 924\, 580 \times 10^8.\end{aligned}$

Therefore, using the data in Problem 6,

$\begin{aligned} r_{\mathrm S} &= \frac{ 2 \cdot 6.674\, 300 \times 10^{-11} \cdot 5.972 \times 10^{24} }{(2.997\, 924\, 580 \times 10^8)^2} \\ &= \frac{ 2 \cdot 6.674\, 300 \cdot 5.972 }{2.997\, 924\, 580^2} \times \frac{10^{-11} \cdot {10}^{24}}{(10^8)^2} \\ &= 8.869\, 805\, 825\, 400 \times 10^{-3} \\ &\approx 8.87 \times 10^{-3} \\ &= 8.87\, \text{mm}. \end{aligned}$

Thus, the black hole would have a radius of 8.87 mm.

Problem 9. The gross domestic product (GDP) of a country is the total market value of all final goods and services in that country. The GDP per capita of a country is the GDP per person in that country. For simplicity, we will measure GDP in US dollars (USD).

Given that Singapore has a GDP of 572.47 billion USD and a population of 6.04 million as of June 2024, calculate Singapore’s GDP per capita as of June 2022.

(Click for Solution)

Solution. Using scientific notation, the required GDP per capita is given by

$\begin{aligned} \frac{572.47 \times 10^9}{6.04 \times 10^6} &= \frac{5.7247}{6.04} \times \frac{10^2 \cdot 10^9}{10^6} \\ &= 0.9477980132 \times 10^5 \\ &= 94\, 779.801\, 320 \\ &\approx 94\, 800\, \text{USD}. \end{aligned}$

In contrast, the GDP per capita of Australia in 2024 is approximately 65 000 USD.

—Joel Kindiak, 30 Mar 26, 0007H

KindiakMath

Scientific Notation

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Scientific Notation

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