Proving Vector Space Sets

1. Mar 3, 2015

B18

1. The problem statement, all variables and given/known data
Determine if they given set is a vector space using the indicated operations.

2. Relevant equations

3. The attempt at a solution
Set {x: x E R} with operations x(+)y=xy and c(.)x=xc
The (.) is the circle dot multiplication sign, and the (+) is the circle plus addition sign.

I tried to use axioms 1,2,3,6,7,8,9,10 to prove it isn't a vector space and have not found anything yet. Im not entirely sure how I can define the 0 vector which is why i haven't tried axioms 4, and 5.

Anyone able to steer me down the right path on this one?

2. Mar 3, 2015

Ray Vickson

Nobody can say, because we do not know what book you are using and do not know exactly what is the content of Axioms 1,2,3, etc. Different books sometimes have slightly different axioms and will often number them differently.

3. Mar 3, 2015

B18

Here is the list of axioms I'm using. I never thought of that. Thanks Ray.

4. Mar 3, 2015

Dick

You say you did axiom 6? $(-1)$ is in $R$. What is $\frac{1}{2} \odot (-1)$? Are you sure you don't mean $R$ to be the positive reals? And why don't you think about trying 1 as a '0 vector'.

Last edited: Mar 4, 2015
5. Mar 4, 2015

HallsofIvy

The "0" vector has the property that v+ 0= v for any vector v. Here, that must be that v times "0" is equal to v. So what number has that property? Of course, every vector must have an "additive inverse" which, since "addition of vectors" is here defined as their product, means there is likely to be a problem with the number 0. That is one reason Dick asked about whether the vectors were not the positive reals rather than all real numbers.

6. Mar 4, 2015

B18

Wow, that one flew right past me. Apparently I wasn't thinking about exponent rules enough. The set is all real numbers though, nothing indicating that it is only positives.

7. Mar 4, 2015

Staff: Mentor

Can you post a photo of the problem? If the set involved is all real numbers, you're going to have problems with things like $.5 \odot (-1)$, which is $(-1)^.5$. Also not stated in this thread is the field that the scalars come from.

8. Mar 4, 2015

Sure thing!

9. Mar 4, 2015

B18

In this case the zero vector would have to be equal to 1?

10. Mar 4, 2015

HallsofIvy

Yes, the "zero vector", with this definition of vector addition, must be the number "1".

11. Mar 4, 2015

Staff: Mentor

So (B18), keeping in mind that 1 in this space acts like 0, what do you have to "add" to a number x so that $x \oplus ? = 1$? IOW, what do you have to "add" to a number x to get "zero"? Hope that doesn't boggle your mind too much!