Standard Form

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7.3Standard Form

In this Math tutorial, you will learn:

  • What is the standard form? Why it is used?
  • What is the relationship between standard form and other forms of expressing numbers?
  • Why standard form is more suitable than the decomposed form when making operations with numbers?
  • How to write decimals in standard form and vice-versa?
  • How to add and subtract numbers in standard form? Why this is much easier than adding or subtracting in the decomposed form?
  • How to multiply and divide numbers in standard form?
  • How to write very big and very small numbers in standard form?
  • How to deal with powers of numbers written in standard form?

Introduction

How do you write the number 35 million? 427 billion? 12 trillion? Is it suitable to write them with all those zeroes, i.e. in the ordinary form? Why?

An atom is about 0.1 nanometre in size. How do you express this number in metres given that 1 metre = 1 billion nanometre?

The above examples involve very large or very small numbers. Obviously, they can be written as they are, i.e. in the ordinary form. However, there is an alternate method which is much more suitable to express such numbers. This method is going to be explained in this tutorial.

The Meaning of Standard Form. The Relationship between Ordinary, Decomposed and Standard Form of Numbers.

In tutorial 1.7, we briefly discussed writing ordinary numbers as powers of ten. This method or writing numbers is useful when we want to highlight various placeholders and the specific weight they "occupy" in a number. Thus, the number 271 is written in the decomposed form as

271 = 2 × 100 + 2 × 10 + 1 × 1

Moreover, we can write 100, 10 and 1 as powers of ten, i.e.

271 = 2 × 102 + 2 × 101 + 1 × 100

Writing numbers in the decomposed form is very useful to understand what value each digit has. However, this method is too long. It is better to write them in a shorter way while preserving some features of expanded or decomposed form discussed above. Thus, we write only the power of the leftmost digit while the original number is written in two parts: one decimal part that gives all digits of the original number and another part that gives the powers of ten. This method of writing numbers is known as the standard form. Therefore, the standard form represents a form of writing numbers in terms of powers of ten.

Any number N written in standard form has the general structure

N = A × 10m

where A is a number between 1 and 10 (without reaching the value 10) while m is an integer that represents the highest power of the decomposed form of that number.

If a number W in the ordinary form has n digits, when written in the standard form it becomes

W = A × 10n - 1 + B × 10n - 2 + ⋯ + (N-1) × 101 + N × 100
= A.BC…N × 10n - 1

where A, B, … , N are the coefficients representing the place values of the original number.

Example 1

Which of the following numbers is written in the correct decomposed form? Then, write all of them in the standard form.

  1. 12,478
  2. 3 × 52 + 2 × 51 + 9 × 50
  3. 7 × 10001/3 + 4 × 1001/2 + 3 × 70
  4. 6 × 104 + 3 × 101 + 9 × 100

Solution 1

  1. This number is not written in the decomposed form but in the ordinary form instead. If decomposed it becomes
    12,478 = 1 × 104 + 2 × 103 + 4 × 102 + 7 × 101 + 8 × 100
    The standard form of this number therefore is
    12,478 = 1.2418 × 104
  2. This number is not written in the decomposed form as it is not written in terms of the powers of ten but of powers of five instead. If we want to write it in the decomposed form, first we must concert it in an ordinary number and only then we can write it in standard form. Thus,
    3 × 52 + 2 × 51 + 9 × 50
    = 3 × 25 + 2 × 5 + 9 × 1
    = 75 + 10 + 9
    = 94
    When written in the decomposed form, this number becomes
    94 = 9 × 101 + 4 × 100
    and in the standard form it becomes
    94 = 9.4 × 101
  3. The number is not in the decomposed form but with a few arrangements it can be written as such. We have
    7 × 10001/3 + 4 × 1001/2 + 3 × 70
    = 7 × ∛1000 + 4 × √100 + 3 × 1
    = 7 × ∛(103 ) + 4 × √(102 ) + 3 × 1
    = 7 × 10 + 4 × 10 + 3 × 1
    = 70 + 40 + 3
    = 113
    When written in the decomposed form, this number becomes
    113 = 1 × 102 + 1 × 101 + 3 × 100
    The standard form of this number therefore is
    113 = 1.13 × 102
  4. This number is already written in the decomposed form as all terms represent powers of ten. We can just write the missing terms for aesthetic purpose although this is not necessary, i.e.
    6 × 104 + 3 × 101 + 9 × 100
    = 6 × 104 + 0 × 103 + 0 × 102 + 3 × 101 + 9 × 100
    This helps to write the number in the standard form without errors. Thus,
    6 × 104 + 3 × 101 + 9 × 100 = 6.0039 × 104

Writing Decimals in Standard and Decomposed Form

In tutorial 3.5 we discussed decimals as an alternative form to fractions for expressing the non-whole numbers. In that tutorial, we explained that a decimal is composed by three parts: a group of digits on the left, which represent the whole part of the number, another group of digits on the right that represents the non-whole part and a decimal point that separates the whole and non-whole parts.

We can extend the reasoning used for numbers written in standard form beyond the decimal point to also include the decimal numbers. This can be easily done now that you are familiar with negative indices. Thus, a decimal that is written in the form

N = abcd.efgh

(where abcd is the whole part while efgh is the decimal part) can be expressed in the decomposed form as

N = a × 103 + b × 102 + c × 101 + d × 100 + e × 10-1 + f × 10-2 + g × 10-3 + h × 10-4

In other words, we have

N = a × 1,000 + b × 100 + c × 10 + d × 1 + e × 1/10 + f × 1/100 + g × 1/1,000 + h × 1/10,000

When written in the standard form, this number becomes

N = a.bcdefgh × 103

For example, 32.7 is written in the standard form as

32.7 = 3 × 101 + 2 × 100 + 7 × 10-1
= 3.27 × 101

Example 2

Write the following decimals in the standard form.

  1. 106.24
  2. 3.089

Solution 2

  1. We have
    106.24 = 1 × 102 + 0 × 101 + 6 × 100 + 2 × 10-1 + 4 × 10-2
    = 1.0624 × 102
  2. 3.089 = 3 × 100 + 0 × 10-1 + 8 × 10-2 + 9 × 10-3
    = 3.089 × 100

Addition and Subtraction with Numbers in Standard and Decomposed Form

If we add or subtract two numbers in the decomposed form, we must be careful to make the proper arrangements if any coefficient becomes less than 1 or more than 10. This is like adding or subtracting like terms in the numerical or algebraic expressions.

For example, if we have the addition

A + B = (1 × 102 + 5 × 101 + 4 × 100 + 9 × 10-1 ) + (6 × 101 + 5 × 100 + 4 × 10-1 )
= 1 × 102 + (5 × 101 + 6 × 101 ) + (4 × 100 + 5 × 100 ) + (9 × 10-1 + 4 × 10-1 )
= 1 × 102 + 11 × 10-1 + 9 × 100 + 13 × 10-1

At this point, we must make the proper corrections in the coefficients to make them smaller than 10 by carrying the excess to the next term. For this, we begin from right to left. Hence, 13 becomes 1; 9 first becomes 10, then 0; 11 first becomes 12, then 2, and 1 becomes 2. In this way, we obtain

A + B = 2 × 102 + 2 × 10-1 + 0 × 100 + 3 × 10-1

As you see, this method is very long and time consuming. Therefore, we can write the numbers in the standard form to ease the operations. We have

A = 1 × 102 + 5 × 101 + 4 × 100 + 9 × 10-1 = 1.549 × 102
B = 6 × 101 + 5 × 100 + 4 × 10-1 = 6.54 × 101

To complete the operations of addition and subtraction, we must first convert all numbers at the same power of ten. Thus, we write the second number B as 0.654 × 102 instead of 6.54 × 101 although this is not the correct form of expressing a number in the standard form. This is done only for the operations sake. Therefore, we obtain

A + B = 1.549 × 102 + 0.654 × 102
= (1.549 + 0.654) × 102
= 2.203 × 102

Proof: Expressing the two numbers in the ordinary form, yields

A = 1 × 102 + 5 × 101 + 4 × 100 + 9 × 10-1 = 154.9
B = 6 × 101 + 5 × 100 + 4 × 10-1 = 65.4

Thus,

154.9+ 65.4
220.3

When expressed in the standard form, this number becomes

220.3 = 2 × 102 + 2 × 10-1 + 0 × 100 + 3 × 10-1
= 2.203 × 102

Since all results are identical, the solution is correct.

On the other hand, in subtraction we often need to borrow values from the next term on the right when dealing with numbers in the decomposed form. For example,

A-B = (4 × 102 + 1 × 101 + 6 × 100 + 7 × 10-1 )-(6 × 101 + 9 × 100 + 2 × 10-1 )
= 4 × 102 + (1 × 101 -6 × 101 ) + (6 × 100 -9 × 100 ) + (7 × 10-1-2 × 10-1 )

At this point, we must do the operations but while taking into account the borrowing method used in subtraction of ordinary numbers. Again, all corrections are made from right to left. Thus, the rightmost coefficient is 7 - 2 = 5; the next one is 7 because after borrowing, we have 16 - 9 = 7; the next one is 4 because 1 became 0 in the previous borrowing, so after a new borrowing we have 10 - 6 = 4. Last, 4 became 3 due to the above borrowing. In this way, we obtain

A-B = 3 × 102 + 4 × 101 + 7 × 100 + 5 × 10-1

Again, this procedure is too long, so it is better to have the numbers written in the standard form. Hence, we write

A = 4 × 102 + 1 × 101 + 6 × 100 + 7 × 10-1 = 4.167 × 102
B = 6 × 101 + 9 × 100 + 2 × 10-1 = 6.92 × 101

We write the second number in such a way to fit the first number although it is not in the correct standard form. Thus, we have

B = 6.92 × 101 = 0.692 × 102

Therefore, we have

A-B = 4.161 × 102 -0.692 × 102
= (4.167-0.692) × 102
= 3.475 × 102

Proof: Expressing the two numbers in the ordinary form, yields

A = 4 × 102 + 1 × 101 + 6 × 100 + 7 × 10-1 = 416.5
B = 6 × 101 + 9 × 100 + 2 × 10-1 = 69.2

Thus,

416.7- 69.2
347.5

When expressed in the standard form, this number becomes

347.5 = 3 × 102 + 4 × 10-1 + 7 × 100 + 5 × 10-1
= 3.475 × 102

This result corresponds to that obtained when subtracting the two numbers A and B written in the standard form.

Example 3

Complete the following operations in the decomposed form and check the results by expressing the numbers in the standard and ordinary form.

  1. (3 × 103 + 6 × 101 + 2 × 10-2) + (4 × 102 + 8 × 101 + 5 × 10-2)
  2. (5 × 103 + 2 × 101 + 1 × 10-2 )-(3 × 102 + 8 × 100 + 6 × 10-2)

Solution 3

  1. It is better to have the numbers written as A and B. Thus, we have
    A = 3 × 103 + 6 × 101 + 2 × 10-2
    B = 4 × 102 + 8 × 101 + 5 × 10-2
    Moreover, we can fill the missing terms writing zero as coefficient to avoid confusion, similar to when placing zeroes before or after numbers added or subtracted in column. This yields for A and B:
    A = 3 × 103 + 0 × 102 + 6 × 101 + 0 × 100 + 0 × 10-1 + 2 × 10-2
    B = 0 × 103 + 4 × 102 + 8 × 101 + 0 × 100 + 0 × 10-1 + 5 × 10-2
    Adding the like terms yields
    A + B = (3 + 0) × 103 + (0 + 4) × 102 + (6 + 8) × 101 + (0 + 0) × 100 + (0 + 0) × 10-1 + (2 + 5) × 10-2 = 3 × 103 + 4 × 102 + 14 × 101 + 0 × 100 + 0 × 10-1 + 7 × 10-2
    = 3 × 103 + 5 × 102 + 4 × 101 + 0 × 100 + 0 × 10-1 + 7 × 10-2
    = 3 × 103 + 5 × 102 + 4 × 101 + 7 × 10-2
    Proof:
    In the standard form:
    A = 3 × 103 + 0 × 102 + 6 × 101 + 0 × 100 + 0 × 10-1 + 2 × 10-2 = 3.06002 × 103
    B = 4 × 102 + 8 × 101 + 0 × 100 + 0 × 10-1 + 5 × 10-2 = 4.8005 × 10-2 = 0.48005 × 103
    Thus,
    A + B = 3.06002 × 103 + 0.48005 × 103
    = (3.06002 + 0.48005) × 103
    = 3.54007 × 103
    In the ordinary form:
    A = 3 × 103 + 0 × 102 + 6 × 101 + 0 × 100 + 0 × 10-1 + 2 × 10-2 = 3060.02
    B = 0 × 103 + 4 × 102 + 8 × 101 + 0 × 100 + 0 × 10-1 + 5 × 10-2 = 480.05
    Adding these numbers in column yields
    3060.02+ 480.05
    3540.07
    This number corresponds to
    A + B = 3 × 103 + 5 × 102 + 4 × 101 + 0 × 100 + 0 × 10-1 + 7 × 10-2 = 3.54007 × 103
    as it should be. Therefore, the solution is correct.
  2. Again, for convenience, we use the letters A and B to represent the given numbers. We have
    A = 5 × 103 + 2 × 101 + 1 × 10-2
    B = 3 × 102 + 8 × 100 + 6 × 10-2
    Including the missing terms to avoid confusion during operations yields
    A = 5 × 103 + 0 × 102 + 2 × 101 + 0 × 100 + 0 × 10-1 + 1 × 10-2
    B = 3 × 102 + 0 × 101 + 8 × 100 + 0 × 10-1 + 6 × 10-2
    Therefore,
    A-B = 5 × 103 + (0-3) × 102 + (2-0) × 101 + (0-8) × 100 + (0-0) × 10-1 + (1-6) × 10-2
    From right to left the coefficients become as follows:
    1 - 6 = -5, so borrowing one ten, yields 11 - 6 = 5.
    Since the borrowing was made possible from the tens, all the other values after the tens become 9. Thus, the coefficient of tenths becomes 9 - 0 = 9.
    For units, we have 9 - 8 = 1.
    Now, the coefficient of tens is 1, not 2 as it initially was, because of the previous borrowing. Hence, the coefficient of tens is 1 - 0 = 1.
    As for hundreds, we have 0 - 3. Thus, we borrow from hundreds and as a result, we obtain 10 - 3 = 7.
    Last, the thousands coefficient became 4 because of the previous borrowing. Thus, the result of the given subtraction is
    A-B = 4 × 103 + 7 × 102 + 1 × 101 + 1 × 100 + 9 × 10-1 + 5 × 10-2
    Proof:
    In standard form:
    A = 5 × 103 + 0 × 102 + 2 × 101 + 0 × 100 + 0 × 10-1 + 1 × 10-2 = 5.02001 × 103
    B = 3 × 102 + 0 × 101 + 8 × 100 + 0 × 10-1 + 6 × 10-2 = 3.0806 × 102 = 0.30806 × 103
    Thus,
    A-B = 5.02001 × 103 -0.30806 × 103
    = (5.02001 - 0.30806) × 103
    = 4.71195 × 103
    In ordinary form:
    A = 5 × 103 + 0 × 102 + 2 × 101 + 0 × 100 + 0 × 10-1 + 1 × 10-2 = 5020.01
    B = 3 × 102 + 0 × 101 + 8 × 100 + 0 × 10-1 + 6 × 10-2 = 308.06
    Thus,
    5020.01- 308.06
    4711.95
    When written in the decomposed form, this number becomes
    A-B = 4711.95
    = 4 × 103 + 7 × 102 + 1 × 101 + 1 × 100 + 9 × 10-1 + 5 × 10-2
    This value is the same as that obtained through subtraction of the two numbers written in the standard and ordinary form. Therefore, the solution was correct.

Multiplication and Division of Numbers in Standard Form

In multiplication and division of numbers in standard form, it is better to not involve the expanded form because of the difficulties in handling such numbers.

When multiplying two numbers in the standard form, the numerical parts before the powers of ten multiply while indices that show the powers of ten add to each other. In symbols, we have

(A × 10m ) ∙ (B × 10n ) = (A ∙ B) × 10m + n

All corrections derived from the fact that A · B may be greater than 10 are made after writing the result of operation in the above form.

For example, if X = 3.24 × 105 and Y = 5.42 × 108, when multiplying them, we obtain

A × B = (3.24 × 105 ) ∙ (5.42 × 108 )
= (3.24 ∙ 5.42) × 105 + 8
= 17.5608 × 101 3

A similar procedure is also used in the division of numbers written in the standard form. Thus, in this operation the numerical parts before powers of ten divide while the powers of ten are subtracted. For example, if A = 4.29 × 107 and B = 3.0 × 104, the result of division between them is

A ÷ B = 4.29 × 107 ÷ 3.0 × 104
= (4.29 ÷ 3.0) × 107-4
= 1.43 × 103

Example 4

Calculate:

  1. (8.62 × 109) · (7.5 × 1013)
  2. (1.31 × 108) ÷ (6.55 × 103)

Solution 4

  1. We have
    (8.62 × 109 ) ∙ (7.5 × 1013)
    = (8.62 ∙ 7.5) × 109 + 13
    = 64.65 × 1022
    = 6.465 × 1023
  2. We have
    (1.31 × 108 ) ÷ (6.55 × 103 )
    = (1.31 ÷ 6.55) × 108 - 3
    = 0.2 × 105
    = 2 × 104

Very Big and Very Small Numbers

Standard numbers are not suitable to use when expressing normal values if we don't have any strong reason for this. For example, it is not necessary to write 38 as 3.8 × 101 or 0.41 as 4.1 × 10-1 unless this is explicitly demanded.

We normally use the standard form for convenience - to express either very large or very small numbers. For example, instead of saying, "the green light has a frequency of about 545 trillion hertz", we say, "the frequency of green light is about 5.45 × 1014 hertz". This saves us the time and effort to write numbers with many zeroes at the end.

The same thing can be said for very small numbers as well. Thus, instead of saying "the diameter of a hydrogen atom is 0.106 nm" (nm means 'nanometre', which is equal to one billionth of one metre), we say "the diameter of a hydrogen atom is 1.02 × 10-10 m". This saves us a lot of previous time in units conversions and other annoying actions deriving from the use of such numbers.

The general rules applied when converting very small or very large ordinary numbers into standard form are as follows:

  1. If a whole number A = abcdef…n is very big, the decimal point shifts n - 1 positions due left and the index of ten also becomes n - 1, where n is the number of digits of the original number. Hence, the number becomes A = a.bcdef…n × 10n-1. For example, in the number A = 327,416, we have n = 6. Thus, we obtain for the standard form of this number: A = 3.27416 × 105.
  2. If a big number contains a decimal part (if it is not whole therefore), the above rule is applied by starting from the decimal point. However, the digits after the decimal point are all written when the number is converted into the standard form. For example, if A = 30,681.27 is converted to standard form (n = 5), it becomes A = 3.068127 × 104.
  3. If a very small decimal number B = 0.0000abcd needs to be converted into standard form, we shift the decimal point from its actual position to after the first non-zero digit (here after a). As for the index, we count how many positions the decimal point has shifted and this number is written as a negative index. In our case, we have a shift by five positions due right, so B = a.bcd × 10-5.

Example 5

The wavelength of yellow light is 580 nm. Given that the speed of light is 300,000 km/s, calculate the frequency of yellow light in hertz. The equation of light waves is

speed = wavelength × frequency

where speed is measured in metres per second.

Solution 5

First, we write all values in standard form. Thus, since nanometre is one billionth of metre or 10-9 m and 1 km = 1000 m = 103 m, we have

Wavelength = 580 nm = 580 × 10-9 m = 5.80 × 10-7 m
Speed = 300,000 km/s = 300,000 × 1000 m/s = 300,000,000 m/s = 3.0 × 108 m/s

Thus, since

speed = wavelength × frequency
frequency = speed/wavelength
= 3.0 × 108 m/s/5.80 × 10-7 m
≈0.517 × 108-(-7) Hz
= 0.517 × 1015 Hz
= 5.17 × 1014 Hz

Powers of Numbers Written in the Standard Form

If a number written in the standard form is raised to a certain power, we raise in that power only the numerical part before the power of ten, while the powers of ten multiply. This is just the application of properties of indices explained in Tutorial 7.1. In symbols, we have

(A × 10n )m = Am × 10n ∙ m

For example,

(2.7 × 105 )3 = (2.73 ) × 105 ∙ 3
= 19.683 × 1015
= 1.9683 × 1016

The same procedure is also used for numbers written in standard form when they are raised in a negative power. You just have to be careful when writing the sign of the result. For example,

(3.09 × 102 )-3 = (3.09-3 ) × 102 ∙ (-3)
≈ 0.0339 × 10-6
= 3.39 × 10-8

Example 6

The side length of a small cube is a = 1.84 mm. What is its volume in cubic metres? Volume of a cube is calculated by the formula V = a3. The conversion factor is 1 m = 1000 m.

Solution 6

First, we express the side length in metres. We have

a = 1.84 mm = 1.84 × 10-3 m

Thus,

V = a3 = (1.84 × 10-3 m)3
= (1.843 ) × (10-3 )3 m3
= 6.229504 × 10(-3) ∙ 3 m3
= 6.229504 × 10-9 m3

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