How to prove this is a rational number I'm not sure how to prove this is a rational number $\frac{q}{m}$, can some one show me?
$$\frac{q}{m}=\frac{(\frac{1+\sqrt5}{2})^n - (\frac{1-\sqrt5}{2})^n}{\sqrt5}$$
 A: You can use the binomial theorem to show that all the rational terms in the numerator (the ones with even powers of $\sqrt 5$ cancel) and all the terms with an odd number of factors of $\sqrt 5$ add.  It takes paying attention to the minus signs.  Then the $\sqrt 5$ in the denominator takes care of the odd $\sqrt 5$, leaving a rational.  
Another approach is induction.  Prove it by hand for $n=0, n=1$, then show that the values satisfy the Fibonacci recurrence.  As the recurrence has rational coefficients, they will all be rational (in fact, natural)
A: Lt me consider $$S(n) = 2\frac{ \left((a+b)^n-(a-b)^n\right)}{b}$$The binomial theorem will be applied to each term and the result will be expanded and simplified. So, the results are successively $$S(1) = 4$$ $$S(2) = 8 a$$  $$S(3)= 4\left(3 a^2+b^2\right)$$ $$S(4)= 16 a \left(a^2+b^2\right)$$ $$S(5)=4 \left(5 a^4+10 a^2 b^2+b^4\right)$$  $$S(6)=8 a \left(3 a^2+b^2\right) \left(a^2+3 b^2\right)$$ $$S(7)=4 \left(7 a^6+35 a^4 b^2+21 a^2 b^4+b^6\right)$$ $$S(8)=32 a \left(a^2+b^2\right) \left(a^4+6 a^2 b^2+b^4\right)$$  
So, you see, as clearly pointed out by the previous comments and answers, that if $a$ is rational and $b$ the square root of a rational, the result is rational. 
A: Use induction to show that the sequence
$$
a_n = \frac{(\frac{1+\sqrt5}{2})^n - (\frac{1-\sqrt5}{2})^n}{\sqrt5}
$$
Satisfies $a_n = a_{n-1} + a_{n-2}$.  Noting that $a_0$ and $a_1$ are rational, we can conclude (again, using induction) that each $a_n = a_{n-1} + a_{n-2}$ is the sum of rational numbers and is hence rational.
Note that we can also write the terms for $n \geq 1$ as
$$
a_n = \sum_{k=0}^{n-1} \frac{\varphi^{n-k}}{(-\varphi)^{k}}
= \sum_{k=0}^{n-1} (-1)^{k}\varphi^{n-2k}
$$
Where $\varphi = \frac{1 + \sqrt5}{2}$ is the golden ratio.  This alternate expression might be easier to deal with for induction.
A: Hint $\ $  It is invariant under conjugation $\,\sqrt{5}\mapsto -\sqrt{5},\,$ so is rational.
