Lesson 31 Solutions

Exercise 1) Let f:\{1, 2, 3, 4\}\rightarrow \{a, b, c\} be the function defined as f(1)=b, f(2)=b, f(3)=c, f(4)=a and let g:\{a,b,c\}\rightarrow \{apple, Kobe, Donkey, fruitcake\} be the function defined as g(a)=Kobe, g(b)=fruitcake, g(c)=Donkey.  Write down (or say out loud, or figure out in your head) the value of g\circ f: \{1, 2, 3, 4\}\rightarrow \{apple, Kobe, Donkey, fruitcake\} for each element in \{1, 2, 3, 4\}.  In other words, calculate g(f(a)) for all a\in A.

Solution:  All we have to do here is “trace through” the various definitions of the functions.  Namely, for figuring out what g(f(a)) is for some a\in \{1, 2, 3, 4\}, we simply need to follow the element a through the two functions.  So here we go.  Let’s start with 1\in \{1, 2, 3, 4\}.  We have f(1)=b, and g(b)=fruitcake, so g\circ f(1)= g(f(1))=fruitcake.  Simple as that!

Similarly, we have f(2)=b and again we have g(b)=fruitcake so we have g\circ f(2)=g(f(2))=fruitcake.  Next, we have f(3)=c and g(c)=Donkey, so g\circ f(3)=Donkey.  Finally we have f(4)=a and g(a)=Kobe, so g\circ f(4)=Kobe.  To recap, then, we have that the composition g\circ f: \{1, 2, 3, 4\}\rightarrow \{apple, Kobe, Donkey, fruitcake\} of f and g is defined by

g\circ f(1)=fruitcake

g\circ f(2)=fruitcake

g\circ f(3)=Donkey

g\circ f(4)=Kobe.

And we’re done!

Exercise 2)  Recall that \mathbb{Z} is the set of integers: \mathbb{Z}=\{..., -3, -2, -1, 0, 1, 2, 3, ...\}.  Let \mathbb{E} denote the set of even integers: \mathbb{E}=\{..., -6, -4, -2, 0, 2, 4, 6, ...\}.  Let \mathbb{T} denote the set of multiples of 3: \mathbb{T}=\{..., -9, -6, -3, 0, 3, 6, 12, ...\}.  Let f:\mathbb{Z}\rightarrow \mathbb{E} be the function that takes each integer to 4 times that integer, so that f(a)=4\times a and for example f(-5)=-20\in \mathbb{E} (note that f is most certainly not surjective).  Now let g:\mathbb{E}\rightarrow \mathbb{T} be the function that sends each element in \mathbb{E} to 3 times itself, so that g(a)=3\times a and for example g(6)=18.  Let g\circ f:\mathbb{Z}\rightarrow \mathbb{T} be the composition of f and g.  Calculate

i) g\circ f(-3)

ii) g\circ f(5)

iii) g\circ f(10).

Solution:  Again, for each of the three things that we need to calculate, we just need to “follow through” the definitions of the functions.  So, for i), we have that f(-3)=-12 since f sends each integer to 4 times itself.  Then g(-12)=-36 since g sends each element in its domain to 3 times itself.  Thus g\circ f(-3)=g(f(-3))=-36.  However, we notice that we can do this much faster.  For let’s choose some arbitrary integer n\in \mathbb{Z}.  Then f(n)=4n by definition of f.  Additionally, by definition of g, g(4n)=3\times 4n=12n.  So effectively what we’re doing by first applying f and then applying g is just multiplying the input integer by 12.  Thus, calculating the three cases that we’re asked to is easy.  Namely,

g\circ f(-3)=-36(=12\times -3)

g\circ f(5)=60(=12\times 5)

g\circ f(10)=120(=12\times 10).

And we’re done!

Back to Lesson 31

On to Lesson 32


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