Download 6 Solved Questions on First Year Interest Group Seminar - Exam 2 | N 1 and more Exams Health sciences in PDF only on Docsity! Examination 2 Solutions 1. [20] Using only Definition 2', show that the set of odd integers (i.e., {…, -3, -1, 1, 3, …} is infinite. Let and consider = + ∈{2 1| }A k k :f A→ A defined by < = + > 0 ( ) 2 0 a a f a a a A for a . To show this function is one-to-one consider a b∈A , ∈ : if a , then either a b or . If b≠ < < <, 0, 0a b, 0,a b > <, 0,a b = ≠ = ( )( )f a a b f b . If < <0 ,a b = <( ) < )+ =2 (f a a b = + ≠ + =( ) 2 2 b f ) b ( . Finally, if , , 0a b > f a a b f b . Thus, f is one-to-one. Lastly, the function f maps into a proper subset of itself since A ∈1 , but if A =( ) 1f a , then either in which case < 0a = < <( ) 0 1f a a , or in which case >a 0 = + > >2 2 1( )f a a . Either of these leads to a contradiction, so we conclude no element of maps to 1. A 2. [20] Suppose the set U is uncountably infinite, the set V is countable and W is the set difference U . Prove or disprove (with a simple counter example): V∼ W is uncountably infinite. Since W U we have U W~V= , V= ∪ . If W were countable then U would be the union of two countable sets. By Corollary 10.1 ,U would also be countable contrary to hypothesis. We conclude that W is uncountably infinite. 3. [20] Prove that the set of complex numbers = + ∈R{ | ,iy x y }C x is uncountably infinite. Consider the mapping : [0,1]f C→ defined by = +( ) 0f x x i . The function is one-to-one since if , ∈[0,1]x = + ≠0 + =0 ( )( )f x x i y i f y . By Theorem 11, C is uncountably infinite since [ is. 0,1] 4. [20] Using only Definition 1, prove that 1 2 . + + = Ο2 23 ( )n n n We use = 6M and . For , = 1N 1n ≥ ≤ ≤ 21 n n , and + + = + ≤1 2 1 6+2 22 3n n =2 6n 2n3n n . Therefore, 1 2 . + + = Ο2 23 ( )n n n 5. [20] Given that a function assumes only positive values (i.e., for all ) and that :f → ) ( ) 0f n > n∈ = Ο2 (f f prove that = Ο(1)f . Since = Ο2 ( )f f we have that for some M and , We also have that , so we may divide by N ( ) 2| ( ) | | ( ) | .n N f n M f n≥ ⇒ ≤ ( ) 0≠f n f n , thus | and we conclude that . | ( ) | |1n N f n M≥ ⇒ ≤ = Ο(1)f
