Continued fraction/Arithmetic: Difference between revisions

and a function <math>G(\mathrm{matrix}</math> <math>\mathit{NG}, \mathrm{Number N}<sub/math>1 </submath>N_1, \mathrm{Number}</math> N<submath>2N_2)</submath>)

: <math>\frac{a_{12}*\times N_1*\times N_2 + a_1*\times N_1 + a_2*\times N_2 + a}{b_{12}*\times N_1*\times N_2 + b_1*\times N_1 + a_2*\times N_2 + b}</math>

: <math>\mathit{NG} = \begin{bmatrix}

\end{bmatrix}</math> adds N₁ to N₂

:<math>\mathit{NG} = \begin{bmatrix}

\end{bmatrix}</math> subtracts N₂ from N₁

:computes <math>\begin{bmatrix}N_1 - N_2</math>

\end{bmatrix}</math> multiplies N₁ by N₂

:computes <math>N_1 \begin{bmatrix}times N_2</math>

\end{bmatrix}</math> divides N₁ by N₂

:computes <math>\begin{bmatrix}N_1 / N_2</math>

\end{bmatrix}</math> calculates (3*N₁ + 4) * (7*N₂ - 5)

:I could define the solution to be N<submath>N_1 = 1</submath> = 1, N<sub>2</submath>N_2 = 1</math> and NG = :

: <math>\mathit{NG} = \begin{bmatrix}

:So I can define arithmetic as operations on this matrix which make a<submath>a_{12}</submath>, a<submath>1a_1</submath>, a<submath>2a_2</submath>, b<submath>b_{12}</submath>, b<submath>1b_1</submath>, b<submath>2b_2</submath> zero and read the answer from <math>a</math> and <math>b</math>. This is more interesting when N<submath>1N_1</submath> and N<submath>2N_2</submath> are continued fractions, which is the subject of the following tasks.