A brief note on Sets, Relations and Functions . Read it before a test. Try each question at the end before you open its answer.
Sets A set of n n n elements has 2 n 2^n 2 n subsets, and 2 n − 1 2^n - 1 2 n − 1 proper subsets. n ( A ∪ B ) = n ( A ) + n ( B ) − n ( A ∩ B ) n(A \cup B) = n(A) + n(B) - n(A \cap B) n ( A ∪ B ) = n ( A ) + n ( B ) − n ( A ∩ B ) . De Morgan: ( A ∪ B ) ′ = A ′ ∩ B ′ (A \cup B)' = A' \cap B' ( A ∪ B ) ′ = A ′ ∩ B ′ and ( A ∩ B ) ′ = A ′ ∪ B ′ (A \cap B)' = A' \cup B' ( A ∩ B ) ′ = A ′ ∪ B ′ . Products and relations ( a , b ) = ( c , d ) (a, b) = (c, d) ( a , b ) = ( c , d ) exactly when a = c a = c a = c and b = d b = d b = d . n ( A × B ) = n ( A ) n ( B ) n(A \times B) = n(A)\,n(B) n ( A × B ) = n ( A ) n ( B ) , and there are 2 n ( A ) n ( B ) 2^{n(A)\,n(B)} 2 n ( A ) n ( B ) relations from A A A to B B B . An equivalence relation is reflexive, symmetric and transitive. Functions Question How to answer it Is it a function? each input has exactly one output Domain remove zero denominators and negatives under even roots Range solve y = f ( x ) y = f(x) y = f ( x ) for x x x , or complete the square Equal functions? same domain and same values
A worked example Find the domain and the range of f ( x ) = 1 x 2 − 9 f(x) = \dfrac{1}{\sqrt{x^2 - 9}} f ( x ) = x 2 − 9 1 .
The root. It must be positive, not zero, since it sits in the bottom: x 2 > 9 x^2 > 9 x 2 > 9 , so ∣ x ∣ > 3 |x| > 3 ∣ x ∣ > 3 . The domain. ( − ∞ , − 3 ) ∪ ( 3 , ∞ ) (-\infty, -3) \cup (3, \infty) ( − ∞ , − 3 ) ∪ ( 3 , ∞ ) . The range. x 2 − 9 \sqrt{x^2 - 9} x 2 − 9 takes every positive value, so its reciprocal does too: ( 0 , ∞ ) (0, \infty) ( 0 , ∞ ) . Practice with new numbers
A program to try The program tests a few values of x x x for the worked example.
Read it, then run it. Which values are not allowed?
1. Change x * x - 9 to x * x - 4. Where is the gap now?
Mistakes to avoid Counting the overlap twice in n ( A ∪ B ) n(A \cup B) n ( A ∪ B ) . Treating ( 2 , 5 ) (2, 5) ( 2 , 5 ) and ( 5 , 2 ) (5, 2) ( 5 , 2 ) as the same pair. Allowing 0 0 0 under a root in the bottom. 1 0 \dfrac{1}{\sqrt{0}} 0 1 is not defined. Calling a relation a function when some input has two outputs. Check yourself n ( A ) = 30 n(A) = 30 n ( A ) = 30 , n ( B ) = 25 n(B) = 25 n ( B ) = 25 and n ( A ∩ B ) = 10 n(A \cap B) = 10 n ( A ∩ B ) = 10 . Find n ( A ∪ B ) n(A \cup B) n ( A ∪ B ) . Stuck? Show a hint Show the answerIs { ( 1 , 1 ) , ( 2 , 1 ) , ( 3 , 1 ) } \{(1, 1), (2, 1), (3, 1)\} {( 1 , 1 ) , ( 2 , 1 ) , ( 3 , 1 )} a function on { 1 , 2 , 3 } \{1, 2, 3\} { 1 , 2 , 3 } ? Stuck? Show a hint Show the answerWhat is the domain of 1 x − 5 \dfrac{1}{x - 5} x − 5 1 ? Stuck? Show a hint Show the answerHow many subsets does a set of 8 8 8 elements have? Stuck? Show a hint Show the answer