Exam 7: Functions

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Is log25=log16625\log _ { 2 } 5 = \log _ { 16 } 625 Why or why not?

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The two quantities are equal. Let x=log16625x = \log _ { 16 } 625 . Then, by definition of logarithm, 16x=62516 ^ { x } = 625 . Now
16x=(24)x=(2x)4 and 625=54.16 ^ { x } = \left( 2 ^ { 4 } \right) ^ { x } = \left( 2 ^ { x } \right) ^ { 4 } \quad \text { and } \quad 625 = 5 ^ { 4 } .
So
(2x)4=54.\left( 2 ^ { x } \right) ^ { 4 } = 5 ^ { 4 } .
Raising both sides to the power 1/41 / 4 gives that
((2x)4)1/4=(54)1/4, and thus 2x=5.\left( \left( 2 ^ { x } \right) ^ { 4 } \right) ^ { 1 / 4 } = \left( 5 ^ { 4 } \right) ^ { 1 / 4 } , \quad \text { and thus } 2 ^ { x } = 5 .
Hence, by definition of logarithm,
log25=x.\log _ { 2 } 5 = x .
Since both quantities equal xx ,
log16625=log25.\log _ { 16 } 625 = \log _ { 2 } 5 .

Let f be a function from a set X to a set Y. Define precisely (but concisely) what it means for f to be one-to-one.

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Sample correct answers: A function ff from a set XX to a set YY is one-to-one if, and only if, given any elements x1x _ { 1 } and x2x _ { 2 } in XX , if f(x1)=f(x2)f \left( x _ { 1 } \right) = f \left( x _ { 2 } \right) then x1=x2x _ { 1 } = x _ { 2 } .

A function ff from a set XX to a set YY is one-to-one if, and only if, given any elements x1x _ { 1 } and x2x _ { 2 } in XX , if x1x2x _ { 1 } \neq x _ { 2 } then f(x1)f(x2)f \left( x _ { 1 } \right) \neq f \left( x _ { 2 } \right) .

A function ff from a set XX to a set YY is one-to-one if, and only if, given any element yy in YY , there is at most one element xx in XX such that f(x)=yf ( x ) = y .

A function ff from a set XX to a set YY is one-to-one if, and only if, each element yy in YY , is the image of at most one element xx in XX .

Let SS be the set of all even integers, and define a function f:ZSf : \mathbf { Z } \longrightarrow \mathbf { S } as follows: f(n)=2n for all integers nf ( n ) = 2 n \quad \text { for all integers } n \text {. } (a) Prove that ff is one-to-one and onto (b) Find a formula for the inverse function f1f ^ { - 1 } . (c) Does the set of all even integers have the same cardinality as the set of all integers? Why or why not?

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a. Proof that ff is one-to-one: Suppose n1n _ { 1 } and n2n _ { 2 } are in SS and f(n1)=f(n2).f \left( n _ { 1 } \right) = f \left( n _ { 2 } \right) . WWe must show that n1=n2./n _ { 1 } = n _ { 2 } . / By definition of f,2n1=2n2f , 2 n _ { 1 } = 2 n _ { 2 } . Dividing both sides by 2 gives n1=n2n _ { 1 } = n _ { 2 } [as was to be shown].

Proof that ff is onto: Let mm be any even integer. [We must show that there is an integer whose image under ff is m.]
Let n=m2n = \frac { m } { 2 } . Then nn is an integer because mm is even, and
f(n)=f(m2)=2(m2)=mf ( n ) = f \left( \frac { m } { 2 } \right) = 2 \left( \frac { m } { 2 } \right) = m
[as was to be shown].
Conclusion: Since ff is both one-to-one and onto, ff is a one-to-one correspondence.
b. Given any even integer m,f1(m)=m2m , f ^ { - 1 } ( m ) = \frac { m } { 2 } .
c. The set of all even integers has the same cardinality as the set of all integers because there is a one-to-one correspondence from the set of all integers to the set of all even integers.

Define a function f:RRf : \mathbf { R } \longrightarrow \mathbf { R } as follows: for all real numbers xx , f(x)=16x5.f ( x ) = 16 x - 5 . Then ff is both one-to-one and onto. Find the inverse function f1f ^ { - 1 } .

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Define F:R×RR×RF : \mathbf { R } \times \mathbf { R } \rightarrow \mathbf { R } \times \mathbf { R } as follows: F(x,y)=(3y1,1x)F ( x , y ) = ( 3 y - 1,1 - x ) for all (x,y)( x , y ) in R×R\mathbf { R } \times \mathbf { R } . (a) F(0,0)=?F(1,4)=F ( 0,0 ) = ? F ( 1,4 ) = ? (b) Is FF one-to-one? Prove or give a counterexample. (c) Is FF onto? Prove or give a counterexample. (d) Is FF a one-to-one correspondence? If not, explain why not. If yes, find F1F ^ { - 1 } .

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Let SS be the set of all nonzero real numbers. Define a function gg from SS to SS by the formula g(x)=1xg ( x ) = \frac { 1 } { x } , for all nonzero real numbers xx . (a) Show that gg is a one-to-one correspondence from SS to SS . (b) Find g1g ^ { - 1 } .

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Let SS be the set of all strings in 0's and 1's, and define a function F:SZnonneg F : S \rightarrow \mathbf { Z } ^ { \text {nonneg } } as follows: for all strings ss in SS , F(s)= the number of 1 ’s in s.F ( s ) = \text { the number of } 1 \text { 's in } s . (a) What is F(001000)?F(111001)F ( 001000 ) ? F ( 111001 ) ? F(10101)F ( 10101 ) ? F(0100)F ( 0100 ) ? (b) Is FF one-to-one? Prove or give a counterexample. (c) Is FF onto? Prove or give a counterexample. (d) Is FF a one-to-one correspondence? If so, find F1F ^ { - 1 } .

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(a) Draw an arrow diagram for a function that is onto but not one-to-one. (b) Draw an arrow diagram for a function that is one-to-one but not onto.

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 Prove that if f:XY and g:YZ are onto, then gf:XZ is also onto. \text { Prove that if } f : X \rightarrow Y \text { and } g : Y \rightarrow Z \text { are onto, then } g \circ f : X \rightarrow Z \text { is also onto. }

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Fill in the blanks: log3(19)=\log _ { 3 } \left( \frac { 1 } { 9 } \right) = ______because _______ .

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 Prove that if f:XY and g:YZ are one-to-one, then gf:XZ is also one-to-one. \text { Prove that if } f : X \rightarrow Y \text { and } g : Y \rightarrow Z \text { are one-to-one, then } g \circ f : X \rightarrow Z \text { is also one-to-one. }

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Let A=B={1,2,3}A = B = \{ 1,2,3 \} , and consider the function f:ABf : A \rightarrow B defined as follows: f(1)=3f ( 1 ) = 3 , f(2)=1,f(3)=3f ( 2 ) = 1 , f ( 3 ) = 3 . Is ff onto? Why or why not?

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Let X={1,2,3,4,5}X = \{ 1,2,3,4,5 \} and Y={u,v,w,x,y}Y = \{ u , v , w , x , y \} , and define h:XYh : X \rightarrow Y as follows: h(1)=v,h(2)=x,h(3)=v,h(4)=v,h(5)=y.h ( 1 ) = v , h ( 2 ) = x , h ( 3 ) = v , h ( 4 ) = v , h ( 5 ) = y . (a) Draw an arrow diagram for hh . (b) Let A={1,2},C={x,v},D={w}A = \{ 1,2 \} , C = \{ x , v \} , D = \{ w \} , and E={w,y}E = \{ w , y \} . Find h(A),h(X),h1(C),h1(D),h1(E), and h1(Y).h ( A ) , h ( X ) , h ^ { - 1 } ( C ) , h ^ { - 1 } ( D ) , h ^ { - 1 } ( E ) , \text { and } h ^ { - 1 } ( Y ) .

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Prove that the set of all integers and the set of all odd integers have the same cardinality.

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Let J5={0,1,2,3,4}J _ { 5 } = \{ 0,1,2,3,4 \} and define a function g:J5×J5J5×J5g : J _ { 5 } \times J _ { 5 } \rightarrow J _ { 5 } \times J _ { 5 } as follows: For all (a,b)( a , b ) \in J5×J5J _ { 5 } \times J _ { 5 } g(a,b)=((5a3)mod5,(4b+2)mod5).g ( a , b ) = ( ( 5 a - 3 ) \bmod 5 , ( 4 b + 2 ) \bmod 5 ) . Find g(3,4)g ( 3,4 ) .

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Is the set of all squares of positive integers countable? That is, is the set S={mZm=k2S = \left\{ m \in \mathbf { Z } \mid m = k ^ { 2 } \right. for some positive integer k}k \} a countable set. Justify your answer.

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Let SS be the set of all strings in 0's and 1's, and define a function gg : SZS \longrightarrow \mathbf { Z } as follows: for each string ss in SS , g(s)=g ( s ) = the number of 1's in ss minus the number of 0 's in s.s . (a) What is g(101011)?g(00100)g ( 101011 ) ? g ( 00100 ) ? (b) Is gg one-to-one? Prove or give a counterexample. (c) Is gg onto? Prove or give a counterexample.

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Let SS be the set of all strings in 0 's and 1's, and define a function g:SZg : S \longrightarrow \mathbf { Z } as follows: for each string ss in SS , g(s)= the number of 0 ’s in s.g ( s ) = \text { the number of } 0 \text { 's in } s . (a) What is g(101011)g ( 101011 ) ? g(00100)g ( 00100 ) ? (b) Is gg one-to-one? Prove or give a counterexample. (c) Is gg onto? Prove or give a counterexample.

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Let f be a function from a set X to a set Y. Define precisely (but concisely) what it means for f to be onto.

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Let X={a,b,c}X = \{ a , b , c \} and Y={u,v}Y = \{ u , v \} . Which of the following arrow diagrams define functions from XX to YY ? a.  Let  X = \{ a , b , c \}  and  Y = \{ u , v \} . Which of the following arrow diagrams define functions from  X  to  Y  ? a.   b.   c.    b.  Let  X = \{ a , b , c \}  and  Y = \{ u , v \} . Which of the following arrow diagrams define functions from  X  to  Y  ? a.   b.   c.    c.  Let  X = \{ a , b , c \}  and  Y = \{ u , v \} . Which of the following arrow diagrams define functions from  X  to  Y  ? a.   b.   c.

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