Question

Let m be an integer, \(m\geq 2\), and let \(Z_m \)be the set of all positive integers less than m, \(Z_m = {0,1,.....,m-1}\) If \(\alpha ~~and~~\beta\) are in \(Z_m\), let \(\alpha +\beta\)be the least positive remainder obtained by dividing the sum of \(\alpha ~~and~~\beta\) by m
Similarly, let \(\alpha\beta\) be the least positive remainder obtained by dividing the product of \(\alpha ~~and~~\beta\) by m.
Prove:
a) \(Z_m\) is a field if and only if m is a prime
b) What is -1 in \(Z_5\) ?
c)What is 1/3 in \(Z_7\) ?
Please, help

Answer 1

i'm not your huckleberry on this stuff. try sattelite

Answer 2

b) it's 4. \[4 \times 1 \equiv 4\equiv -1 \mod 5\]\[4 \times 2 \equiv 3\equiv -2 \mod 5\]\[4 \times 3 \equiv 2\equiv -3 \mod 5\]\[4 \times 4 \equiv 1\equiv -4 \mod 5\]

Answer 3

I got this , friend. In general, it is a multiple of 4, right?

Answer 4

i dont know this, sorrry. Thanks for helpinggg

Answer 5

c. 3x+7y=1 => 3(5)+7(-2)=1 =>
\(3\cdot5\equiv 1 \mod 7\), thus multiplying by 5 is equivalent to dividing by 3, mod 7.

Ex.
\( 3\cdot 5\equiv15\equiv1\mod 7 \) and \(3\div 3=1\)
\( 6\cdot 5\equiv30\equiv2\mod 7 \) and \(6\div 3=2\)
\( 9\cdot 5\equiv45\equiv3\mod 7 \) and \(9\div 3=3\)
\( 12\cdot 5\equiv60\equiv 4\mod 7 \) and \(12\div 3=4\)
and so on...

Answer 6

I am reading, not digest yet, hihihi... spending time to help other, that's why. I am sorry for that

Answer 7

I have a question there: 3*5 =1 (mod7)
but 7*(anything) = 0 (mod 7)
why do you choose -2?

Answer 8

Where are you stuck?

Answer 9

Beats me.

Answer 10

a)

Answer 11

for c) I was taught the remainder in Z_7 is an integer. I am not sure how to get 1/3

Answer 12

Let \(n=1/3\) that means \(3n=1\) and we know the definition for multiplication.

Answer 13

got you,

Answer 14

For a, you probably need to use the division algorithm
\[
n = dq+r
\]

Answer 15

what can you use for your proof?

Answer 16

I don't know, it's my lecture for next semester, my prof asks us preview before the course. I read the book and try some exercises

Answer 17

oh sorry i came late
you are figuring out what \(\frac{1}{3}\) is in \(\mathbb{Z}_7\)

Answer 18

i have never seen it written like this before, usually it is written \(3^{-1}\) i.e. the multiplicative inverse of 3
there are no fractions in \(\mathbb{Z}_7\)

Answer 19

since you only have 7 elements, the best method is to grind it till you find it

Answer 20

\(1\times 3=3\)
\[2\times 3=6\]\[3\times 3=2\]\[4\times 3=5\]\[5\times 3=1\] got it on the fifth try

Answer 21

you said that there is no fraction in Z_7??

Answer 22

yes, that is what i said, and i think i will stick by it

Answer 23

but i guess it is just a matter of notation is all
since \(\mathbb{Z}_7\) is a field every non zero element has a multiplicative inverse
but it is usually denoted as \(a^{-1}\) not \(\frac{1}{a}\)

Answer 24

so, it can be the answer, right? we can have "NO SOLUTION" as a solution, right?

Answer 25

no no, the answer is 5

Answer 26

\(5\times 3=1\) so \(5=3^{-1}\)

Answer 27

\(\color{blue}{\text{Originally Posted by}}\) @wio
Let \(n=1/3\) that means \(3n=1\) and we know the definition for multiplication.
\(\color{blue}{\text{End of Quote}}\)
Definition of multiplication is used to evaluate division.

Answer 28

what @wio said

Answer 29

just never seen it written as \(\frac{1}{3}\) before is all

Answer 30

\(1/3\) is not a number, but an algebraic expression to be parsed.

Answer 31

kk got it, just never saw it

Answer 32

For the proof, you probably want to factor \(m\) into its prime decomposition and consider the consequences.

Answer 33

Proving that something is a field can be pretty tedious, though.

Answer 34

oh, I used to learn this concept from my discrete math 2 years ago but just a little bit.

Answer 35

snap proof, in \(\mathbb{Z}\) the ideal \(p\mathbb{Z}\) is maximal, making \(\mathbb{Z}_p\cong \mathbb{Z}/p\mathbb{Z}\) a field
but i have a feeling that is not what you are being asked to do
what you have to show is that any \(n\in\mathbb{Z}_p\) has a multiplicative inverse

Answer 36

can you please help me @satellite73

Answer 37

all the other properties of being a field are routine, this is the only one that needs work, and i would suggest "bezout" or whatever it is called

Answer 38

@satellite73 I know Bezout theorem

Answer 39

proof is something like
if \(m\in \mathbb{Z}_p\) then since \(p\) is prime, \(m\) and \(p\) are relatively prime
therefore there are \(x,y\) with \(1=xm+yp\)
this gives
\[xm=1-yp\] etc

Answer 40

there is more to it, but that is the basic idea
all other field properties are inherited from the regular properties of integers

Answer 41

yeah bezout. same thing as finding a unit fraction mod n.

ax+by=1
=> ax = 1 mod y
=> a = 1/x mod y

that's all i did.

you were specifically looking for the inverse of 3 so x = 3.

Answer 42

Thanks all, now I know where I can start learning.