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$$ Let \ b \ be \ a \ natural \ number \ and \ q_1, \ q_2, \ r_1, \ r_2 \ integers \ with \ 0 \le r_1 \lt b \ and \ 0 \le r_2 \lt b \ such \ that \ q_1b + r_1 = q_2b + r_2 \ then \ q_1 = q_2 \ and \ r_1 = r_2 $$

I'm assuming the division algorithm will come into play here, but where do I start to prove something like this? I don't think I even fully understand how to prove the division algorithm yet. A walkthrough would be greatly appreciated..

dendritic
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1 Answers1

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The aim of this result is to prove the $q$ and $r$ in the euclidean division are unique.

Suppose $r_1\ge r_2$. Rewrite the equality as $\;r_1-r_2=(q_2-q_1)b$. Then $r_1-r_2$ is a non-negative multiple of $b$. However, since $\;0\le r_2\le r_1<1$, we have $\;0\le r_1-r_2<b$. There's only one such multiple of $b$ smaller than $b$: $0$. Thus $r_1=r_2$, and $(q_2-q_1)b=0$, whence $q_2-q_1=0$.

Bernard
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