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2.5 Let A and B be sets, and let a ∈ A and b ∈ B . Deﬁne their ordered pair
as follows:
(a , b) = {a , {a , b}}.
If a ∈ A and b ∈ B , prove that (a , b ) = (a , b) if and only if a = a and
b = b.
Solution. The result is obviously true if a = a and b = b.
For the converse, assume that
{a {a , b}} = {a {a , b }}
There are two cases:
a = a and {a , b} = {a , b };
a = {a , b } and {a , b} = a .
If a = a , we have {a , b} = {a , b } = {a , b }. Therefore,
{a , b} − {a } = {a , b } − {a }.
If a = b, the left side is empty, hence the right side is also empty, and so
a = b ; therefore, b = b . If a = b, the the left side is {b}, and so the right
side is nonempty and is equal to {b }. Therefore, b = b , as desired.
In the second case, a = {a , b } = {{a , b}b }. Hence,
a ∈ {a , b}
and
{a , b} ∈ {{a , b}, b } = a ,
contradicting the axiom a ∈ x ∈ a being false. Therefore, this case cannot
occur.
2.6 Let
= {(x , x ) : x ∈ R}; thus, is the line in the plane which passes
through the origin and which makes an angle of 45◦ with the x axis.
(i) If P = (a , b) is a point in the plane with a = b, prove that is
the perpendicular bisector of the segment P P having endpoints
P = (a , b) and P = (b, a ).
Solution. The slope of
is 1, and the slope of P P is
(b − a )/(a − b) = −1. Hence, the product of the slopes is −1,
and so
is perpendicular to the P P . The midpoint of P P is
M = ( 1 (a + b), 1 (a + b)), which lies on , and
2
2
P M = [a − 1 (a + b)]2 + [b − 1 (a + b)]2 =  M P .
2
2 ...
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.
 Fall '11
 KeithCornell

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