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Suppose A and B be two non-empty sets then every subset of A Χ B describes a relation from A to B and each relation from A to B is subset of AΧB.
Consider R : AΧB and (a, b) € R. then we can declare that a is associated to b by the relation R and write it as
a R b. If (a, b) ¢R, we write it as a b.
Example Let A {1, 2, 3, 4, 5}, B = {1, 3}
We set a relation from A to B as: a R b iff a ≤ b; a € A, b € B. Then
R ={(1, 1), (1, 3), (2, 3), (3, 3)}AΧB
i need solution manual of "calculus and analytic geometry thomas 6th edition book "
1/a+b+x =1/a+1/b+1/x a+b ≠ 0 Ans: 1/a+b+x =1/a+1/b+1/x => 1/a+b+x -1/x = +1/a +1/b ⇒ x - ( a + b + x )/ x ( a + b + x ) = + a + b/ ab ⇒
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