Question

Abstract algebra

Answer 1

Show that one cannot define division in Z_n for any arbitrary positive integer n. When division is possible in Z_n ?

Answer 2

@Kainui

Answer 3

@pooja195

Answer 4

@ganeshie8

Answer 5

@lochana

Answer 6

you can divide by \(a\) only if the inverse exists for \(a\) under modulus \(n\)

Answer 7

Its
\[Z_n\]

Answer 8

\(\mathbb{Z}_n = \{0,1,2,\ldots, n-1\}\)

is the set of least nonnegative residues under modulus \(n\) right ?

Answer 9

yeah correct

Answer 10

\(\dfrac{b}{a} = ba^{-1}\)

First, notice that division by \(a\) is well defined only if the multiplicative inverse of \(a\) exists under modulus \(n\)

Answer 11

yeah......

Answer 12

Next, how do you find the inverse of \(a\) under modulus \(n\) ?

Answer 13

\[[1/2]=[0]\]

Answer 14

???

Answer 15

lets take Z_6 here

Answer 16

you get the multiplicative inverse of \(a\) by solving :
\[ax\equiv 1\pmod{n}\]

yes ?

Answer 17

how good are you with congruence s?

Answer 18

I know the meanings and few theorems.

Answer 19

that should do :)

Answer 20

ok..:)

Answer 21

lets take \(\mathbb{Z}_6\) maybe, just for working an example

Answer 22

ok..:)

Answer 23

let \(a=5\)
whats the inverse of \(a\) in \(\mathbb{Z}_6\) ?

Answer 24

w.r.t multi[lication ?? or division?

Answer 25

i never heard of an inverse with respect to division before

Answer 26

w.r.t multiplication, yes :)

Answer 27

its 5 itself

Answer 28

Yes, what about the multiplicative inverse of 4 ?

Answer 29

4 dont have any multiplicative inverse since the multiplicative invers exists iff gcd(a,n)=1 where a belongs to Zn

Answer 30

so you do know the criterion for existence of inverses

Answer 31

yeah, for multiplication i know. but the question is asked about division

Answer 32

we define division using multiplication :

\(\dfrac{b}{a} = ba^{-1}\)

Answer 33

ohkay,,,:)

Answer 34

let me ask you a question

Answer 35

consider two cases : 1) \(n\) is composite, 2) \(n\) is prime

Answer 36

For each case,
what can you say about the existence of inverse for each of the elements in \(\mathbb{Z}_n\) ?

Answer 37

2nd case inverse will exist for every element of Z_n\{0}
1st case inverse will exist for those elemnts whose gcd(a,n)=1

Answer 38

Yes, that means we are sure that few elements will not have inverses when \(n\) is composite.

Answer 39

yeah...

Answer 40

Since division is defined based on inverses, we cannot define division for arbitrary \(n\).

Answer 41

division operation is definied on the set \(\mathbb{Z}_n\) only when \(n\) is prime

Answer 42

oh and about the composite one, do you have counter example?

Answer 43

or will we say that since multiplicative inverse doesnot exist in Z_n for all n for that reason

Answer 44

there exists at least one element without an inverse when \(n\) is composite

Answer 45

to define the division operation on the entire set, you must have inverse for each and every element

Answer 46

since at least one element has no inverse when \(n\) is composite, you cannot define division operation on the set

Answer 47

oh got it thanks.. but in Z_p we dont have inverse of 0, so we will exclude it?

Answer 48

in Z_n w.r.t multiplication we also excluded 0

Answer 49

Oh I completely forgot about that!
0 has no inverse, so we cannot define division operation over the set \(\mathbb{Z}_n\) for any \(n\)

Answer 50

If you exclude 0, it is no longer the set \(\mathbb{Z}_n\)

Answer 51

correct

Answer 52

so we will say that since 0 dont have any multiplicative inverse, we cannot define division on Z_n and to define division on z_n we will have to bring restrictions:
case 1) compositive: those gcd(a,n)=1
case 2) prime : exclude 0

Answer 53

Btw, I'm not good in abstract algebra...
do let me know if im wrong.. :)

Answer 54

no its ok.. I too am learning

Answer 55

we don't need cases like composite/prime
saying gcd(a,n)=1 is sufficient i guess

Answer 56

yeah correct

Answer 57

when \(n\) is a prime, we have \(\gcd(a,n)=1\) for all \(1\le a\lt n\)

Answer 58

gcd(a,0)=0 not equals 1

Answer 59

so 0 will be automatically excluded

Answer 60

right

Answer 61

https://s3.amazonaws.com/upload.screenshot.co/587aea3343

Answer 62

so I think we are basically done with this problems.. Let me ask you one thing.
I was reading Theory of Elasticity and it was written that
\[e _{ij}x _{i}x _{j}=\pm k^2\]

Answer 63

is a qudaric surface

Answer 64

can you demonstrate how its a quadric surface?

Answer 65

sorry i may not be useful here..

Answer 66

ohkay.. anyways thanks.. I always think that you have knowledge in every field.:)