Abstract algebra

- jango_IN_DTOWN

Abstract algebra

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- jango_IN_DTOWN

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

- jango_IN_DTOWN

@Kainui

- jango_IN_DTOWN

@pooja195

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## More answers

- jango_IN_DTOWN

@ganeshie8

- jango_IN_DTOWN

@lochana

- ganeshie8

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

- jango_IN_DTOWN

Its
\[Z_n\]

- ganeshie8

\(\mathbb{Z}_n = \{0,1,2,\ldots, n-1\}\)
is the set of least nonnegative residues under modulus \(n\) right ?

- jango_IN_DTOWN

yeah correct

- ganeshie8

\(\dfrac{b}{a} = ba^{-1}\)
First, notice that division by \(a\) is well defined only if the multiplicative inverse of \(a\) exists under modulus \(n\)

- jango_IN_DTOWN

yeah......

- ganeshie8

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

- jango_IN_DTOWN

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

- jango_IN_DTOWN

???

- jango_IN_DTOWN

lets take Z_6 here

- ganeshie8

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

- ganeshie8

how good are you with congruence s?

- jango_IN_DTOWN

I know the meanings and few theorems.

- ganeshie8

that should do :)

- jango_IN_DTOWN

ok..:)

- ganeshie8

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

- jango_IN_DTOWN

ok..:)

- ganeshie8

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

- jango_IN_DTOWN

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

- ganeshie8

i never heard of an inverse with respect to division before

- ganeshie8

w.r.t multiplication, yes :)

- jango_IN_DTOWN

its 5 itself

- ganeshie8

Yes, what about the multiplicative inverse of 4 ?

- jango_IN_DTOWN

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

- ganeshie8

so you do know the criterion for existence of inverses

- jango_IN_DTOWN

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

- ganeshie8

we define division using multiplication :
\(\dfrac{b}{a} = ba^{-1}\)

- jango_IN_DTOWN

ohkay,,,:)

- ganeshie8

let me ask you a question

- ganeshie8

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

- ganeshie8

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

- jango_IN_DTOWN

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

- ganeshie8

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

- jango_IN_DTOWN

yeah...

- ganeshie8

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

- ganeshie8

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

- jango_IN_DTOWN

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

- jango_IN_DTOWN

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

- ganeshie8

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

- ganeshie8

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

- ganeshie8

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

- jango_IN_DTOWN

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

- jango_IN_DTOWN

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

- ganeshie8

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

- ganeshie8

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

- jango_IN_DTOWN

correct

- jango_IN_DTOWN

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

- ganeshie8

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

- jango_IN_DTOWN

no its ok.. I too am learning

- ganeshie8

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

- jango_IN_DTOWN

yeah correct

- ganeshie8

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

- jango_IN_DTOWN

gcd(a,0)=0 not equals 1

- jango_IN_DTOWN

so 0 will be automatically excluded

- ganeshie8

right

- ganeshie8

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

- jango_IN_DTOWN

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
\[e _{ij}x _{i}x _{j}=\pm k^2\]

- jango_IN_DTOWN

is a qudaric surface

- jango_IN_DTOWN

can you demonstrate how its a quadric surface?

- ganeshie8

sorry i may not be useful here..

- jango_IN_DTOWN

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

- jango_IN_DTOWN

|dw:1447430541710:dw|

- jango_IN_DTOWN

@ganeshie8 see what I got U(10) is not forming a group w.r.t division.

- jango_IN_DTOWN

@Kainui

- ganeshie8

what makes you think it is a not a group ?
@jango_IN_DTOWN

- jango_IN_DTOWN

|dw:1447514054756:dw|

- jango_IN_DTOWN

@ganeshie8

- ganeshie8

that table looks fine to me ?

- jango_IN_DTOWN

no identity element.. even though every column and every element contains distinct element

- jango_IN_DTOWN

* every column and every row

- ganeshie8

Oh right !
\(1/a \ne a/1\)

- ganeshie8

thats the reason right ?

- ganeshie8

\(a/1 = a\)
but \(1/a\) is not \(a\)

- jango_IN_DTOWN

1 is the right identity.. but not the left identity

- ganeshie8

thank you, makes sense :)

- jango_IN_DTOWN

but I tested this on many groups and got an interesting observation

- ganeshie8

what is it

- jango_IN_DTOWN

Z_n will form a group under ab^-1 mod n if we consider the subset of Z_n in which every element is its own inverse

- jango_IN_DTOWN

For in that case ab^-1=ab

- jango_IN_DTOWN

I mean ab^-1mod n = ab mod n if b^-1=b

- ganeshie8

nice, nice
like \(U_4\) and \(U_6\) ?

- jango_IN_DTOWN

WE can say that" it is not necessary for a given composition table to form a group even if every row and every column of it contains distinct elements" as we saw in this case.
If we take members from U(8), then we can see that the members of it forms a group under the composition ab^-1 mod n , but we cannot name it U(8) since the definition of U(8) is different.. We have to name it something different.. See the observation is really interesting

- jango_IN_DTOWN

the example I gave, if you take only 1 and 9, see it will form a group.. see the order of the group is 2< order of U(10)

- jango_IN_DTOWN

I will be back after an hour.:)

- ganeshie8

Very interesting indeed ! we can show that there exist at least two elements in \(Z_n\) such that \(xx^{-1}=e\)

- ganeshie8

okay, im also going for dinner... cya :)

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