Question

16.
Let g : [ 0, ∞ ) → [ 0, ∞ ) be defined by g(x) = ⌈x^2⌉.
Let A = { x ∈ [ 0, ∞ ) | 3.2 < x < 8.9 } .
(a) Find g (A).
(b) Find g^-1(A).

17. Let E = { 4n | n ∈ ℕ } and consider the characteristic function

E : ℤ → ℤ
What is the . . .
(a) domain
(b) codomain
(c) range
(d)

E ( { 2n | n ∈ ℕ } )
(e)

E ( { 2n | n ∈ ℕ } )

Answer 1

16, Part B is the one I don't know how to solve.
17, I have no idea what to do at all.

Answer 2

Alright, let's start with 16.b. then. When they say \(g^{-1}(A)\), what they mean is the following.\[g^{-1}(A)=\{x\in[0,\infty)\,|\,g(x)\in A\}.\]Since \(g\) is defined as \(g(x)=\lceil x^2\rceil\), that means that \(g\) will always take integer values. The only integers in the set \(A\), are \(4,5,6,7,8\). So to find \(g^{-1}(A)\), you need to find what intervals map to one of those integers.

Did this make sense to you?

Answer 3

I am digesting. GIve me one sec please

Answer 4

ok so for A, we are uising ANY positive real number correct?
Then for B, our DOMAIN is any real number that has been true the function g(A) which makes it an integer due to the ceiling? right?

Answer 5

What was range is now domain? Is that correct to assume?

Answer 6

Ok, so for 17, it wants you to find a bunch of stuff. In general, suppose you have a function, any function, between two sets. Then it's commonly denoted as\[f:X\to Y\]where \(X\) and \(Y\) are the sets. In some cases \(X\) and \(Y\) may be the same. However, when it asks you for the domain, the domain is the set \(X\), and the codomain is the set \(Y\).

In your case, both the domain and the codomain are both \(\mathbb{Z}\).

Answer 7

The range, is a little different. The range is defined as everything in the codomain that is mapped to. So if \(x\in X\), and \(f(x)=y\) where \(y\in Y\), then \(y\) is in the range. Conversely, if \(y\) is not in the range, then there is no \(x\in X\) such that \(f(x)=y\).

Answer 8

Are you familiar with what a characteristic function is?

Answer 9

A little bit. Isn't it one that only has a range set of {0,1} ?

Answer 10

Correct. The exact definition of \(\chi_E\) is as follows:\[\chi_E:X\to\{0,1\}\]where \(E\) is a subset of \(X\), \(\chi_E(x)=1\) if \(x\in E\), and \(\chi_E(x)=0\) if \(x\notin E\).

Answer 11

like a true / false function

Answer 12

But you have to be careful, because I've written it slightly differently than what your problem wrote it as. In the problem, it has \(\chi_E\) going from \(\mathbb{Z}\) to \(\mathbb{Z}\), so that the codomain is \(\mathbb{Z}\). The way I described it, it merely has a codomain of \(\{0,1\}\). So you do have to take some care.

In the end, go with the way the problem wrote it, so that you have a codomain of \(\mathbb{Z}\).

Answer 13

Cool. Thanks.
How would I solve 16b. Do I have plug in the limits ?

Answer 14

I think it is better called the inequality boundaries?

Answer 15

Yeah, there shouldn't be any limits involved. However, you have a fixed set of integers that you want to backtrack to the original intervals they could have come from. For example, 4 is one of the aforementioned integers. The first step you need to backtrack, is the ceiling function. This tells you that \(x^2\) must lie in the interval \(3<x^2\le4\). Now, you also have to notice that \((3,3.2]\) is an interval that is NOT contained in the set \(A\). Thus, you really have that \(3.2<x^2\le4\).

Now, you take the square root of everything, to get the \(\sqrt{3.2}<x\le2\). So any \(x\) that is contained in this interval, is sent to \(4\) by \(g(x)\).

Answer 16

I am sorry, that is what I don't follow. How do you obtain the 3 and 4?

Answer 17

I got the \(3<x^2\le 4\) from the fact that if \(x^2\) is in that interval, then the ceiling function will round it up to 4. However, if it isn't in that interval, it will be rounded up to something completely different. So for example, \(\lceil2.5\rceil=3\), \(\lceil4.1\rceil=5\), etc.

In the particular case I was looking at, I only wanted values that rounded up to 4.

Answer 18

ahh but why tho? That is what I fail to see in this cases in the exercises

Answer 19

why is the 4 significant ?

Answer 20

Why do I care just about 4? Why it's always rounded up? Why what?

Answer 21

yessir!

Answer 22

I may be confused from algebra when I understood the inverse function just meant using the range to get the domain.

Answer 23

That is essentially what I'm tying to do with the inverse function. It just so happens that the range of \(g(x)\) is only \(4,5,6,7,8\). And so we can go back one by one to see what intervals get mapped to each integer, and then take the union of the intervals. For now, I only care about 4 since I wanted to start the problem doing it case by case. We will eventually have to care about 5,6,7, and 8 as well.

It's always rounded up, because that's what the ceiling function does. In general, if \(x\) is a real number in the interval \(n<x\le n+1\), then \(\lceil x\rceil=n+1\).

Answer 24

ahh ok cool I got that part!

Answer 25

so that is where we get the 4 from.

Answer 26

and we go from 4 as a range to whatever is preimage is

Answer 27

Yes! That's exactly it!

So now we've done the case for \(4\). We only have 5,6,7, and 8 left. But instead of doing this case by case, we just have to do it for \(8\), and then make sure to include everything between what we got for 8 and what we got for 4. This is OK because both \(x^2\) and the ceiling function are non-decreasing, and so intervals won't start doing anything too strange.

Answer 28

yeah so... we need to look at what interval of real numbers, once ceilinged (lol) will give us 4!

Answer 29

so we know [3,4] will give us 4, BUT there is a catch

Answer 30

the pieces that we can work with.. do not include [3,3.2]

Answer 31

so for the first " step " , we have (3.2,4] right?

Answer 32

Right so far.

Answer 33

for 5, 6 and 7 we are cool, because those are " smoothly" flowing into the end side of inequality.

Answer 34

ahhh I SEE WHAT HAPPENS!! YAY we have a square!

Answer 35

so I have to account that..!

Answer 36

we took care of assuming for x, but we are doing x^2... so hmmm... we have to take the square root of (3.2,4]

Answer 37

because once we square that, we will get our desire ceiling (x^2)

Answer 38

{sqrt(3.2) , 2 ] right??

Answer 39

Perfect so far.

Answer 40

then we now look at the end.. our last integer is 8... so hmmm we need a range of numbers that will take as to 8 , namely, [7, 8]
However... we can go all the way to 8.9... so

Answer 41

but we don't really care.. because 9 is not an integer we were able to create from the original instruction.

Answer 42

Still looking good.

Answer 43

so I think [sqrt(7) , squrt{8) ]

Answer 44

which using our concept of UNION OF SETS, we get ( sqrt(3.2) , sqrt (8) ] ?

Answer 45

There's one catch. It's technically \[(\sqrt7,\sqrt8].\]With a parenthese instead of a bracket on the left side. But in the end, it doesn't matter, because as you said, after we union everything together, we get\[(\sqrt{3.2},\sqrt8]\]

Answer 46

So that's the end of 16.b. We have that \(g^{-1}(A)=(\sqrt{3.2},\sqrt8]\).

Answer 47

cool!

Answer 48

in our solution however, they say it is

Answer 49

they use sqrt(3) rather than sqrt (3.2)

Answer 50

Oh. Duh. After we get the preimage of the floor, we're no longer directly working with the set \(A\), so we don't care at all about whether or not it's in between \(3\) and \(3.2\). So really we should have left it at 3.

That was my bad. I should have caught that earlier.

Answer 51

Great! lol sorry I was flossing really quick

Answer 52

That settles it for 16b. Thanks it is much clearer now!

Answer 53

You're very welcome.

Answer 54

I am still truly lost about #17. Because I am kinda bad with symbols.

Answer 55

This is the answer for 17

Answer 56

For 17, we already went over parts a, b, and c right? Those were \(\mathbb{Z},\mathbb{Z}\), and \(\{0,1\}\).

OK, good. The solutions agree.

Answer 57

Yupe. Those two are clear.

Answer 58

it is (d) and (e) that are a but alien.

Answer 59

Alright. Then let's look at parts d and e. Part d is actually remarkably easy. It's asking for the image of the set\[\{2n\,|\,n\in\mathbb{N}\}.\]But we know the image (or the range) of \(\mathbb{Z}\) is \(\{0,1\}\). So our answer for d must be a subset of that. Then by simply checking that \(2,4\in\{2n\,|n\in\mathbb{N}\}\) and noticing that \(\chi_E(2)=0\) (since \(2\notin E\)) and \(\chi_E(4)=1\) (since \(4\in E\)), it must be that our answer is \(\{0,1\}\).

Answer 60

hmm how do you get the 1 and 4 sorry?

Answer 61

ohh we are just plugging the set N... starting with n = 1? to get {2 , 4, 6, 8 .... } and those get either a 0 or 1 ?

Answer 62

Precisely. And by simply checking a few values, we see that we get both 0 and 1.

Answer 63

There is pattern here ! awesome

Answer 64

I think they made a typo on (e) I think they mean to ask for the inverse

Answer 65

That would certainly make more sense. Part e is also the hardest I think, but their answer is correct.

Answer 66

We we need to find the preimages and I think we are going to need to find a way to express it in an arbitrary fashion

Answer 67

So to find the inverse, they really want to find the preimage. But remember that the image of \(\chi_E\) is \(\{0,1\}\). But only \(0\in\{2n\,|\,n\in\mathbb{Z}\}\). That means that we're really only finding the preimage of \(\{0\}\).

Following so far?

Answer 68

why are we ditching 1?

Answer 69

oohhh because 0, 4, 8 and such right?

Answer 70

1 gives us 2.. and it is not represented there.

Answer 71

That's not quite it. You're looking at the wrong set.

We exclude 1 because \[1\notin\{2n\,|\,n\in\mathbb{Z}\}\]That set is only the even integers, and 1 is not an even integer. 0 on the other hand, is even, and so we do include it.

Answer 72

ahh ok got it!

Answer 73

Great. So we only need to find the preimage of \(\{0\}\) now.

Answer 74

But we know that the definition of \(\chi_E(x)\) includes the fact that \(\chi_E(x)=0\) if and only if \(x\notin E\). So we only need to find a way to write "every integer that isn't in \(E\)." What the solution has, is just as good a way to write it as any.

Answer 75

{k ∈ ℤ | ¬∃n ∈ N (k = 4n)}
k is an integer suc that there is not an N that belong to the Natural numbers )?

Answer 76

*such that

Answer 77

What they say, is \[\{k\in\mathbb{Z}\,|\,\neg\exists n\in\mathbb{N}(k=4n)\}.\]In words, this translates to "the set of integers \(k\) such that there is no natural number \(n\) so that k=4n."

Answer 78

Dear Heavens. lol. I am not a theorist

Answer 79

I really suck at mathematical theory.. in calculus the epsilon proofs made me cry lol

Answer 80

It takes practice :P

Although I hate the delta-epsilon proofs as well. I much prefer algebra proofs.

Answer 81

is this an algebraic proof? lol

Answer 82

This is more of a set-theory kind of thing. Not my favorite, but I'm familiar enough with the style that I'm usually fine with it.

Answer 83

This class is making me stress a lot

Answer 84

Some things feel very natural , but others don't

Answer 85

What's the class called?

Answer 86

Discrete Mathematics 1, Computer Science major

Answer 87

Discrete math can honestly be a "make it or break it" kind of class. With a good professor, you'll likely love this stuff later, but with a bad professor, it can really hurt the way you think about math.

Answer 88

But you shouldn't worry if some things seem a little unintuitive at first. That feeling happens more and more often as you take more and more advanced classes.

Answer 89

My class is online..... I am destroyed inside

Answer 90

:(

Well, I'll always be happy to help if I'm online, and there are plenty of others here that are more than capable.

Answer 91

you were great! thanks!

Answer 92

I am doing the review for the midterm

Answer 93

Good luck!

Answer 94

Thanks a lot!