Unions,

Intersections, Compliments: Direct and

Contradiction in Proofs Edwin

LopezAmerican

Public University #1) Prove the following

by using: a direct argument; the definition for union; the definition for

intersection; and the definition for set complement. Proof: If x?A?B then,

x?UB. And by definition of the unition

x?A or x?B. But x?UB means that x?U and

x?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 then

x?UB and x?A?B, so x?UB. Hence

UB?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 be

disjoint. So, because A?B=?, then there

is nothing that connects A and B except the empty set. Since, U/B=A?B^c, that is x?A or x?B, then

A=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 following

by using: a direct argument; and definition 2.1.16. Consider practice problem

2.1.14 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 and

x?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: a

direct argument; definition 2.2.9; definition of a rational number; sum and

product of integers are integers.By defintion the only way for there to be

an equivalence relation iff P is symmetric, reflexive, and transitive. Therefore, we must prove all three conditions

for this to me true. First argument: Suppose x, y ?R and xPy then x-y is an integer.

Since, y-x=-(x-y), this would

make y-x symmetric. Thus yPx.

y-x is also an integer, Which

clearly 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 is

true. 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. Since

this is the case it would make this condition true. Thus, R is an equivelance relation on R. #4) Prove the following

by using: a contradiction argument; definition 2.3.1; definition and notation

2.3.13; and Theorem 2.3.16. Suppose f:A?B, S?A, and

T?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?T

s. 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 a

contradiction. ?Work Cited 1 Lay, S. R. (2012). Analysis with an Introduction to Proof, 5th Edition, Bookshelf

Online, Retrieved from

https://bookshelf.vitalsource.com/#/books/9781269603232/ (“When Do You Use “Suppose” and When

“Let?”” 2012), Retrieved January 27, 2018 from

https://english.stackexchange.com/questions/68498/when-do-we-use-suppose-and-when-let