places as there are decimal places both in 17·036 and 4·27, because the product of two decimal numbers will contain as many ciphers as there are ciphers in
- This question now arises: What if there should not be as many figures in the product as there are decimal places in the multiplier and multiplicand together? To see what must be done in this case, multiply ·172 by ·101, or ¹⁷²/₁₀₀₀ by ¹⁰¹/₁₀₀₀. The product of these two is ¹⁷³⁷²/₁₀₀₀₀₀₀, or ·017372 (135). Therefore, when the number of places in the product is not sufficient to allow the rule of the last article to be followed, as many ciphers must be placed at the beginning as will make up the deficiency.
ADDITIONAL EXAMPLES.
| ·001 × ·01 is | ·00001 |
|---|---|
| 56 × ·0001 is | ·0056. |
EXERCISES.
Shew that
| 3·002 × 3·002 | = | 3 × 3 + 2 × 3 × ·002 + ·002 × ·002 |
|---|---|---|
| 11·5609 × 5·3191 | = | 8·44 × 8·44 - 3·1209 × 3·1209 |
| 8·217 × 10·001 | = | 8 × 10 + 8 × ·001 + 10 × ·217 + ·001 × ·217. |
| Fraction. | Square. | Cube. |
|---|---|---|
| 82·92 | 6875·7264 | 570135·233088 |
| ·0173 | ·00029929 | ·000005177717 |
| 1·43 | 2·0449 | 2·924207 |
| ·009 | ·000081 | ·000000729 |
| 15·625 | × | 64 | = | 1000 |
|---|---|---|---|---|
| 1·5625 | × | ·64 | = | 1 |
| ·015625 | × | ·0064 | = | ·0001 |
| ·15625 | × | ·64 | = | ·1 |
| 1562·5 | × | ·064 | = | 100 |
| 15625000 | × | ·064 | = | 1000000 |
- The division of a decimal by a decimal number, such as 10, 100, 1000, &c., is performed by moving the decimal point as many places to the left as there are ciphers in the decimal number. If there are not places enough in the