Is Binet's formula for the Fibonacci numbers exact? Is Binet's formula for the Fibonacci numbers exact?
$F_n = \frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n \sqrt{5}}$
If so, how, given the irrational numbers in it?
Thanks.

As others have noted, the $\sqrt 5$ parts cancel, leaving an integer. We can recover the Fibonacci recurrence formula from Binet as follows:
$$F_n+F_{n-1} = \frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n \sqrt{5}}+\frac{(1+\sqrt{5})^{n-1}-(1-\sqrt{5})^{n-1}}{2^{n-1} \sqrt{5}}=$$$$\frac{(1+\sqrt{5})^{n-1}(1+\sqrt 5+2)-(1-\sqrt{5})^{n-1}(1-\sqrt5+2)}{2^n \sqrt{5}}$$
Then we notice that $(1\pm\sqrt5)^2=6\pm2\sqrt 5=2(3\pm\sqrt5)$
And we use this to simplify the final expression to $F_{n+1}$ so that $F_n+F_{n-1} =F_{n+1}$
And the recurrence shows that if two successive $F_r$ are integers, every Fibonacci number from that point on is an integer. Choose $r=0,1$. This is another way of proving that the cancellation happens.
