# Thread: Vector space and 10 axioms related question

1. ## Vector space and 10 axioms related question

I'm stuck at a problem related to vector space detection. I know how to solve it but don't understand a particular information.

We know that if a set of objects follow 10 axioms of vector space than that defines a vector space.

Say we have a problem:

The set of all positive real numbers with operations
$
x+y = xy\,\,\,and\,\,\, k.x = x^k
$

And asked to verify if the set is vector space or not.

Now one of the axioms to be satisfied is:
For each $u$ in $V$, there is an object $-u$ in $V$, called a negative of $u$, such that
$
u + (-u) = (-u) + u = 0
$

For the above problem an object $u = x$ then $-u = \frac{1}{x}$
So, $u + (-u) = x.\frac{1}{x} = 1 = \mathbf0 = 1 = \frac{1}{x}.x = -u + u$
Now my question: why's $1 = 0$ How's that possible? In what way $1 = 0$?

Shouldn't we say that this set is not a vector space because it violates one of the axioms?

I can't seem to find the answer. Can anyone help me on this?

2. Originally Posted by x3bnm
I'm stuck at a problem related to vector space detection. I know how to solve it but don't understand a particular information.

We know that if a set of objects follow 10 axioms of vector space than that defines a vector space.

Say we have a problem:

The set of all positive real numbers with operations
$
x+y = xy\,\,\,and\,\,\, k.x = x^k
$

And asked to verify if the set is vector space or not.

Now one of the axioms to be satisfied is:
For each $u$ in $V$, there is an object $-u$ in $V$, called a negative of $u$, such that
$
u + (-u) = (-u) + u = 0
$

For the above problem an object $u = x$ then $-u = \frac{1}{x}$
So, $u + (-u) = x.\frac{1}{x} = 1 = \mathbf0 = 1 = \frac{1}{x}.x = -u + u$
Now my question: why's $1 = 0$ How's that possible? In what way $1 = 0$?

Shouldn't we say that this set is not a vector space because it violates one of the axioms?

I can't seem to find the answer. Can anyone help me on this?
$x+0=x(0)=0=(0)x=0+x$

$-x=0$

I think you can do this.

3. Originally Posted by x3bnm
I'm stuck at a problem related to vector space detection. I know how to solve it but don't understand a particular information.

We know that if a set of objects follow 10 axioms of vector space than that defines a vector space.

Say we have a problem:

The set of all positive real numbers with operations
$
x+y = xy\,\,\,and\,\,\, k.x = x^k
$

And asked to verify if the set is vector space or not.

Now one of the axioms to be satisfied is:
For each $u$ in $V$, there is an object $-u$ in $V$, called a negative of $u$, such that
$
u + (-u) = (-u) + u = 0
$

For the above problem an object $u = x$ then $-u = \frac{1}{x}$
So, $u + (-u) = x.\frac{1}{x} = 1 = \mathbf0 = 1 = \frac{1}{x}.x = -u + u$
Now my question: why's $1 = 0$ How's that possible? In what way $1 = 0$?

Shouldn't we say that this set is not a vector space because it violates one of the axioms?

I can't seem to find the answer. Can anyone help me on this?
x + y = xy
so we know that u + 1 = u(1) = u.

Thus 1 is the zero element for addition. Not to be confused by 0 which is the usual zero element for addition in the real numbers. We can verify this with another problem:
k*x = x^k. So -u = (-1)u = 1/u. Thus we have u + (-u) = u*1/u = 1 as required.

-Dan

4. Thank you dwsmith and topsquark for taking the time explaining this.

I understand both of your solutions. Again thanks a lot.