Let O be the set of all odd integers, and let 2Z be the set of all even integers. Prove that O has the same cardinality as 2z. Proof: In order to show that O has the same cardinality as 22 we must show that there is a well-defined function f: 0 -→ 2Z that is both one-to-one and onto. We will show that the following is a function from 0 to 2Z that satisfies these requirements. (Choose one definition for f and use it for the rest of the proof.) O f(n) = ", for each odd integer n in o O f(n) = Inl, for each odd integer n in o O f(n) = n - 1, for each odd integer n in o O f(n) = n+ 2, for each odd integer n in o O f(n) - 3n, for each odd integer n in o Well-Defined One-to-One Onto Proof that f is onto: To show that f is onto, let m be any even integer in 2z. By definition of even, there exists an integer k such that m = 2k. On a separate piece of scratch paper, find an odd integer in 0, written as an expression using the variable k, with the property that when fis applied to it, the result is 2k. Write the expression in the box below. By construction, the quantity in the box is an odd integer, and when the function fis applied to it, the result is the even integer 2k, which equals m. Thus, we have shown that there exists an element of O that is sent to m by f, and so f is onto. Conclusion: Since f is a well-defined function from O to 2Z that is one-to-one and onto, we conclude that O and 22 have the same cardinality.

true or false? PROVE YOUR ANSWER b) If f :N→N and g :N→N are functions such that f is not onto but  fg is onto, then g is onto. The claim is of answer:                           prove of answer

3-part question: Suppose f: X --> Y and g: Y --> Z are functions, and g of f: X --> Z is injective.  (a) Show that f is injective. (b) Provide an example where f is not injective.  (c) Impose a condition on f so that it, together with the assumption that g of f is injective, implies that g is injective.

Direct Proof for F → A, ~J • ~K, H → (G → F), ~K → (~J → H), therefore G → A.