Question

let R non commutative ring define x*y=y.x
show that the set R forms a ring using the operation * and + instead of . and +

i have to prove closer,and distributive
like x*(a+b)=(a+b).x=a.x+b.x=(x*a)+(x*b)
(a+b)*x=x.(a+b)=x.a+x.b=a*x+b*x
so is the distributive law check?

Answer 1

what is the dot supposed to mean?

Answer 2

multiple

Answer 3

just i worry about the second part .is still have distributed

Answer 4

i am lost still
looks like you are just changing times to *

Answer 5

no,,it is like the question in my book,i have to prove closer,associative law and distributive but i answer the question like:it is clear closer and associative but i still not make sure about distributed since they are not equal ,

Answer 6

x*a)+(x*b) does not equal a*x+b*x

Answer 7

oooh i see i should learn to read
it is \(x*y=y\times x\)

Answer 8

and you want to show that \[x*(y+z)=x*y+x*z\]right?

Answer 9

i get it
so it is really a matter of interpreting what those things actually are

Answer 10

the left hand side \(x*(y+z)\) is by definition \((y+z)\times x\) and the right side \(x*y+x*z\) is by definition \(y\times x+z\times x\)

Answer 11

you get it right away by the distributive property of \(\times\) in \(R\)

Answer 12

do u think it is closer

Answer 13

do u thing is the ring is closer and why?

Answer 14

yes, if is closed
what you are doing is defining \(R^{op}\) read "R op" which makes any right module of \(R\) into a left module of \(R^{op}\)

Answer 15

same underlying set
only difference is if \(xy=z\) in \(R\) then \(yx=z\) in \(R^{op}\)

Answer 16

it is closed under multiplication because \(R\) is
that is, if \(x, y \in R\) then by closure of \(R\) so is \(xy\)
in \(R^{op}\) if \(x, y\in R\) then \(x*y=y\times x\) and since \(R\) is closed, then \(y\times x\in R\) also