Unions, B are said to be disjoint. So, because

Intersections, Compliments:  Direct and
Contradiction in Proofs Edwin
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