-1

Want help in investigating the various geometrical reasons associated with finding the $\gcd$ by Euclidean algorithm, as follows:

(i) Why the slope and intercept of the linear Diophantine equation(LDE) remains the same under the Euclidean algorithm, i.e. under recursive repeated subtraction. The term 'recursive' refers to the exchange of divisor and dividend with remainder and divisor respectively in the next step.

The underlying assumption is that for a LDE $ax+by =c$, the steps involved in finding $\gcd$ must be keeping the slope and intercept same, i.e. not affecting the character of the straight line represented by the LDE. This assumption is derived from the fact that the algebraic interpretation of the Euclidean algorithm considers $\gcd$ as invariant for both sets of pairs of values for any given step: (dividend, divisor), (divisor, remainder).

Based on the assumption, want to know how the given subtraction achieves it geometrically.

(ii) Also for a LDE $ax+by =c$, know that for it to have solution $(a,b)\mid c$. But, what does this condition (i.e. $\gcd$ being a factor of $c$) imply geometrically.

(iii) As $\gcd$ is a linear combination that yields the smallest positive integer value. So, is it possible to express $\gcd$ as a parameter of the straight line in geometrical terms.
Elaborating the above, algebraically, $\gcd$ is the largest common factor of $a,b$; while LDE has two geometric parameters in form $y = mx+c =\frac{c}{b} - \frac{a}{b}x$: (i) slope $(m)= \frac{a}{b}$, (ii) y-intercept $(c)=\frac{c}{b}$.
So how to interpret $\gcd$ in terms of the slope, intercept of the given LDE.

jitender
  • 707
  • @downvoters I have not found an intuitive proof of $\gcd$ till date, & hence feel that geometric one will only serve it. So, please tell a reason to contradict, or upvote it back. Read Griffin Harriet, Gareth Jones, et. all; but none satisfied me; and it is definite that here would be a geometric interpretation, then why not I should ask for fear of downvoting. Knowledge does not come from fear. – jitender Feb 25 '18 at 12:30
  • Also, it is true that the Euclidean geometry concerns the slope of a line, and the intercept. So, why not view the $\gcd$ in terms of the slope - intercept form of line; although it is true that the algorithm is known before Euclid. – jitender Feb 25 '18 at 13:21

1 Answers1

1

(i) Consider a particular example,

$$16x+10y=26$$

We have $16=10+6$,

hence $$(10+6)x+10y=26$$

We just represented $16$ as $10+6$ without changing the equation at all, hence that is why the equation remains the same. We do have $\gcd(16,10)=\gcd(10,6)$, but it doesn't change the equation, it just change the representation of the coefficient.

Similarly, for the general case when we have $a=bq+r$.

$$ax+by=c$$ becomes $$(bq+r)x+by=c$$

(ii) Consider that $b \neq 0$,

By Bezout Theorem, we know that there are integer solutions to $ax+by = \gcd(a,b)=d$, that is there are integer solutions to $y=-\frac{a}bx+\frac{d}{b}$. We know that there are solutions to $ax+by=c$ if $d$ divides $c$, geometrically, the new line corresponds to $y=-\frac{a}bx+\frac{c}{b}$, hence we know that there are solutions to the system, if the line $y=-\frac{a}bx+\frac{d}{b}$ is translated vertically by integer multiple of $\frac{d}{b}$.

(iii)

Suppose $\gcd(a,b)=d$ and $c=kd$ where $k \in \mathbb{Z}$, and suppose we have found that $a \hat{x} + b \hat{y} = d$

it is known that the general solution to the linear diophantine equation $ax+by=c$ can be written as

$$(x,y)= \left( k\hat{x}+ \frac{mb}d , k\hat{y}- \frac{ma}d \right)$$

where $m \in \mathbb{Z}$. You can either use that expression or use the intuition of part $(ii)$ to deduce that the solution are on the line $$y=-\frac{a}bx + \frac{kd}b=-\frac{a}bx + \frac{c}b$$

Siong Thye Goh
  • 155,038
  • Thanks a lot for the kindest help. It has opened my mind for the missing steps to understand the same. – jitender Feb 26 '18 at 01:40