Let \(a\) and \(b\) be two inetgers with \(b\gt 0\).
  To prove there exists two unique integers \(q\) and \(r\) such that \begin{align*} a=bq+r \text{ where } 0\leq r \lt b \end{align*}   Let us construct a set \begin{align*} S=\set{a-bt:t\in \mathbb{Z},~a-bt\geq 0 } \end{align*}   we have \begin{align*} & b\gt 0 \text{ and } b\in \mathbb{Z}\\ \implies & b\geq 1 \\ \implies & b\big|a\big|\geq \big|a\big|\\ \implies & a+ b\big|a\big|\geq a+\big|a\big|\\ \implies & a- b\big(-\big|a\big|\big)\geq 0 ~\because a+\big|a\big|\geq 0\\ \implies & a- b\big(-\big|a\big|\big)\in S \\ \implies & S\ne \phi \end{align*}   Since \(a,b\) and \(t\) are integers and \(a-bt\geq 0\)
  \(\therefore~ S\) contains non-negetive integers.
  Then \(S\) has a least element.
  Let \(r\) be the least element of \(S\).
  Then \(\exists\) an integer \(q\) such that \begin{align*} & a-bq=r \text{ where } 0\leq r \\ \implies & a=bq+r \text{ where } 0\leq r \end{align*}   If possible let \(r\geq b\).
  Then \begin{align*} & r-b \leq r ~\because b\gt 0\\ \end{align*}   Again \begin{align*} & r\geq b \\ \implies & r-b \geq 0\\ \implies & a-bq-b \geq 0\\ \implies & a-b\big(q+1\big) \geq 0\\ \implies & a-b\big(q+1\big) \in S\\ \end{align*}   \(\therefore\) \(r-b \leq r\) and \(r-b \in S\).
  A contradiction since \(r\) is the least element of \(S\).
  Therefore our assumption is wrong.
  \(\therefore r\ngeq b \implies r\lt b\).
  Hence \begin{align*} a=bq+r \text{ where } 0\leq r \lt b \end{align*}   If possible there exists another two integers \(q_{1},r_{1}\) sucht that \begin{align*} a=bq_{1}+r_{1} \text{ where } 0\leq r_{1} \lt b \end{align*}   Now \begin{align*} & 0\leq r \lt b \text{ and } 0\leq r_{1} \lt b \\ \implies & 0\leq \big|r-r_{1}\big| \lt b \end{align*}   Also \begin{align*} & a=bq+r \text{ and } a=bq_{1}+r_{1} \\ \implies & bq_{1}+r_{1}=bq+r \\ \implies & bq_{1}-bq=r-r_{1} \\ \implies & b\big(q_{1}-q\big)=r-r_{1} \\ \implies & b\big|q_{1}-q\big|=\big|r-r_{1}\big| ~\because b\gt 0\\ \implies & b\big|q_{1}-q\big|\lt b\\ \implies & \big|q_{1}-q\big|\lt 1~\because b\gt 0\\ \implies & 0\leq \big|q_{1}-q\big|\lt 1\\ \implies & \big|q_{1}-q\big|=0~\because q,~q_{1}\text{ are integers}\\ \implies & q_{1}=q \\ \implies & r_{1}=r \end{align*}   Hence \(\exists\) unique integers \(q\) and \(r\) such that \begin{align*} & a=bq+r \text{ where } 0\leq r \lt b \end{align*}