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Question 999118: The set X consists of all numbers from x+y√3 where x and y are integers.
Show the following.
A) X is closed under addition and multiplication of real numbers.
B) there is an identity in x for addition.
C) there is an identity in x for multiplication.
D) not every element of x has an inverse with respect to multiplication.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! I'll do part A to get you started
A)
Closure under addition
Let
z = a+b*sqrt(3)
w = c+d*sqrt(3)
where a,b,c,d are integers
We need to show that z+w can be written in the form x+y*sqrt(3) where x and y are integers.
z+w = (z) + (w)
z+w = (a+b*sqrt(3)) + (c+d*sqrt(3))
z+w = a+b*sqrt(3) + c+d*sqrt(3)
z+w = a+c+b*sqrt(3)+d*sqrt(3)
z+w = (a+c)+(b*sqrt(3)+d*sqrt(3))
z+w = (a+c)+(b+d)*sqrt(3)
The last equation is in the form x+y*sqrt(3) where x = a+c and y = b+d
a+c and b+d are integers, so x & y are integers.
This proves the set X is closed under addition.
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Closure under multiplication
Let
z = a+b*sqrt(3)
w = c+d*sqrt(3)
where a,b,c,d are integers
We need to show that z*w can be written in the form x+y*sqrt(3) where x and y are integers.
z*w = (z)*(w)
z*w = (a+b*sqrt(3))*(c+d*sqrt(3))
z*w = a*(c+d*sqrt(3))+b*sqrt(3)*(c+d*sqrt(3))
z*w = a*c+a*d*sqrt(3)+b*sqrt(3)*c+b*sqrt(3)*d*sqrt(3)
z*w = a*c+a*d*sqrt(3)+b*sqrt(3)*c+b*d*sqrt(3)*sqrt(3)
z*w = a*c+a*d*sqrt(3)+b*sqrt(3)*c+b*d*sqrt(3*3)
z*w = a*c+a*d*sqrt(3)+b*sqrt(3)*c+b*d*sqrt(9)
z*w = a*c+a*d*sqrt(3)+b*sqrt(3)*c+b*d*3
z*w = a*c+a*d*sqrt(3)+b*sqrt(3)*c+3bd
z*w = (ac+3bd)+(ad*sqrt(3)+bc*sqrt(3))
z*w = (ac+3bd)+(ad+bc)*sqrt(3)
The last equation is in the form x+y*sqrt(3) where x = ac+3bd and y = ad+bc
ac+3bd and ad+bc are integers, so x & y are integers.
This proves the set X is closed under multiplication.
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