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# Division algorithm -Doubled ,tripled

I can't prove it.Can you?

Note by Shivamani Patil
3 years, 3 months ago

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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 $$a=qb+r$$

- 3 years, 3 months ago

Use \leq for $$\leq$$. Nice proof.

- 3 years, 3 months ago

Thanks! Had been trying to find that LaTeX.

- 3 years, 3 months ago

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

Staff - 3 years, 3 months ago

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...

- 3 years, 3 months ago