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Unformatted text preview: Modern Analysis, Homework 2 Possible solutions 1. (2 pts) Suppose f : A → B is a function. Prove that f is a bijection if and only if it is both injective (onetoone) and surjective (onto). Solution: Suppose first that f is a bijection, ie there exists a function g : B → A such that f ◦ g = id B and g ◦ f = id A (here id S denotes the identity function on a set S , ie id S ( s ) = s for all s ∈ S ). We can first argue that f is injective: let x, y ∈ A . Suppose that f ( x ) = f ( y ). Then g ( f ( x )) = id A ( x ) = x and g ( f ( y )) = id A ( y ) = y so x = y and f is injective. Now we argue surjectivity: let z ∈ B . Consider g ( z ). This is an element in A , and f ( g ( z )) = id B ( z ) = z , so f is surjective. Now suppose that f is both injective and surjective. Let z ∈ B . Then by the surjectivity of f , there exists x z ∈ A such that f ( x ) = z . By the injectivity of f , this is the unique such x z , and it therefore makes sense to define g : B → A by g ( z ) = x z . Then clearly f ( g ( z )) = z so f ◦ g = id B . Similarly, let x ∈ A . Then, in the notation from above, x = x f ( x ) (ie x is the unique element in A that gets mapped to f ( x ) under f ). So by definition of g , g ( f ( x )) = x and therefore g ◦ f = id A . 2. (2 pts) Prove by induction that for all positive integers n , n summationdisplay i =1 i = n ( n + 1) 2 . Solution: Let P ( n ) be the predicate (logical statement, logical proposition, whatever): “the formula ∑ n i =1 i = n ( n +1) 2 is true”. Clearly P (1) is true since both sides of the equa tion compute the value 1. Now suppose that P (1) , P (2) , ··· , P ( k ) hold for some k (this is called the inductive hypothesis ). In particular, we have ∑ k i =1 i = k ( k +1) 2 . Add k + 1 to both sides of this equation. On the left we get ∑ k +1 i =1 i and on the right hand side we get k ( k +1) 2 +( k +1) = ( k +1) ( k 2 + 1 ) = ( k +1)( k +2) 2 , and therefore P ( k +1) holds! By the principle of mathematical induction, it follows that P ( n ) holds for any n ∈ N . 3. (5 pts) Write down an explicit bijection N × N → N . Hint: count by “diagonals” and use problem 2. Solution: Picture N × N as the points in the first quadrant with integral coordinates. Given ( m, n ) ∈ N × N , there is some maximal number of completely filled “diagonals” to the lower left of ( m, n ) (see Figure 1). In fact, there are m + n − 2 such diagonals. The total number of dots on these diagonals is a “triangular” number, and is equal to ∑ m + n 2 i =0 i = ( m + n 1)( m + n 2) 2 by problem 2 . Now count the number of dots along the diagonal from the line y = 1 to the dot ( m, n ). This number is m . Therefore the number f ( m, n ) = ( m + n 1)( m + n 2) 2 + m com pletely encodes the pair ( m, n ) (ie, f is injective). Further, it is “clear” that f is surjective: by counting dots along the diagonals, we can get every number in N ....
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This note was uploaded on 10/18/2011 for the course MATH S4061X taught by Professor Peters during the Spring '11 term at Columbia.
 Spring '11
 Peters

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