sol2 - MATH 235 – Algebra I Solutions to assignment 2...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 235 – Algebra I Solutions to assignment 2 September 23, 2007 Solution of 1. Let f : N × N → N be the function described in the question; we’re asked to come up with a formula for f . For k ∈ N , let D k denote the k th (anti)diagonal in N × N , that is, D k = { (0 ,k ) , (1 ,k- 1) , (2 ,k- 2) ,..., ( k, 0) } . Remark that we can write any element of D k as ( i,k- i ) with 0 ≤ i ≤ k . Also note that f increases by 1 whenever you increase i by one within a diagonal, or move from the last element of one diagonal to the first element of the next. First let’s find a formula for f (0 ,k ), k ∈ N . As (0 ,k ) is the initial element in the k th diagonal D k , its value by f is the total number of elements in the previous diagonals: f (0 ,k ) = | D | + | D 1 | + ··· + | D k- 1 | = 1 + 2 + ··· + k = k ( k + 1) 2 . Note that for the i th element ( i,k- i ) of D k , we should have f ( i,k- i ) = f (0 ,k ) + i = k ( k + 1) 2 + i. Now consider an element ( m,n ) ∈ N × N . To find out which element of which diagonal it is, we solve the equation ( m,n ) = ( i,k- i ) for k and i and find out that i = m and k = m + n . Plugging this in the last formula, we get f ( m,n ) = ( m + n )( m + n...
View Full Document

This note was uploaded on 04/18/2008 for the course MATH 235 taught by Professor Goren during the Fall '07 term at McGill.

Page1 / 3

sol2 - MATH 235 – Algebra I Solutions to assignment 2...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online