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

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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 0 | + | 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 + 1) 2 + m = m 2 2 + mn + 3 m 2 + n 2 2 + n 2 , which is a polynomial expression in m and n .
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