# Example prove that n k 1 k n n 1 2 solution we pose

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Example. Prove that n k =1 k = n ( n +1) 2 . Solution. We pose the following counting question. How many ways are there to choose two numbers from S = { 0 , 1 , 2 , . . . , n } ? By definition, there are ( n +1 2 ) = n ( n +1) 2 distinct pairs of S . This gives us the RHS. We can also compute the answer as follows. Let P be the set of all pairs of S . P can be partitioned into S 1 , S 2 , . . . , S n , where S i , 1 i n , is the set of pairs in which i is the bigger element in the pair. Clearly, | P | = | S 1 | + | S 2 | + . . . + | S n | = 1 + 2 + . . . + n = n summationdisplay k =1 k = LHS This proves the claim. Functions Definition. Let A and B be sets. A function f is a mapping from A to B such that each element in A is mapped to exactly one element in B . We write f ( x ) = y where y B is the unique element that a A is mapped to. We denote such a function by f : A B . Definition. Given a function f : A B , A is called the domain of f and B is called the co-domain of f . If f ( x ) = y then y is called the image of x and x is called the preimage of

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January 10, 2011 Lecture Outline 3 y . The range of f is the set of all images of elements in A . Definition. A function f is a one-to-one function if f ( x ) = f ( y ) implies x = y . For exam- ple, the function f ( x ) = x 2 from the set of integers to the set of integers is not one-to-one because f ( x 1 ) = f ( x 2 ) when x 1 = 2 and x 2 = 2, however x 1 negationslash = x 2 . On the other hand the function f : R R , defined by the rule f ( x ) = 4 x 1, for all real numbers x , is one-to-one. Definition. A function f is a onto function if its range equals its co-domain. For example, the function f ( x ) = x + 1 from the set of integers to the set of integers is an onto function. Definition. A function f is a one-to-one correspondence , if it is both one-to-one and onto. Definition. Let f be a one-to-one correspondence from the set A to the set B . The inverse function of f is the function that assigns to an element b belonging to B the unique element a in A such that f ( a ) = b . The inverse function of f is denoted by f 1 . Hence, f 1 ( b ) = a when f ( a ) = b . The Pigeonhole Principle If k +1 or more objects are distributed among k bins then there is at least one bin that has two or more objects. For example, the pigeon hole principle can be used to conclude that in any group of thirteen people there are at least two who are born in the same month. Example. There are n pairs of socks. How many socks must you pick without looking to ensure that you have at least one matches pair? Solution. The pigeonhole principle can be applied by letting n bins correspond to the n pairs of socks. If we select n + 1 socks and put each one in the box corresponding to the pair it belongs to then there must be at least one box containing a matched pair. The Generalized Pigeonhole Principle If n objects are placed into k boxes, then there is at least one box containing at least n/k objects. Proof: Assume otherwise, i.e., each box contains at most n/k ⌉ − 1 objects. Then, the total number of objects is at most k parenleftBigceilingleftBig n k ceilingrightBig 1 parenrightBig < k parenleftBig n k + 1 1 parenrightBig = n This is a contradiction as there are n objects.
4 Lecture Outline January 10, 2011

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