jango_IN_DTOWN
  • jango_IN_DTOWN
Abstract algebra
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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jango_IN_DTOWN
  • 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
  • jango_IN_DTOWN
@Kainui
jango_IN_DTOWN
  • jango_IN_DTOWN
@pooja195

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jango_IN_DTOWN
  • jango_IN_DTOWN
@ganeshie8
jango_IN_DTOWN
  • jango_IN_DTOWN
@lochana
ganeshie8
  • ganeshie8
you can divide by \(a\) only if the inverse exists for \(a\) under modulus \(n\)
jango_IN_DTOWN
  • jango_IN_DTOWN
Its \[Z_n\]
ganeshie8
  • ganeshie8
\(\mathbb{Z}_n = \{0,1,2,\ldots, n-1\}\) is the set of least nonnegative residues under modulus \(n\) right ?
jango_IN_DTOWN
  • jango_IN_DTOWN
yeah correct
ganeshie8
  • 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
  • jango_IN_DTOWN
yeah......
ganeshie8
  • ganeshie8
Next, how do you find the inverse of \(a\) under modulus \(n\) ?
jango_IN_DTOWN
  • jango_IN_DTOWN
\[[1/2]=[0]\]
jango_IN_DTOWN
  • jango_IN_DTOWN
???
jango_IN_DTOWN
  • jango_IN_DTOWN
lets take Z_6 here
ganeshie8
  • ganeshie8
you get the multiplicative inverse of \(a\) by solving : \[ax\equiv 1\pmod{n}\] yes ?
ganeshie8
  • ganeshie8
how good are you with congruence s?
jango_IN_DTOWN
  • jango_IN_DTOWN
I know the meanings and few theorems.
ganeshie8
  • ganeshie8
that should do :)
jango_IN_DTOWN
  • jango_IN_DTOWN
ok..:)
ganeshie8
  • ganeshie8
lets take \(\mathbb{Z}_6\) maybe, just for working an example
jango_IN_DTOWN
  • jango_IN_DTOWN
ok..:)
ganeshie8
  • ganeshie8
let \(a=5\) whats the inverse of \(a\) in \(\mathbb{Z}_6\) ?
jango_IN_DTOWN
  • jango_IN_DTOWN
w.r.t multi[lication ?? or division?
ganeshie8
  • ganeshie8
i never heard of an inverse with respect to division before
ganeshie8
  • ganeshie8
w.r.t multiplication, yes :)
jango_IN_DTOWN
  • jango_IN_DTOWN
its 5 itself
ganeshie8
  • ganeshie8
Yes, what about the multiplicative inverse of 4 ?
jango_IN_DTOWN
  • 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
  • ganeshie8
so you do know the criterion for existence of inverses
jango_IN_DTOWN
  • jango_IN_DTOWN
yeah, for multiplication i know. but the question is asked about division
ganeshie8
  • ganeshie8
we define division using multiplication : \(\dfrac{b}{a} = ba^{-1}\)
jango_IN_DTOWN
  • jango_IN_DTOWN
ohkay,,,:)
ganeshie8
  • ganeshie8
let me ask you a question
ganeshie8
  • ganeshie8
consider two cases : 1) \(n\) is composite, 2) \(n\) is prime
ganeshie8
  • 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
  • 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
  • ganeshie8
Yes, that means we are sure that few elements will not have inverses when \(n\) is composite.
jango_IN_DTOWN
  • jango_IN_DTOWN
yeah...
ganeshie8
  • ganeshie8
Since division is defined based on inverses, we cannot define division for arbitrary \(n\).
ganeshie8
  • ganeshie8
division operation is definied on the set \(\mathbb{Z}_n\) only when \(n\) is prime
jango_IN_DTOWN
  • jango_IN_DTOWN
oh and about the composite one, do you have counter example?
jango_IN_DTOWN
  • jango_IN_DTOWN
or will we say that since multiplicative inverse doesnot exist in Z_n for all n for that reason
ganeshie8
  • ganeshie8
there exists at least one element without an inverse when \(n\) is composite
ganeshie8
  • ganeshie8
to define the division operation on the entire set, you must have inverse for each and every element
ganeshie8
  • ganeshie8
since at least one element has no inverse when \(n\) is composite, you cannot define division operation on the set
jango_IN_DTOWN
  • 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
  • jango_IN_DTOWN
in Z_n w.r.t multiplication we also excluded 0
ganeshie8
  • 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
  • ganeshie8
If you exclude 0, it is no longer the set \(\mathbb{Z}_n\)
jango_IN_DTOWN
  • jango_IN_DTOWN
correct
jango_IN_DTOWN
  • 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
  • ganeshie8
Btw, I'm not good in abstract algebra... do let me know if im wrong.. :)
jango_IN_DTOWN
  • jango_IN_DTOWN
no its ok.. I too am learning
ganeshie8
  • ganeshie8
we don't need cases like composite/prime saying gcd(a,n)=1 is sufficient i guess
jango_IN_DTOWN
  • jango_IN_DTOWN
yeah correct
ganeshie8
  • ganeshie8
when \(n\) is a prime, we have \(\gcd(a,n)=1\) for all \(1\le a\lt n\)
jango_IN_DTOWN
  • jango_IN_DTOWN
gcd(a,0)=0 not equals 1
jango_IN_DTOWN
  • jango_IN_DTOWN
so 0 will be automatically excluded
ganeshie8
  • ganeshie8
right
ganeshie8
  • ganeshie8
https://s3.amazonaws.com/upload.screenshot.co/587aea3343
jango_IN_DTOWN
  • 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
  • jango_IN_DTOWN
is a qudaric surface
jango_IN_DTOWN
  • jango_IN_DTOWN
can you demonstrate how its a quadric surface?
ganeshie8
  • ganeshie8
sorry i may not be useful here..
jango_IN_DTOWN
  • jango_IN_DTOWN
ohkay.. anyways thanks.. I always think that you have knowledge in every field.:)
jango_IN_DTOWN
  • jango_IN_DTOWN
|dw:1447430541710:dw|
jango_IN_DTOWN
  • jango_IN_DTOWN
@ganeshie8 see what I got U(10) is not forming a group w.r.t division.
jango_IN_DTOWN
  • jango_IN_DTOWN
@Kainui
ganeshie8
  • ganeshie8
what makes you think it is a not a group ? @jango_IN_DTOWN
jango_IN_DTOWN
  • jango_IN_DTOWN
|dw:1447514054756:dw|
jango_IN_DTOWN
  • jango_IN_DTOWN
@ganeshie8
ganeshie8
  • ganeshie8
that table looks fine to me ?
jango_IN_DTOWN
  • jango_IN_DTOWN
no identity element.. even though every column and every element contains distinct element
jango_IN_DTOWN
  • jango_IN_DTOWN
* every column and every row
ganeshie8
  • ganeshie8
Oh right ! \(1/a \ne a/1\)
ganeshie8
  • ganeshie8
thats the reason right ?
ganeshie8
  • ganeshie8
\(a/1 = a\) but \(1/a\) is not \(a\)
jango_IN_DTOWN
  • jango_IN_DTOWN
1 is the right identity.. but not the left identity
ganeshie8
  • ganeshie8
thank you, makes sense :)
jango_IN_DTOWN
  • jango_IN_DTOWN
but I tested this on many groups and got an interesting observation
ganeshie8
  • ganeshie8
what is it
jango_IN_DTOWN
  • 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
  • jango_IN_DTOWN
For in that case ab^-1=ab
jango_IN_DTOWN
  • jango_IN_DTOWN
I mean ab^-1mod n = ab mod n if b^-1=b
ganeshie8
  • ganeshie8
nice, nice like \(U_4\) and \(U_6\) ?
jango_IN_DTOWN
  • 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
  • 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
  • jango_IN_DTOWN
I will be back after an hour.:)
ganeshie8
  • ganeshie8
Very interesting indeed ! we can show that there exist at least two elements in \(Z_n\) such that \(xx^{-1}=e\)
ganeshie8
  • ganeshie8
okay, im also going for dinner... cya :)

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