--- id: 5900f3a51000cf542c50feb8 challengeType: 5 title: 'Problem 57: Square root convergents' forumTopicId: 302168 --- ## Description
It is possible to show that the square root of two can be expressed as an infinite continued fraction.
$\sqrt 2 =1+ \frac 1 {2+ \frac 1 {2 +\frac 1 {2+ \dots}}}$
By expanding this for the first four iterations, we get: $1 + \frac 1 2 = \frac 32 = 1.5$ $1 + \frac 1 {2 + \frac 1 2} = \frac 7 5 = 1.4$ $1 + \frac 1 {2 + \frac 1 {2+\frac 1 2}} = \frac {17}{12} = 1.41666 \dots$ $1 + \frac 1 {2 + \frac 1 {2+\frac 1 {2+\frac 1 2}}} = \frac {41}{29} = 1.41379 \dots$ The next three expansions are $\frac {99}{70}$, $\frac {239}{169}$, and $\frac {577}{408}$, but the eighth expansion, $\frac {1393}{985}$, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator. In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?
## Instructions
## Tests
```yml tests: - text: squareRootConvergents() should return a number. testString: assert(typeof squareRootConvergents() === 'number'); - text: squareRootConvergents() should return 153. testString: assert.strictEqual(squareRootConvergents(), 153); ```
## Challenge Seed
```js function squareRootConvergents() { return true; } squareRootConvergents(); ```
## Solution
```js // solution required ```