What is the Zero Vector in a Vector Space with Unconventional Operations?

[B18]

Homework Statement

Determine if they given set is a vector space using the indicated operations.

Homework Equations

The Attempt at a Solution

Set {x: x E R} with operations x(+)y=xy and c(.)x=x^c
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. I am 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?

 

[Ray Vickson]

Homework Statement

Determine if they given set is a vector space using the indicated operations.

Homework Equations

The Attempt at a Solution

Set {x: x E R} with operations x(+)y=xy and c(.)x=x^c
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. I am 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?

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.

 

[B18]

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

 

[Dick]

View attachment 79922
Here is the list of axioms I'm using. I never thought of that. Thanks Ray.

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'.

 

[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.

 

[B18]

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'.

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.

 

[Mark44]

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.

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.

 

[B18]

Sure thing!

 

[B18]

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.

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

 

[HallsofIvy]

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

 

[Mark44]

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

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!