Dividend vs Remainder

Number Theory Level 3

For positive integers a a , b b , q q and r r , where a > b a>b , a = b q + r , 0 ≤ r < b a=bq+r, 0\leq r <b .

What is the exact relation between a a and r r ?

r = 1 2 a r=\dfrac{1}{2}a r ≤ 1 2 a r\leq\dfrac{1}{2}a r > 1 2 a r>\dfrac{1}{2}a r < 1 2 a r<\dfrac{1}{2}a r ≥ 1 2 a r\geq\dfrac{1}{2}a

Muhammad Rasel Parvej by Muhammad Rasel Parvej , 27, Bangladesh

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

The correct answer is r < 1 2 a \boxed{r<\dfrac{1}{2}a} .

Let me demonstrate a proof by contradiction. Let's assume, r ≥ 1 2 a r\geq \dfrac{1}{2}a . Then, b > 1 2 a b>\dfrac{1}{2}a , as b > r b>r . So, b + r > 1 2 a + 1 2 a = a b+r>\dfrac{1}{2}a+\dfrac{1}{2}a=a .

But, a a

= b q + r =bq+r

≥ b + r \geq b+r (As q ≥ 1 q\geq1 )

> a >a

That means, a > a a>a .

That means, r ≥ 1 2 a ⟹ a > a r\geq \dfrac{1}{2}a \implies a>a , which is a contradiction.

So, r < 1 2 a \boxed{r<\dfrac{1}{2}a}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...