Question

I am confused with proving the vector space axiom of c(u+v) = cu + cv for the set V of positive real numbers with addition and scalar multiplication defined as follows: x+y = xy and cx=x^c

The proof for that axiom is:

c*(x + y) = c*(xy) = (xy)^c = (x^c)(y^c) = (c*x)(c*y) = c*x + c*y

I got lost after the (c*x)(c*y) point... How does that translate into the final step? Shouldn't it look like: c^2(xy)?

Answer 1

Comment: a friend pointed out that I may be confused with scalar/vector... But since c is actually a exponent, you can't turn the expression of (c*x)(c*y) into c*x + c*y using addition function, can you?

Answer 2

are these 2 separate problems? or is that the conditions which must be met fro x and y

Answer 3

x+y = xy and cx=x^c

x+y-xy = 0

x(1-y)+y = 0

x(1-y)=-y

x = y/(y-1) ; y not= 1

x^c = cy/(y-1)

hmm

Answer 4

c is considered a scalar while x and y are vectors

Answer 5

oh, i wasnt playing with them as vectors was i :)

Answer 6

then xy means dot product right?

Answer 7

x.y = x1y1 + x2y2 + ... + xnyn

but that wouldnt make sense either since dot product produces a scalar while addition produces another vector

Answer 8

oh well, i cant make sense out of it, sorry :)

Answer 9

maybe i am making it too hard
heres the website. look at example 5
http://tutorial.math.lamar.edu/Classes/LinAlg/VectorSpaces.aspx

Answer 10

part g, that is... i am confused with the definition of addition step.... how can we go from (x^c)(y^c) to (x^c) + (y^c)? doesnt that violate the standard addition rule?

Answer 11

ok, so its a self defined addition and scalar multiplication for the vector space.

Answer 12

but i am confused about that because it is defined as x+y=xy, not x^c + y^c = (xy)^c

Answer 13

and your concern is in with g specificaly? right

Answer 14

yes

Answer 15

g is the distributive property c(x+y) = cx + cy in general

Answer 16

that i agree with. but i dont see how its applicable when you have (x^c)(y^c)...... we have exponents there

Answer 17

is cx defined as x^c?

Answer 18

yeah

Answer 19

then by the definition that they provide; its consistent

Answer 20

when x^c stands alone, i can see that. but when it is with y^c, i dont see how we can turn it into an expression that looks like addition as shown in part g

Answer 21

c(x+y) = c(xy); say xy = n

cn = n^c

(xy)^c

x^c y^c

x^c is a; y^c = b

ab = a+b
=x^c + y^c

Answer 22

these relate to some real positive number; in the scheme provided at the start of example 5

Answer 23

they are NOT operations that have anything to do with what math is normally defined as

Answer 24

but we could argue that they are rules of logs

Answer 25

log(a) + log(b) = log(ab)

c log(a) = log a^c

Answer 26

work it out like that and see if it makes any sense to you :)

Answer 27

and also, logs only are defined for postive real numbers greater than 0 as well :)

Answer 28

c ( log(x)+log(y)) = c log(xy)

c log(xy) = log(xy)^c = log(x^c y^c) = log(x^c) + log(y^c)

log(x^c) + log(y^c) = c log(x) + c log(y)

Answer 29

remove logs and youve got example 5

Answer 30

I follow your steps but when you turned x^c to a and y^c to b to perform the addition, I don't buy that. I mean it runs against logic. At least in the way I see it

Answer 31

Any better way to explain it? Thanks for your patience!

Answer 32

You also said that they are not standard math operations. How can I truly understand that?

Answer 33

they mimic the math of logs .....

Answer 34

:how so?

Answer 35

c*(x + y) = c*(xy) = (xy)^c = (x^c)(y^c) = (c*x)(c*y) = c*x + c*y
The last equal sign is by virtue of your definition of addition