Euclidean Algorithm Question

Question

So I have been asked to find $d=(a,b)$ when $a=1109$ and $b=4999$ and express $d$ as a linear combination of $a$ and $b$

Well I have worked out that $d=1$ but I am struggling to express $d$ as a linear combination of $a$ and $b$

$4999=1109*4+563$
$1109=1*563+546$
$563=1*546+17$
$546=17*32+2$
$17=2*8+1$
$2=2*1$

then i have $d=1$ to work out $d$ as a linear combination i have $563=4999-4*1109$
$546=1109-1*563$
$546=1109-1(4999-1109*4)$
$546=5*1109-4999$
$17=563-546$
$17=4999-4*1109-1109-563$
$17=4999-4*1109-1109+4999-4*1109$
$17=2*4999-9*1109$
$2=546-17*32$
$2=...$
$2=-65*4999+293*1109$
$1=17-8*2$
$1=...$
$1=522*4999-2353*1109$

What I need to know is how do I work out $2=-65*4999+293*1109$ and $1=522*4999-2353*1109$ without pages of working out? also is what i have done trying to express $d$ as a linear combination of $a$ and $b$ correct? Is there a quicker way of working it out?

Many thanks for your help

Answer 1

http://en.wikipedia.org/wiki/Continued_fraction#Finite_continued_fractions

The cross product of consecutive convergents in simple continued fractions are either $\pm 1.$ The first in the string is always $\frac{0}{1}.$ The second in the string is always the nonsensical $\frac{1}{0}.$ The convergents for $\frac{4999}{1109}$ are $$ \frac{0}{1}, \; \frac{1}{0}, \; \frac{4}{1}, \; \frac{5}{1}, \; \frac{9}{2}, \; \frac{293}{65}, \; \frac{2353}{522}, \; \frac{4999}{1109}.$$

We get little 2 by 2 determinant $$ 2353 \cdot 1109 - 522 \cdot 4999 = -1. $$

So $$ -2353 \cdot 1109 + 522 \cdot 4999 = 1. $$

This is not a difficult method to learn and has other uses. Note that the letters $a_i,$ called quotients on wikipedia, are $$ \langle 4; 1,1,32,8,2 \rangle. $$ These are $ \langle a_0; a_1,a_2,a_3,a_4,a_5 \rangle. $ I did it with a calculator. Beginning with the number $x_0 = \frac{4999}{1109} \approx 4.5077664563,$ we begin with $$ a_0 = \lfloor x_0 \rfloor = 4.$$ The fractional part is $x_0 - \lfloor x_0 \rfloor \approx 0.5077664563,$ and we take the reciprocal to get $$ x_1 = \frac{1}{x_0 - \lfloor x_0 \rfloor} \approx \frac{1}{0.5077664563} \approx 1.969804617 $$

Then $$ a_1 = \lfloor x_1 \rfloor = 1.$$ The fractional part is $x_1 - \lfloor x_1 \rfloor \approx 0.969804617,$ and we take the reciprocal to get $$ x_2 = \frac{1}{x_1 - \lfloor x_1 \rfloor} \approx \frac{1}{0.969804617} \approx 1.031135532. $$

Then $$ a_2 = \lfloor x_2 \rfloor = 1.$$ The fractional part is $x_2 - \lfloor x_2 \rfloor \approx 0.031135532,$ and we take the reciprocal to get $$ x_3 = \frac{1}{x_2 - \lfloor x_2 \rfloor} \approx \frac{1}{0.031135532} \approx 32.11764617. $$

Then $$ a_3 = \lfloor x_3 \rfloor = 32.$$ The fractional part is $x_3 - \lfloor x_3 \rfloor \approx 0.11764617,$ and we take the reciprocal to get $$ x_4 = \frac{1}{x_3 - \lfloor x_3 \rfloor} \approx \frac{1}{0.11764617} \approx 8.500064218. $$

Then $$ a_4 = \lfloor x_4 \rfloor = 8.$$ The fractional part is $x_4 - \lfloor x_4 \rfloor \approx 0.500064218,$ and we take the reciprocal to get $$ x_5 = \frac{1}{x_4 - \lfloor x_4 \rfloor} \approx \frac{1}{0.500064218} \approx 1.999743161. $$ it is here that experience becomes critically important. We are seeing the effect of tiny calculator roundoffs. The actual value is $x_5 = 2,$ from which we get $a_5 = 2,$ and that concludes the s.c.f.

Let us put in the "quotients" where they belong in the calculation: $$ \frac{0}{1}, \; \frac{1}{0}, \; \color{magenta}{\frac{4}{}}, \; \frac{4}{1}, \; \color{magenta}{\frac{1}{}}, \; \frac{5}{1}, \; \color{magenta}{\frac{1}{}}, \; \frac{9}{2}, \; \color{magenta}{\frac{32}{}}, \; \frac{293}{65}, \; \color{magenta}{\frac{8}{}}, \; \frac{2353}{522}, \; \color{magenta}{\frac{2}{}}, \;\frac{4999}{1109}.$$

In every case, given two consecutive numerators (in black) followed by a quotient (in magenta), the first numerator plus the second numerator times the quotient is the following numerator. So, $$ 0 + 1 \cdot 4 = 4, \; 1 + 4 \cdot 1 = 5, \; 4 + 5 \cdot 1 = 9, \; 5 + 9 \cdot 32 = 293, \; 9 + 293 \cdot 8 = 2353, \; 293 + 2353 \cdot 2 = 4999. $$

Same for denominators, $$ 1 + 0 \cdot 4 = 1, \; 0 + 1 \cdot 1 = 1, \; 1 + 1 \cdot 1 = 2, \; 1 + 2 \cdot 32 = 65, \; 2 + 65 \cdot 8 = 522, \; 65 + 522 \cdot 2 = 1109. $$

In case you are worried by this: the exact version, with fractions instead of calculator decimal approximations, is exactly what you have been doing: $$ x_0 = \frac{4999}{1109} = 4 + \frac{563}{1109}, $$ $$ x_1 = \frac{1109}{563} = 1 + \frac{546}{563}, $$ $$ x_2 = \frac{563}{546} = 1 + \frac{17}{546}, $$ $$ x_3 = \frac{546}{17} = 32 + \frac{2}{17}, $$ $$ x_4 = \frac{17}{2} = 8 + \frac{1}{2}, $$ $$ x_5 = \frac{2}{1} = 2. $$

Answer 2

$4999=4*1109+563$ so $563=4999-4*1109=b-4a$

$1109=1*563+546$ so $546=1109-563=a-(b-4a)=5a-b$

$563=1*546+17$ so $17=563-546=(b-4a)-(5a-b)=2b-9a$

$546=32*17+2$, so $2=546-32*17=(5a-b)-32(2b-9a)=293a-65b$

$17=8*2+1$ so $1=17-8*2=(2b-9a)-8(293a-65b)=522b-2353a=522*4999-2353*1109$

It's even faster and easier if you don't insist on using positive remainders all the time:

$4999=5*1109-546$ so $546=5*1109-4999=5a-b$

$1109=2*546+17$ so $17=1109-2*546=a-2(5a-b)=2b-9a$

$546=32*17+2$, so $2=546-32*17=(5a-b)-32(2b-9a)=293a-65b$

$17=8*2+1$ so $1=17-8*2=(2b-9a)-8(293a-65b)=522b-2353a=522*4999-2353*1109$

Answer 3

$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\pp}{{\cal P}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}$ \begin{align} 4999 &= 4\times 1109 + 563\tag{1} \\ 1109 &= 1\times 563 + 546\tag{2} \\ 563 &= 1\times 546 + 17\tag{3} \\ 546 & = 32 \times 17 + 2\tag{4} \\ 17 &= 8\times 2 + \color{#ff0000}{\large 1}\tag{5} \\ 2 &= 2\times 1 + 0\tag{6} \end{align}

With $\pars{4}$ and $\pars{5}$ we eliminate the factor $2$: $$ 8\times 546 - 17 = 256\times 17 - \color{#ff0000}{\large 1} \quad\imp\quad 8\times 546 = 257\times 17 - \color{#ff0000}{\large 1}\tag{7} $$

With $\pars{3}$ and $\pars{7}$ we eliminate the factor $17$: $$ 257\times 563 - 8\times 546 = 257\times 546 + \color{#ff0000}{\large 1} \quad\imp\quad 257\times 563 = 265\times 546 + \color{#ff0000}{\large 1}\tag{8} $$

With $\pars{2}$ and $\pars{8}$ we eliminate the factor $546$: $$ \!\!\!\!\!\!\!1109\times 265 - 257\times 563 = 265\times 563 - \color{#ff0000}{\large 1} \quad\imp\quad 1109\times 265 = 522\times 563 - \color{#ff0000}{\large 1}\tag{9} $$

With $\pars{1}$ and $\pars{9}$ we eliminate the factor $563$: $$ 4999\times 522 - 1109\times 265 = 2088\times 1109 + \color{#ff0000}{\large 1} \quad\imp\quad \color{#ff0000}{\large1 = 522\times \color{#0000ff}{4999} - 2353\times \color{#0000ff}{1109}} $$

Answer 4

Using the Euclid-Wallis Algorithm, $$ \begin{array}{r} &&4&1&1&32&8&2\\\hline 1&0&1&-1&2&-65&522&-1109\\ 0&1&-4&5&-9&293&-2353&4999\\ 4999&1109&563&546&17&2&1&0\\ \end{array} $$ The column with the $1$ at its base says that $(4999,1109)=1$ and that for any $k$, $$ (522-1109k)4999+(-2353+4999k)1109=1 $$ In particular, we have for $k=0$ $$ 522\cdot4999-2353\cdot1109=1 $$ or for $k=1$ $$ 2646\cdot1109-587\cdot4999=1 $$

Answer 5

Your general methodology seems correct. You have indeed worked backwards through your Euclidean Algorithm result and found coefficients for a and b. It is not entirely clear however why you began with 546 = 1109-1*563. You may want to begin with 4999-4*1109 = 563 and then perform a similar method as you have been. The work on these can be long unfortunately.

Answer 6

Write down the quotients backwards 8 32 1 1 4, leaving out the last quotient 8. Then write 8 32 1 1 4 1 8 Then multiply the first 32 in the top row by the 8 in the bottom row and add the 1 in the second row, getting 257, and write 8 32 1 1 4 1 8 257 Then multiply the first 1 in the first row by the 257 in the second row and add the 8 in the second row, getting 265, and write 8 32 1 1 4 1 8 257 265
Multiply the second 1 in the first row by the 265 and add 257, getting 522, and write 8 32 1 1 4 1 8 257 265 522
Finally, multiply the 4 in the first row by 522 and add 265, getting 2353, and write 8 32 1 1 4 1 8 257 265 522 2353 Now write a - sign under the 8 in the second row, a + under the 257, etc, getting 8 32 1 1 4 1 8 257 265 522 2353 - + - + - Finally, write x under the 2353 and y under the 522, the result of which is x = -2253, y - 522, and we have solved 1109x + 4999y = 1.

Art DuPre'