Set down the number whose square root is wanted, say 43. This square root is 6 and a fraction. Set down the integer 6 in the first and third row, and 1 in the second row always. Form the successive rows each from the one before, in the following manner:
| One row being | The next row has b′ , a′ , c′ , formed in this order, thus, |
|---|---|
| a | a′ = excess of b′c′ , already formed, over a . |
| b | b′ = quotient of 43 - a ² divided by b . |
| c | c′ = integer in the quotient of 6 + a divided by b′ . |
| Thus the second row is formed from the first, as under: | |
| 6 | 1 = excess of 7 × 1 (both just found) over 6. |
| 1 | 7 = 43 - 6 × 6 divided by 1. |
| 6 | 1 = integer of 6 + 6 divided by 7 (just found). |
| The third row is formed from the second, thus: | |
| 1 | 5 = excess of 1 × 6 over 1. |
| 7 | 6 = 43 - 1 × 1 divided by 7. |
| 1 | 1 = integer of 6 + 1 divided by 6; |
and so on. In process of time the second column, 1, 7, 1, occurs again, after which the several columns are repeated in the same order. As a final process, take the set in the lowest line (excluding the first, 6), namely, 1, 1, 3, 1, 5, 1, 3, &c. and use them by the rule given at the beginning of this article, as follows:
| 1 | 1 | 3 | 1 | 5 | 1 | 3 | 1 | 1, | &c. |
|---|---|---|---|---|---|---|---|---|---|
| 1 | 1 | 4 | 5 | 29 | 34 | 131 | 165 | 296 | |
| 1 | 2 | 7 | 9 | 52 | 61 | 235 | 296 | 531 |
Hence, 6¹⁶⁵/₂₉₆ is very near the square root of 43, not erring by so much as
| 1 | . |
|---|---|
| 296 × 531 |
If we try it, we shall find (⁶¹⁶⁵/₂₉₆) to be ¹⁹⁴¹/₂₉₆, the square of which is ³⁷⁶⁷⁴⁸¹/₈₇₆₁₆, or 43⁷/₈₇₆₁₆.
This rule is of use when it is frequently wanted to use one square root, and therefore desirable to ascertain whether any easy approximation exists by means of a common fraction. For example, √2 is often used.
| √2 | = 1 + ... |
|---|---|
| 1 | 1 1 |
| 1 | 1 1 |
| 1 | 2 2 2 2 2 2 |
| 1 | 2 | 5 | 12 | 29 | 70 | &c. |
|---|---|---|---|---|---|---|
| 2 | 5 | 12 | 29 | 70 | 169 |
Here it appears that