I'm trying to study Number theory on my own, I encountered this problem. I knew how to solve it. the problem is
Let $a,b,c \in Z$ such that: $$a = b*c*q1 + r1$$ and $$a = b*q2 + r2 $$ and $$ q2 = c*q3 + r3 $$
the question is, show that q1 = q3
to solve it I concidered for the first expression that $$0 \leq r1 < |b*c|$$
and for the second expression first I substituted q2 by it's expression so it became: $$ a = b*c*q3 + (b*r3 + r2) $$ now I need to show that: $$ 0 \leq b*r3+r2 < |b*c|$$ If I show that, then it means that r1 = b*r3 + r2 which means that q1=q3
so I did this: from the third expression I have $$ 0 \leq r3 \leq |c| - 1$$ so by multiplying by |b| it became: $$ 0 \leq |b|*r3 \leq |b*c| - |b|$$ and I have from the second expression that: $$ 0 \leq r2 \leq |b| $$
so from the above two inequalities I add their expressions and I get: $$ 0 \leq |b|*r3 + r2 \leq |b*c|$$
here is my problem then: If it was $b \in N$ then I could simply remove the absolute value, now because $b \notin N$ so it's hard for me to solve it from here, if you can show me the other steps with clear explanations I would appreciate it. Please answer only if you know the real answer I'm someone who study maths on his own without teacher so please don't give me bad or wrong things, I don't have money to study in my county or close friends that are good at math to ask
Edited: Here is what I need from the last step to become $$ 0 \leq b*r3+r2 < |b*c|$$ so I need to remove that absolute value from the |b| to become b
I think this works. We take $a,b,c \in \mathbb{Z}$ such that $a = bcq_1+r_1$, $a=bq_2+r_2$, and $q_2=cq_3+r_3$ where, $0 \leq r_1 <bc$, $0 \leq r_2 <b$ , and $0 \leq r_3<c$. Then we have that, $$ \begin{aligned} bq_2+r_2&=bcq_1+r_1 \\ b(cq_3+r_3)+r_2&=bcq_1+r_1 \\ bc(q_3-q_1) &=r_1-br_3-r_2 \end{aligned} $$ Note that $r_1-br_3-r_2 < bc$ because $r_1 <bc$ and $0 \leq r_2<b$ and $0 \leq r_3 <c$. So we have that $bc|(r_1-br_3-r_2)$ which is strictly less than $bc$. This implies that $r_1-br_3-r_2 = 0$ hence $q_3-q_1=0$, and so $q_3=q_1$.