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Welcome to our Math lesson on Addition and Subtraction with Numbers in Standard and Decomposed Form, this is the third lesson of our suite of math lessons covering the topic of Standard Form, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.
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.
- (3 × 103 + 6 × 101 + 2 × 10-2) + (4 × 102 + 8 × 101 + 5 × 10-2)
- (5 × 103 + 2 × 101 + 1 × 10-2 )-(3 × 102 + 8 × 100 + 6 × 10-2)
Solution 3
- 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. - 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.
More Standard Form Lessons and Learning Resources
Powers and Roots Learning Material| Tutorial ID | Math Tutorial Title | Tutorial | Video
Tutorial | Revision
Notes | Revision
Questions |
|---|
| 7.3 | Standard Form | | | | |
| Lesson ID | Math Lesson Title | Lesson | Video
Lesson |
|---|
| 7.3.1 | The Meaning of Standard Form. | | |
| 7.3.2 | Writing Decimals in Standard and Decomposed Form | | |
| 7.3.3 | Addition and Subtraction with Numbers in Standard and Decomposed Form | | |
| 7.3.4 | Multiplication and Division of Numbers in Standard Form | | |
| 7.3.5 | Very Big and Very Small Numbers | | |
| 7.3.6 | Powers of Numbers Written in the Standard Form | | |
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