Adv Alegbra HW Solutions 40

Adv Alegbra HW Solutions 40 - 40 2.5 Let A and B be sets,...

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Unformatted text preview: 40 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.

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