# Math Help - Fraction problem

1. ## Fraction problem

1) Under what conditions will F1 and F2 be equal?

2) If F1 and F2 are equal, what is their value?

2. I'll rename, for simplicity: the first continued fraction I'll call "x", and the other I'll call "y".

Note that x = A + B/(A + (B/(A + B/...)))) = A + B/[A + B/(A + (B/(A + (B/...)))) = A + B/x. Similarly, y = B + A/y.

For these to be equal, we must have A + B/x = B + A/y. What can you get from this?

3. Hello, dxdy!

I have a good start on it . . . I'll let you finish it.

Let $A$ and $B$ be distinct positive real numbers and consider the continued fractions:

. . ${\color{blue}[1]}\;\;F_1 \;=\; A + \frac{B}{A + \dfrac{B}{A + \hdots}}\qquad\qquad{\color{blue}[2]}\;\; F_2 \;=\;B + \frac{A}{B + \dfrac{A}{B + \hdots}}$

1) Under what conditions will $F_1 = F_2$ ?

From [1]: . $F_1 \;=\; A+ \frac{B}{F_1} \quad\Rightarrow\quad F_1^2 - AF_1 - B \:=\:0$

. . Quadratic Formula: . $F_1 \;=\;\frac{A \pm\sqrt{A^2+4B}}{2}$

From [2]: . $F_2 \;=\;B + \frac{A}{F_2} \quad\Rightarrow\quad F_2^2 - BF_2 - A \:=\:0$

. . Quadratic Formula: . $F_2 \;=\;\frac{B \pm\sqrt{B^2 + 4A}}{2}$

$\text{Since }F_1 = F_2\!:\;\;\frac{A \pm\sqrt{A^2+4B}}{2} \;=\;\frac{B \pm\sqrt{B^2+4A}}{2}$

. . . . . . $A - B \;=\;\pm\sqrt{B^2+4A} \mp \sqrt{A^2+4B}$

. . . . . . $A - B \;=\;\pm\left(\sqrt{B^2+4A} - \sqrt{A^2+4B}\right)$

Square both sides:

. . $A^2 - 2AB + B^2 \;=\;B^2 + 4A - 2\sqrt{A^2+4B}\sqrt{B^2+4A} + A^2 + 4B$

. . . . $\sqrt{A^2+4B}\sqrt{B^2+4A} \;=\;2A + 2B + AB$

Square both sides:

. . $(A^2+4B)(B^2+4) \;=\;(2A+2B+ AB)^2$

. . $A^2+B^2 + 4A^3 + 4B^3 + 16AB \;=\;4A^2 + 8AB + 4A^2B + 4B^2 + 4AB^2 + A^2B^2$

This simplifies to: . $A^3 - A^2B - AB^2 + B^3 \;=\;A^2 - 2AB + B^2$

. . Factor: . $A^2(A-B) - B^2(A-B) \;=\;(A-B)^2 \quad\Rightarrow\quad (A-B)(A^2-B^2) \;=\;(A-B)^2$

. . Factor: . $(A-B)(A-B)(A+B) \:=\:(A-B)^2 \quad\Rightarrow\quad (A-B)^2(A+B) \:=\:(A-B)^2$

Since $A \neq B$, we can divide by $(A-B)^2\!:\;\;\boxed{A + B \;=\;1}$

Can you finish it?

4. F1 = 1/2
f2 = 1/2