Not the division algorithm (but fairly similar)

**Sigyn3** · August 2nd 2008, 11:13 AM

Prove that for any positive integer $n$ , there exist unique positive integers $k$ and $r$ such that $n = \frac{k(k-1)}{2}+r$ with $1\le r\le k$ .

**ThePerfectHacker** · August 2nd 2008, 05:39 PM

Originally Posted by Sigyn3

Prove that for any positive integer $n$ , there exist unique positive integers $k$ and $r$ such that $n = \frac{k(k-1)}{2}+r$ with $1\le r\le k$ .

Consider the set $\{ n - \tfrac{x(x-1)}{2}\geq 0 | x\in \mathbb{Z}^+ \}$ . There exists a smallest integer $n - \tfrac{k(k-1)}{2} = r$ . We claim that $r\leq k$ . Assume not, then $r>k$ and we can write $r=r_0+k$ where $r_0<r$ . Thus, we have $n - \tfrac{k(k-1)}{2} = r_0+k$ which gives $n-\tfrac{(k+1)k}{2} = r_0$ . And this is a contradiction since $r$ was supposed to be minimal.

**Sigyn3** · August 2nd 2008, 06:10 PM

Does this also prove uniqueness?

Thread: Not the division algorithm (but fairly similar)

LinkBack

Thread Tools

Display

Not the division algorithm (but fairly similar)

Similar Math Help Forum Discussions

The Division Algorithm 2

The Division Algorithm

Division algorithm/mod

The Division Algorithm

Division Algorithm

Search Tags