text: 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
  So LHS=RHS
  
  Hence (R,o) is a gorup.
  
  sciencesolve | Teacher | (Level 3) Educator Emeritus
  
  Posted on
  
  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".
  
  We’ve answered 317,947 questions. We can answer yours, too.
explanation: The document contains a math question and a detailed answer. The question asks whether the set (R, o) with the operation xoy = x + y + sqrt(3) forms a group by checking the group axioms. The answer provides a step-by-step verification of the group axioms.
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distilabel_metadata.role: system
distilabel_metadata.content: 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
  So LHS=RHS
  
  Hence (R,o) is a gorup.
  
  sciencesolve | Teacher | (Level 3) Educator Emeritus
  
  Posted on
  
  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".
  
  We’ve answered 317,947 questions. We can answer yours, too.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and a detailed answer. The question asks whether the set (R, o) with the operation xoy = x + y + sqrt(3) forms a group by checking the group axioms. The answer provides a step-by-step verification of the group axioms.
  
  Label: question-and-answer