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Which irrational number represents the infinite simple continued fraction [0;7]?

-So from my current understanding [o;7] can be represented as the following:

$ = \frac{1}{7 + \frac{1}{7 + \frac{1}{7 +...}}}$

So, x would be $x = \frac{1}{7 + \frac{1}{7 + x}}$

$\frac{7(7+x)}{7+x} + \frac{1}{7+x}$

$\frac{49 + 7x}{7+x} + \frac{1}{7+x}$

$ \frac{50 + 7x}{7+x}$

From here we get:

$x(50 + 7x) = 7 + x$

$50x = 7x^2 = 7 + x$

$49x + 7x^2 = 7$

$ 49x = 7x^2 - 7 = 0$

$x = \frac{-49\pm \sqrt{2205}}{14}$

  • UPDATED, the square root looked a bit weird to me wanted to confirm my answer or suggestions, any help is appreciated.
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    $\begingroup$ Write it as $x = \frac{1}{7+x}$ instead. $\endgroup$ – Cameron Williams May 1 '16 at 0:42
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    $\begingroup$ Cameron showed you the easy way, but even if you didn't figure out that easy way yourself, why are you having difficulty with "your" way? It's simple fraction algebra: $$\frac 1 {7 + \frac 1 {7+x}} = \frac 1 {\frac {50+ 7x}{7+x}} = \frac {7+x}{50 + 7x}$$ Can you continue from here? For practice, you should solve it both ways (finish this approach, then do it Cameron's easier way) and convince yourself you get the same answer. If you don't get the same answer, try to find where you made a mistake (or where I made a mistake!) - it will be good practice. $\endgroup$ – mathguy May 1 '16 at 0:47
  • $\begingroup$ Your solution is right up to the last step. When you solve for $x$ in the quadratic equation $7x^2+49x-7=0$, you should get $\frac{-49\pm \sqrt{2597}}{14}$ which simplifies to $\frac{ -7\pm \sqrt{53} }{2}$. Your value in your root is 2205, but it should be 2597 since $b^2-4ac=49^2-(4)(7)(-7)=2597$. $\endgroup$ – M47145 May 1 '16 at 1:08
  • $\begingroup$ It should be noted that the general infinite continued fraction $[0; n]$ for $n\geq 1$ is equal to $$-\frac 12\left( n \ ± \sqrt{n^2 + 4}\right) $$ $\endgroup$ – MathematicsStudent1122 May 1 '16 at 1:09
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Let $x$ be the value of the continued fraction $[0;7]$. We have that $$ x = \frac{1}{7 + \frac{1}{7 + \frac{1}{7 +...}}}$$

Now we notice that $x$ appears in this continued fraction, so we can write it as: $$x = \frac{1}{7 + x}$$

This simplifies to $x^2+7x-1=0$, and solving for $x$ gives $x=\frac{-7}{2}\pm \frac{\sqrt{53}}{2}$

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