Question

Determine the inverse (if an inverse exists) of each of the numbers 1 through 6 in Z_7.

Answer 1

Well, since this is \(\mathbb{Z}_7\), the fastest way would still be to just list out all the possibilities. I should make sure it's clear before I start, that an inverse of \(x\pmod{7}\) is a number \(y\) such that \(xy\equiv 1\pmod{7}\).

Answer 2

So for example for 1: 1(x)-1=7q and just check 1 thru 6?

Answer 3

First, and easiest, is \(1\). Since \(1*1=1\), 1 is its own inverse.

Answer 4

you sure it means inverse under multiplication?

Answer 5

The question doesn't specify any more than what I posted.

Answer 6

Next is \(2\). \[2*1=2\]\[2*2=4\]\[2*3=6\]\[2*4=8\equiv1\pmod{7}\]So \(4\) is the inverse of 2 mod 7.

@satellite73 I would assume it's multiplication since @succubusismydogsname's last question dealt with multiplication, and it doesn't ask for an inverse of 0.

Answer 7

kk, just wasn't sure if this was the group \(\mathbb{Z}_7, +\)

Answer 8

And you could just repeat this process for all the numbers 1-6. However, if you're dealing with numbers larger than 7, this may be a very tedious and time consuming process. There are better methods, but I think you might need a bit more familiarity with some of the necessary concepts first.

Answer 9

Sorry for the off subject question but there can only be on inverse? So for an inverse to exist it must be bijective?

Answer 10

*one

Answer 11

you should prove this to your self
start by assuming there are two, then show that they are equal

Answer 12

now how about helping me with my question
jeez

Answer 13

In this case, every number 1-6 has exactly one inverse. However, I don't believe that is necessarily the case. If you had number that wasn't prime (such as 8), then some numbers don't have inverses.

Answer 14

ah proof by contradiction. k will do.

Answer 15

k... what's your question?

Answer 16

It didn't ask for Z_7+ group.

Answer 17

..and they are gone.. ) :

Answer 18

To be even more specific, if \(n\) is prime, then every number in the set \(\{1,...,n-1\}\) has a unique inverse (so there is a bijective map). If \(n\) is not prime, then not every number in the set will have an inverse. BUT, if you take the set \[\{x|x\in\{1,...,n-1\}\text{ and }x\text{ has an inverse modulo }n\}\]then every \(x\) has a unique inverse.

Answer 19

I hope that made sense.

Answer 20

Yeah, so in order to have an inverse for all integers, there must be a bijective map?

Answer 21

Thanks George. Did Satellite get upset and leave or was he being sarcastic?

Answer 22

I think he was being sarcastic :P

And yes, to have an inverse for all the integers (except 0), it must, by definition, be bijective (a function has an inverse if and only if it's bijective). And when \(n\) is prime, you get the nicest bijection.

Answer 23

Splendid!

Answer 24

@kinggeorge

Answer 25

Just to reclarify. When something asks to "show that there exists..." all you need to do is show an example, Correct?

Answer 26

That's one way to do it, and usually the most straightforward way. There are some circumstances where you aren't able to find an example, but you are able to show existence.

But in this case, where you're given specific numbers, finding an example is usually the best strategy.

Answer 27

last question I swear! but with relations can you make up ridiculous things like R={r{1}={5}, w{2}={5}}?

Answer 28

I'm not quite sure I understand what you're saying there. Could you clarify the notation?

Answer 29

Set A={1,2,3,4} and B={5,6,7,8} and R is a relation from A to B. R={r(1)=(5), r(2)=5, r(3)=6, r(4)=7}.

Answer 30

Question: Let A and B be nonempty sets and let R be a nonempty relation from A to B. Show that there exists a subset A′ of A and a subset f of R such that f is a function from A′ to B.

Answer 31

What I would do for this (this works for infinite sets as well), is to divide \(R\) into a bunch of different subsets \(f_a,f_b,...\) where \(a,b,...\in A\) and we have exactly \(|A|\) number of subsets of \(R\). Then if we have \(x\in A\), \(f_x\) would be defined as the subset of \(R\) that only takes \(x\) to various elements in \(B\).

Then, choose exactly one element from \(f_a,f_b,...\) what you get should be a function.

Answer 32

You'll probably need to write that your own way, and in a more formal way, but I think that argument works.

Answer 33

uh gonna have to ruminate on your idea for a bit. Although my idea isn't very creative/impressive doesn't it technically work?

Answer 34

or it would be difficult to make a function from it?

Answer 35

The most important thing in making a function, is make sure any element is only sent to one place. So as long as if \(f(a)=b\) and \(f(a)=c\), then \(b=c\) is true, you probably have a function. So if you can make this,you should be fine.

Answer 36

Yes, i see, i see.