Division Algorithm states that \(a=q_{1}b+r_{1}\) where \(\leq r_{1} < b\) so we need to have \(r=r_{1} + 2b\) to get the wanted inequality. Now substituting \(r_{1}\) in terms of \(r\); \[a=q_{1}b+r-2b\] with which we can get the value of \(q\); \[a=(q_{1}-2)b+r\] Hence, \(q=q_{1}-2\), satisfying the equation \(a=qb+r\)
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Marc Vince Casimiro
·
2 years, 9 months ago

@Marc Vince Casimiro
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With the new display Latex tools, you just need to find some problem / note / solution with the corresponding latex that you need :)
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Calvin Lin
Staff
·
2 years, 9 months ago

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qb+2b<=qb+r<qb+3b ,
b(q+2)<=qb+r<b(q+3) ,
b(q+2)<=a<b(q+3),divide by b>0,
q+2<=a/b<q+3,
This means that rational number a/b is between two integers...
–
Nikola Djuric
·
2 years, 9 months ago

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TopNewestDivision Algorithm states that \(a=q_{1}b+r_{1}\) where \(\leq r_{1} < b\) so we need to have \(r=r_{1} + 2b\) to get the wanted inequality. Now substituting \(r_{1}\) in terms of \(r\); \[a=q_{1}b+r-2b\] with which we can get the value of \(q\); \[a=(q_{1}-2)b+r\] Hence, \(q=q_{1}-2\), satisfying the equation \(a=qb+r\) – Marc Vince Casimiro · 2 years, 9 months ago

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– Siddhartha Srivastava · 2 years, 9 months ago

Use \leq for \( \leq \). Nice proof.Log in to reply

– Marc Vince Casimiro · 2 years, 9 months ago

Thanks! Had been trying to find that LaTeX.Log in to reply

– Calvin Lin Staff · 2 years, 9 months ago

With the new display Latex tools, you just need to find some problem / note / solution with the corresponding latex that you need :)Log in to reply

qb+2b<=qb+r<qb+3b , b(q+2)<=qb+r<b(q+3) , b(q+2)<=a<b(q+3),divide by b>0, q+2<=a/b<q+3, This means that rational number a/b is between two integers... – Nikola Djuric · 2 years, 9 months ago

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