For (R,o), where xoy=x+y+sqrt3 deretmine wheter or not it forms a group, by checking the group axioms.Can somebody just check it for me please. Not really sure if I`m right in that what i have done...
For (R,o), where xoy=x+y+sqrt3 deretmine wheter or not it forms a group, by checking the group axioms.
Can somebody just check it for me please. Not really sure if I`m right in that what i have done (:
I found that:
G1- xoy belogns to R, for any x,y in R so xoy is closed under o .
G2. x o e: x+e+sqrt3=x that is e=-sqrt 3
e o x: e+x+sqrt3=x
, so -sqrt3 is an identity. --- not relly sure if that correct.
G3. x o y= e so x+y+sqrt3=-sqrt3
thus inverse of x = y= -x-2sqrt3
and g o g^(-1)=e=g(-1) o g
G4 LHS= x o(yoz)= xo(y+z+sqrt3)=z+y+z+2sqrt3
RHS-(x o y)o z= (x+y+sqrt3)oz=x+y+z+2sqrt3
Hence (R,o) is a gorup.
You need to check if the 4 axioms hold such that:
1. Closure axiom: For all `g_1, g_2 in R =gt g_1og_2 in R`
Hence, for all `x,y in R` yields `xoy=x + y + sqrt3 ` `in R` (the addition of three real numbers, x+y+sqrt3, is also a real number)
2. Identity axiom: There is an identity element `e in R` such that:
`xoe = eox = x`
`xoe = x+e+sqrt3`
You need to set the equations `x+e+sqrt3` and x equals such that:
`x+e+sqrt3 = x =gt e+sqrt3 = 0 = gt e = -sqrt 3`
3. Inverses axiom: For each `x in R` , there is an inverse `x^(-1)``in R` such that:
`x o x^(-1) = x^(-1) o x = e`
`x o x^(-1) = x + x^(-1) + sqrt 3`
`x + x^(-1) + sqrt 3 = -sqrt 3 =gt x^(-1) = -x - 2sqrt3`
4. Associativity axiom: For all `x,y,z in R` yields:
`xo(yoz) = (xoy)oz`
`x + (yoz) + sqrt 3 = (xoy) + z + sqrt 3`
Reducing `sqrt 3` both sides yields:
`x + (yoz) = (xoy) + z`
You need to substitute `y+z+sqrt3` for (yoz) and `x+y+sqrt3 ` for (xoy) such that:
`x + y + z + sqrt3 = x + y + sqrt3 + z`
Notice that both sides are equal.
Hence, since the 4 axioms hold, then the set R forms a group with binary operation "o".