I am reading this proof of Bézout's identity. It starts as:
For given nonzero integers $a$ and $b$ there is a nonzero integer $ax + by$, $x$ and $y$ are also integers. The minimum absolute value of $ax+by$ is supposed to be $as+bt=d$($d$ can be made positive by changing the signs of $s$ and $t$). Then they say that dividing $a$ or $b$ with $d$ gives the reminder which is of the form $ax+by$.
By Euclid division we have:
$a=q_1d+r_1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ and $\ \ \ \ \ \ b=q_2d+r_2$
$a=q_1(as+bt)+r_1\ \ \ \ \ \ \ \ \ \ \ $ and $\ \ \ \ \ \ b=q_2(as+bt)+r_2$
$a(1-q_1s)+b(-q_1t)=r_1\ \ $ and $\ \ \ \ \ a(-q_2s)+b(1-q_2s)=r_2$
So, $r_1$ and $r_2$ are of form $ax+by$. Then they say $r_1$ and $r_2$ must be less than $d$, acc. to Euclid's division. Upto now everything is clear to me but not the next step. In the next step they say that the only way $r_1$ and $r_2$ can be smaller that $d$ is that they become $0$. How?
Why $r_1$ and $r_2$ have to be $0$?
In the proof, it is stated that $d$ is the minimum absolute value of the form $as + bt$. And, as the remainder of division of $a$ by $d$ must be less than $d$, and also must be of the form $ax+by$ (because $d$ and $a$ are both of the same form), hence the only option which remains for the remainder to be is $0$ (because $d$ is the smallest positive value, hence there cant be any other positive number of the same form less than $d$). Thus, $r_1$ is 0. Similar argument for $r_2$.