Problem 2 Recall the toBinary algorithm from Tutorial 7 problem 3, which finds the binary representation of a given integer n > 0. For example, toBinary (13) = (1, 1, 0, 1), and 1-2³ +1.2² +0-2¹ +1-2⁰ = 8 +4+0+1 = 13. toBinary (n) : 1: if n ≤ 1 then return (n) 2: 3: 4: 5: 6: else (nd,...,no) I = parity (n) return (nd,..., no, x) toBinary ([n/2]) In Tutorial 7 we showed by induction that for any n ≥ 0, if toBinary (n) d>0, then we have on,2² = n. = (nd,...,no) for some Note that toBinary takes in any integer n ≥ 0 and spits out a binary number, which is an element of {0,1}* (that is, an element of Uzo{0,1}). In other words, toBinary is a function from the domain of Z2º to the codomain of {0, 1}*. Use the result from Tutorial 7 to prove that toBinary : Z20 ➜ {0,1}* is one-to-one via a proof by contrapositive.

(a) Write (152) (341) (6352) ≤ S6 in array form. Fill in the blank row below. 1 2 3 4 5 6 (b) Express [ 1 2 3 4 5 6 7 8 9 1 4 2 97 8 5 3 6 as a product of disjoint cycles. (c) How many elements of order 10 does S7 have?