Question

show me the set G = { x+y√3:x,y=Q} is a group w.r.t addition​

Answer 1

$\sf { \underline{SOLUTION}}$

TO PROVE

G = { x+y√3 : x , y ∈ Q } is a group under addition

PROOF

• CLOSURE PROPERTY

Take x + y√3 and u + v√3 ∈ G

Where x, y, u, v ∈ Q

Then

(x + y√3) + ( u + v√3) = ( x + u ) + ( y + v )√3

Since Q is a group under addition

So (x + y√3) + ( u + v√3) ∈ G

So G is closed under addition

• ASSOCIATIVE PROPERTY

Take (x + y√3), (a+b√3), (u + v√3) ∈ G

Then

(x + y√3) + [(a+b√3)+ (u + v√3)]

= (x + y√3) + (a+u) +(b+v) √3

= (x + a + u) +( y+ u + v) √3

And

[(x + y√3) + (a+b√3)] + (u + v√3)]

= [(x + a) +(y+b) √3) + (u + v√3)

= (x + a + u) +( y+ u + v) √3

∴ (x + y√3) + [(a+b√3)+ (u + v√3)]

= [(x + y√3) + (a+b√3)] + (u + v√3)]

G is associative under addition

• EXISTENCE OF IDENTITY ELEMENT

Take (x + y√3) ∈ G

Also 0 ∈ G

Such that

(x + y√3) + 0 = ( x + y√3) = 0 + ( x + y√3)

0 is the identity element

• EXISTENCE OF INVERSE OF AN ELEMENT

Take (x + y√3) ∈ G

Then ( - x - y√3) ∈ G

Such that

(x + y√3) + ( - x - y√3) = 0 = ( -x - y√3) + ( x + y√3)

So ( - x - y√3) is the inverse of ( x + y√3)

Hence G is a group under addition

Hence proved

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