. Assign a grade of A (correct), C (partially correct), or F (failure) to each.Justify assignments of grades other than A. (a) Claim. If A is denumerable, then is denumerable.Proof. Assume A is denumerable.Case 1. If then which is denumerable byhypothesis.Case 2. Assume Since A is denumerable, there existsDefine g by setting Thenso Therefore, isdenumerable. (b) Claim. If A and B are denumerable, then is denumerable.Proof. Assume A and B are denumerable, but that is notdenumerable. Then is finite. Since A and B are denumerable, theyare not empty, so we can choose and Then,and Since is finite, the subsets andare finite. Therefore, A and B are finite. This contradicts thestatement that A and B are denumerable. We conclude that isdenumerable.(c) Claim. The set of positive rationals is denumerable.Proof. Consider the positive rationals in the array in Figure 5.3.1.Order this set by listing all the rationals in the first row, then the secondrow, and so forth. Omitting fractions that are not in lowest terms, wehave an ordering of in which every positive rational appears. Therefore,is denumerable.(d) Claim. If A and B are infinite, thenProof. Suppose A and B are infinite sets. Letand Define as shown:Then, since we never run out of elements in either set, f is one-to-oneand onto B, so(e) Claim. is uncountable.Proof. is uncountable and is a subset of R. Every subset ofan uncountable set is uncountable, so is uncountable.5

Wednesday, August 23, 2017 1:37 PM Definition 1: A ▯ ▯ ▯ linear system has the form: ▯▯▯+ ▯ ▯▯+ ▯ ▯▯+ ⋯+ ▯ ▯▯ ▯= ▯ ▯ ▯▯▯+ ▯ ▯▯+ ▯ ▯▯+ ⋯+ ▯ ▯▯ ▯= ▯ ▯ . . . ▯▯▯ ▯+ ▯ ▯▯ ▯ + ▯ ▯▯ ▯ + ⋯+ ▯ ▯▯ ▯ = ▯ ▯ m is the number of equations in the system, and n is the number of variables in the system....