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

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