Question

Define the following vectoraddition and scalarmultiplication on R^2

Answer 1

\[(x_1,y_1)+(x_2,y_2)= (x_1+x_2+1,y_1+y_2)\]

Answer 2

and

Answer 3

\[a(x,y)=(ax,ay)\]

Answer 4

\[(x_1,y_1)+(x_2,y_2)= (x_1+x_2+1,y_1+y_2)\] ^^??

Answer 5

Do these mathematical applications determine a vectorspace on R2?

Answer 6

the +1 is kinda weird to me

Answer 7

a vector space must have an indentity, an inverse, and a zero vector; right? and be closed

Answer 8

an identity?

Answer 9

a + i = a ; and element that produces no results

Answer 10

think of 0 for addition ,a dn 1 for multiplication

Answer 11

oh yes

Answer 12

(a,b) + (xi, yi)

a + xi + 1 = a; xi = -1
b + yi = b; yi = 0

Answer 13

since the multiplication is defined as usual for a vector space ... there is no mystery about it

Answer 14

(a,b) + (xz,yz) = 0,0

a + xz + 1 = 0 xz = -(a+1)
b + yz = 0; yz = -b

Answer 15

with i, you just mean a random integer that doesn't change the vector?

Answer 16

xi is an indentity element ... x sub i, or we could have said ix, its just convenient letters to describe the term employed

Answer 17

lost me there

Answer 18

googled it, i get it

Answer 19

:)

Answer 20

so how..i don't know how to solve this, how to think of a solution even

Answer 21

i just demonstrated how to determine identity and inverse. Since the elements to pull from come from the set of real numbers then all these operations are possible; the multiplication stuff is just the usual vector scalar definition so there is no new information to add to that part

Answer 22

might want to see if commutative and associate work out

Answer 23

alright

Answer 24

let x = (xa, xb), let y = (ya,yb)

x+y = (xa+ya+1 , xb + yb)

does y + x equal the same thing?

y+x = (xa+ya+1 , xb + yb) yes

for some z = (za,zb) does the associative property hold?

Answer 25

(x+y)+z = ((xa+ya+1)+za+1 , xb + yb + zb)

x+(y+z) = the same?

Answer 26

y+z = (ya+za+1, yb + zb)

x + (y+z) = (xa +1 + (ya+za+1), xb + yb + zb)

Answer 27

z..?

Answer 28

to prove asso. ?

Answer 29

yeah, the associative property requires at least 3 elements to assess

if: (x+y)+z = x+(y+z), then the associative property holds

Answer 30

all this proving is still kinda difficult for me

Answer 31

i'll write down what you wrote and try and understand

Answer 32

you just gotta keep the definitions that they use in defining the operations; if you can produce the same "form" each time then you have a closed operation

Answer 33

hmm ok

Answer 34

good luck

Answer 35

thanks

Answer 36

i really don't get it