dis6-sol.pdf - CS 70 Fall 2019 1 Discrete Mathematics and Probability Theory Alistair Sinclair and Yun S Song DIS 6 Count It For each of the following

dis6-sol.pdf - CS 70 Fall 2019 1 Discrete Mathematics and...

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CS 70 Discrete Mathematics and Probability Theory Fall 2019 Alistair Sinclair and Yun S. Song DIS 6 1 Count It! For each of the following collections, determine and briefly explain whether it is finite, countably infinite (like the natural numbers), or uncountably infinite (like the reals): (a) Z , the set of all integers. (b) Q , the set of all rational numbers. (c) The integers which divide 8. (d) The integers which 8 divides. (e) The functions from N to N . (f) Numbers that are the roots of nonzero polynomials with integer coefficients. Solution: (a) Countable and infinite. See Lecture Note 10. (b) Countable and infinite. See Lecture Note 10. (c) Finite. They are {- 8 , - 4 , - 2 , - 1 , 1 , 2 , 4 , 8 } . (d) Countably infinite. We know that there exists a bijective function f : N Z . Then the function g ( n ) = 8 f ( n ) is a bijective mapping from N to integers which 8 divides. (e) Uncountably infinite. We use Cantor’s Diagonalization Proof: Let F be the set of all functions from N to N . We can represent a function f F as an infinite sequence ( f ( 0 ) , f ( 1 ) , ··· ) , where the i -th element is f ( i ) . Suppose towards a contradiction that there is a bijection from N to F : 0 ←→ ( f 0 ( 0 ) , f 0 ( 1 ) , f 0 ( 2 ) , f 0 ( 3 ) ,... ) 1 ←→ ( f 1 ( 0 ) , f 1 ( 1 ) , f 1 ( 2 ) , f 1 ( 3 ) ,... ) 2 ←→ ( f 2 ( 0 ) , f 2 ( 1 ) , f 2 ( 2 ) , f 2 ( 3 ) ,... ) 3 ←→ ( f 3 ( 0 ) , f 3 ( 1 ) , f 3 ( 2 ) , f 3 ( 3 ) ,... ) . . . CS 70, Fall 2019, DIS 6 1
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Consider the function g : N N where g ( i ) = f i ( i )+ 1 for all i N . We claim that the function g is not in our finite list of functions. Suppose for contradiction that it were, and that it was the n -th function f n ( · ) in the list, i.e., g ( · ) = f n ( · ) . However, f n ( · ) and g ( · ) differ in the n -th argument, i.e. f n ( n ) 6 = g ( n ) , because by our construction g ( n ) = f n ( n )+ 1. Contradiction! (f) Countably infinite. Polynomials with integer coefficients themselves are countably infinite. So let us list all polynomials with integer coefficients as P 1 , P 2 ,... . We can label each root by a pair ( i , j ) corresponding to the j -th root of polynomial P i (we can have an arbitrary ordering on the roots of each polynomial). This means that the roots of these polynomials can be mapped in an injective manner to N × N which we know is countably infinite. So this set is either finite or countably infinite. But every natural number
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