Given the operation in RR^2: (x_1,y_1)+(x_2,y_2)=(x_1x_2,y_1y_2) I would like to find whether this is a vector space in RR. Looking at the Additive Zero Axiom, we get: (x_1,y_1)+0=(x_1(0),y_1(0))=0 To satisfy the Additive Zero Axiom, (x_1,y_1)+0=(x_1,y_1) must be true. For this to be true, 0 would have to be (1,1) Is this possible, or would we be able to say this is not a vector space?

beninar6u 2022-09-29 Answered
Given the operation in R 2 :
( x 1 , y 1 ) + ( x 2 , y 2 ) = ( x 1 x 2 , y 1 y 2 )
I would like to find whether this is a vector space in R . Looking at the Additive Zero Axiom, we get:
( x 1 , y 1 ) + 0 = ( x 1 ( 0 ) , y 1 ( 0 ) ) = 0
To satisfy the Additive Zero Axiom, ( x 1 , y 1 ) + 0 = ( x 1 , y 1 ) must be true. For this to be true, 0 would have to be (1,1)
Is this possible, or would we be able to say this is not a vector space?
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Answers (2)

Toby Barron
Answered 2022-09-30 Author has 7 answers
Is not a vector space, for instance let us try find the zero element in R 2 with the given operation.
Let P = ( x , y ) R 2
( x , y ) + ( e 1 , e 2 ) = ( x , y ) implies ( x e 1 , y e 2 ) = ( x , y ) and then e 1 = 1 and e 2 = 1 it is the zero element must be (1,1).
But in this case (0,0) isn´t invertible since (0,0)+(a,b)=(1,1) implies (0,0)=(1,1) which is a contradiction.
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priscillianaw1
Answered 2022-10-01 Author has 1 answers
The additive identity is indeed ( 1 , 1 )
Let's check for inverse of ( 0 , 0 )
For any x , y R
( 0 , 0 ) + ( x , y ) = ( 0 , 0 ) ( 1 , 1 ) .
Hence it can't be a vector space.
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