# The Critical Thinking Assignment For Module One

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CT #1 Option #2: Sets #2 The Critical Thinking assignment for Module One, Option Two, is a series of five questions pertaining to Set Theory and Venn diagrams. In short, Set Theory is the notion in which items (e.g., numbers, colors, people), referred to in mathematics as “elements” can be grouped into “sets” in order to organize and understand elements (Jech, 1971). In relation, Venn diagrams are circular illustrations utilized to visually express relationships between sets (Grünbaum, 1975). It is imperative to understand the relation between sets and Venn diagrams in order to fully comprehend Set Theory. Problem Number One Accompanied by a detailed Venn diagram, question number one prompts “Which region represents (B ∩ C)∩A’?”.…show more content…
Let U = {a, b, c, d, e, f, g}, A= {a, b, c}, B= {c, d, e, f} Find A’-B’. There are three necessary steps to solve. First, one must find the complement of A, A’ = {d, e, f, g}. Next, one must solve for the complement of B, B’ = {a, b, g}. Now, the problem solver has the critical information to solve for the difference (A’ = {d, e, f, g}) - (B’ = {a, b, g}). Finally, one can deduce A’-B’ = {d, e, f}. Problem Number Four Similarly to problem number one, problem number four returns to the use of a Venn diagram to solve the problem. However, this time there is an additional complement within the brackets. Let U = {a, b, c, d, e, f, g}, A= {a, b, c, f}, B= {c, d}, C= {a, d, e, f, g}, find (‘B∩C)∩A’. Step one, fill out a three circle Venn diagram including all sets, U, A, B, C. Next, find the complement of B, B’= {a, b, e, f, g}. Then, find the intersection of B’ and C, (‘B∩C) = {a, d, e, f, g}. Next, one must solve for A’, A’ = {e,g} Lastly, solve for pug in set values and solve, {a, d, e, f, g} ∩ {e, g} = {e. g}. Therefore, one can now resolve, (‘B∩C)∩A’ = {e, g}. Problem Number Five Lastly, problem number five again addresses complements. The equation is as follows, U= {2, 4, 6, 8, 10, 12, 14, 16, 18,...}, A= {6, 8, 10} Find A’. Remembering the complement is the set of elements of U that are not elements of A, one must solve for A’. Therefore, one