Proof: Positive Real Numbers as Vector Space with Modified Operations

[MurdocJensen]

Homework Statement

Show that the set of all positive real numbers, with x+y and cx redefined to equal the usual xy and x^c, is a vector space. What is the zero vector?

The Attempt at a Solution

My attempt stops at me trying to decipher the problem. Are they asking me to take particular vector space rules and change them and show that, given the change in the rules, the set of all real positive numbers is a vector space?

I'm also confused as to what they mean by xy and x^c, in that x and y are both vectors and I'm not sure what kind of multiplication they want me to do.

 

[Mark44]

Homework Statement

Show that the set of all positive real numbers, with x+y and cx redefined to equal the usual xy and x^c, is a vector space. What is the zero vector?

The Attempt at a Solution

My attempt stops at me trying to decipher the problem. Are they asking me to take particular vector space rules and change them

No, you have the same axioms (10 of them I believe).

and show that, given the change in the rules, the set of all real positive numbers is a vector space?

A vector space is not just a set of things (positive reals in this case); it is a set, together with two operations, + and *, that satisfy the standard vector space axioms.

I'm also confused as to what they mean by xy and x^c, in that x and y are both vectors and I'm not sure what kind of multiplication they want me to do.

x and y are positive real numbers.
To minimize confusion, I'll use $\oplus$ to represent addition and $\odot$ to represent multiplication in this vector space.

For example, $2 \oplus 5$ = $2 \cdot 5$ = 10, and $2 \odot 3$ = $2^3$ = 8

 

[HallsofIvy]

A vector space is always a space over some scalar field. x+ y is defined for x and y vectors, ax is defined for a a scalar and x a vector. In this particular case, both scalars and vectors are numbers but you will still need to distinguish between them. For example, one of the axioms for vector spaces is that scalar multiplication "distributes" over addtion: a(x+ y)= ax+ ay. Here, x, y, and a are all numbers and "a(x+ y)" is $(xy)^a$ while "ax+ ay" is $(x^a)(y^a)$. Are those the same?

 

[MurdocJensen]

Mark44: Yes they are the same, but I went about that part differently. I got (xy)^a = x^a + y^a, but I guess we can simplify to your version because these are just 'numbers' being raised to a power, which means xxxx + yyyy is the same as xxxxyyyy or x⁴y⁴, which is just (xy)⁴. 4 is replacing c in this particular case.

This is the first time in my life I am writing as mathematically as this. I suck at it so far.