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 \(a=qb+r\)

@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 :)

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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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 \(a=qb+r\)

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Use \leq for \( \leq \). Nice proof.

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Thanks! Had been trying to find that LaTeX.

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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...

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