Unions,Intersections, Compliments: Direct andContradiction in Proofs EdwinLopezAmericanPublic University #1) Prove the followingby using: a direct argument; the definition for union; the definition forintersection; and the definition for set complement. Proof: If x?A?B then,x?UB. And by definition of the unitionx?A or x?B. But x?UB means that x?U andx?B by definition of complements. Since, x?U and x?B, we have x?UB as required.

Thus A?B?UB. Conversely, we must show that UB?A?B. If x?UB, then x?U and x?B.

Since x?A?B, we have x?A or x?B, so x?A?B. But then x?UB. Thus x?U and x?B, so x?UB. But thenx?UB and x?A?B, so x?UB. HenceUB?A?B. If U, then A?B. Since A?B, then x?A or x?B. Therefore U is also x?A or x?B.

If A?B, then x?A and x?B. Therefore, {?} is also x?A and x?B. If A?B=?, then A and B are said to bedisjoint. So, because A?B=?, then thereis nothing that connects A and B except the empty set. Since, U/B=A?B^c, that is x?A or x?B, thenA=A?A, then A=UB That is A is x?A and Thus, A?B^cB=A?UB; A?UB=A?A=A therefore, A=UB.?#2) Prove the followingby using: a direct argument; and definition 2.1.16.

Consider practice problem2.1.14 as an outline for your proof. Proof: We begin by showing that B?(?_(j?J)??A_j)??_(j?J)??(B?A_j ).? ? If x?B?(?_(j?J)??A_j)?, then x????:??J s.t. x?A_j ?.

If x?A_j, then certainly x?B?(???A_j)? and x??_(j?J)?( B?A_j). Thus x??_(j?J)??(B?A_j).? On the other hand, if x??B?A?_(j ) then x?B andx?A_j. But this implies that x?(?_(j?J)??B)? and x?(?_(j?J)??A_j) ?, so x?(?_(j?J)??B)?(?_(j?J)??A_j)??.

Hence, B?(?_(j?J)?A_j )??_(j?J)?(?B?A?_j ) . Conversely, if y?(?_(j?J)??B)?(?_(j?J)??A_j)??, then y?(?_(i?J)??B)? and y?(?_(i?J)??A_j)?. There are two cases to consider: when y?B and y?B. If y?B, then B?(?_(j?J)?A_j ) and this part is done. On the other hand, if y?B, then since B?(?_(j?J)??A_j)?, we must have y??_(j?J)?A_j this implies that y?B?(?_(j?J)??A_j).

? Hence, ?_(j?J)?(B?A) ?B?(?_(j?J)?A_j ). #3) Prove the following by using: adirect argument; definition 2.2.9; definition of a rational number; sum andproduct of integers are integers.By defintion the only way for there to bean equivalence relation iff P is symmetric, reflexive, and transitive. Therefore, we must prove all three conditionsfor this to me true.

First argument: Suppose x, y ?R and xPy then x-y is an integer. Since, y-x=-(x-y), this wouldmake y-x symmetric. Thus yPx. y-x is also an integer, Whichclearly proves the condition is true. Second argument: Suppose x?R,then x-x=0, Thus, x is symmetric, xPx.

Since zero is an integer this condition istrue. Third Argument: Suppose x,y?R, xPy and yPz. Then x-y and y-z are integers. Thus, the sum of (x-y)+(y-z)=x-z.

Thus, xPz. x-z is also an integer. Sincethis is the case it would make this condition true.

Thus, R is an equivelance relation on R. #4) Prove the followingby using: a contradiction argument; definition 2.3.1; definition and notation2.3.13; and Theorem 2.3.16.

Suppose f:A?B, S?A, andT?A.S?T then f(S)?f(T).Proof: f:A?B, S and T are subsets of A and S?T.

Let y?f(S). Then there is x?S s. t. y=f(x). Since S?T, x?T. So there is x?Ts. t.

y=f(x), and therefore y?f(T) by definition of image. Since, we have f(S)?f(T) . Therefore, this makes the statement acontradiction. ?Work Cited 1 Lay, S. R.

(2012). Analysis with an Introduction to Proof, 5th Edition, BookshelfOnline, Retrieved fromhttps://bookshelf.vitalsource.com/#/books/9781269603232/ (“When Do You Use “Suppose” and When”Let?”” 2012), Retrieved January 27, 2018 fromhttps://english.stackexchange.com/questions/68498/when-do-we-use-suppose-and-when-let