0

On the internet, I've seen the greatest common divisor with the notation $a \cdot s + b \cdot t = ggT(a,b)$

So for $a = 121$ and $b = 33$, the table is

\begin{array}{|c|c|c|c|} \hline r& q & s & t \\ \hline 121 & - & 1 & 0\\ \hline 33 & 3 & 0 &1 \\ \hline 22 & 1 & 1& -3\\ \hline 11 & 2 & -1& 4\\ \hline 0 & & & \\ \hline \end{array}

I don't understand how the values of $s$ and $t$ are formed in the table.

For example, I can see that $4 = 1-(-3)$, but why is then $0 - 1 = -3$?

Henno Brandsma
  • 242,131
Math Dummy
  • 53

1 Answers1

1

The first two rows $s$ and $r$ are always chosen we have $1\ 0$ and $0\ 1$, because we always want to have

$$121s+ 33t = r\tag{1}$$

in every row. And taking these coefficients for the first two rows will do that trivially. Make sure that the largest of the two numbers we want to have the gcd of is first, the smaller one second. We always want the first column to be decreasing.

Now, from row 2 onwards $q$ is the quotient of the previous row by the current row, and in your example we see that $$121=3\times 33 + 22$$ so $q=3$ and the new $r$ (remainder) is $22$. To get $22$ we have thus subtracted $3$ ($q$) times the smaller one from the larger one, and to preserve the relation (1) we have to compute the $r$ and $s$ from the previous rows in the same way:

$$(1\times 121 + 0 \times 33) + (-3)\times(0 \times 121 + 1\times 33)$$

which comes down to (just looking at the coefficients):

$$(1-3 \times 0)\times 121 + (0 - 3 \times 1) \times 33$$

and so the new $s$ for row 3 is $1-3 \times 0 = 1$ and the new $t$ is $0-3=-3$.

You can check that indeed $$121 + (-3)\times 33 = 22$$ so relation (1) has been preserved.

Now $22$ goes once into $33$ with remainder $11$, so the next $q$ is $1$ indeed, and we do a simple subtraction to get from $33$ and $22$ to $11$, and the $s$ and $t$ follow $0-1=-1$ and $1-(-3)=4$.

The final remainder is $0$ and we stop. So $11$ is the gcd and the coefficients are $-1$ and $4$.

Henno Brandsma
  • 242,131