prove that a set V is a vector space

Hello,

I am trying to prove that a set V is a vector space, I managed to prove most of the process, but stuck with a couple of things...

The set V is the set of positive real numbers, and the field is R.

the two operations are:

x++y = xy

a**x = x^a

(where ++ is the addition, ** is the scalar multiplication, and a is a real scalar)

I find it hard to prove that for 2 scalars r and s, and a member of V, x

(r++s)**x = (r**x) + (s**x)

basically what I don't understand, is what do to when having

r**s, in other words, when I have to multiply 2 scalars, do I do it using the regular operator, or using the new X*scalar operator

any help would be appreciated...

Re: prove that a set V is a vector space

Quote:

Originally Posted by

**WeeG** Hello,

I am trying to prove that a set V is a vector space, I managed to prove most of the process, but stuck with a couple of things...

The set V is the set of positive real numbers, and the field is R.

the two operations are:

x++y = xy

a**x = x^a

(where ++ is the addition, ** is the scalar multiplication, and a is a real scalar)

I find it hard to prove that for 2 scalars r and s, and a member of V, x

(r++s)**x = (r**x) + (s**x)

basically what I don't understand, is what do to when having

r**s, in other words, when I have to multiply 2 scalars, do I do it using the regular operator, or using the new X*scalar operator

any help would be appreciated...

$\displaystyle (r++s)**x=x^{rs}$

$\displaystyle (r**x)++(s**x)=x^rx^s=x^{r+s}$

$\displaystyle (r++s)**x\neq(r**x)++(s**x)$

So, $\displaystyle V$ is not a vector space.

If $\displaystyle a**x=a^x$, then $\displaystyle V$ would be a vector space. Are you sure you've copied the question correctly?

Re: prove that a set V is a vector space

Quote:

Originally Posted by

**WeeG** I find it hard to prove that for 2 scalars r and s, and a member of V, x (r++s)**x = (r**x) + (s**x)

The equality is not correctly written. Take into account that $\displaystyle ++$ is the sum of vectors (not of scalars). You have to prove $\displaystyle (r+s)^{**}x=r^{**}x++s^{**}x$ . Then,

$\displaystyle (r+s)^{**}x=x^{r+s}=x^rx^s=x^r++x^s=r^{**}x++s^{** }x$

Re: prove that a set V is a vector space

Quote:

Originally Posted by

**alexmahone** So, $\displaystyle V$ is not a vector space.

Look at my previous post.

Re: prove that a set V is a vector space

Quote:

Originally Posted by

**FernandoRevilla** $\displaystyle (r+s)^{**}x=r^{**}x++s^{**}x$

How come you're using 2 different kinds of addition in your equation? The OP's symbol for addition is ++, so why shouldn't the equation be $\displaystyle (r++s)^{**}x=r^{**}x++s^{**}x$?

Re: prove that a set V is a vector space

Quote:

Originally Posted by

**alexmahone** How come you're using 2 different kinds of addition in your equation. The OP's symbol for addition is ++, so why shouldn't the equation be $\displaystyle (r++s)^{**}x=r^{**}x++s^{**}x$?

Let us see, the field is $\displaystyle (\mathbb{R},+,\cdot)$ with the usual operations sum and product. The operation defined on $\displaystyle V$ is $\displaystyle ++$ . So, for two scalars $\displaystyle r,s$ its sum is $\displaystyle r+s$ , not $\displaystyle r++s$ .

Re: prove that a set V is a vector space

thank you both !

let me get is straight, you are saying, that the operators ++ and ** are between a scalar and a member of V (even if V is number too), and when dealing with 2 scalars, I use the "regular" operators ?

sounds logical I suppose

Re: prove that a set V is a vector space

Quote:

Originally Posted by

**WeeG** thank you both !

let me get is straight, you are saying, that the operators ++ and ** are between a scalar and a member of V (even if V is number too), and when dealing with 2 scalars, I use the "regular" operators ?

sounds logical I suppose

++ is between two members of V.

** is between a scalar and a member of V.

Re: prove that a set V is a vector space

yes, and what happen when I have 2 scalars, do I use the "regular" operators ?

Re: prove that a set V is a vector space

Quote:

Originally Posted by

**WeeG** yes, and what happen when I have 2 scalars, do I use the "regular" operators ?

Yes, notice that the problem says *and the field is $\displaystyle \mathbb{R}$* without mentioning "strange operations", so the interpretation must be the usual operations.

Re: prove that a set V is a vector space

Quote:

Originally Posted by

**WeeG** Hello,

I am trying to prove that a set V is a vector space, I managed to prove most of the process, but stuck with a couple of things...

The set V is the set of positive real numbers, and the field is R.

the two operations are:

x++y = xy

a**x = x^a

(where ++ is the addition, ** is the scalar multiplication, and a is a real scalar)

I find it hard to prove that for 2 scalars r and s, and a member of V, x

(r++s)**x = (r**x) + (s**x)

basically what I don't understand, is what do to when having

r**s, in other words, when I have to multiply 2 scalars, do I do it using the regular operator, or using the new X*scalar operator

any help would be appreciated...

How can V as defined above be a vector space? The zero vector is missing from set V...

Have i misunderstood the question?

(Wondering)

Re: prove that a set V is a vector space

Quote:

Originally Posted by

**agentmulder** How can V as defined above be a vector space? The zero vector is missing from set V... Have i misunderstood the question?

For all $\displaystyle x\in V$ we have $\displaystyle x++1=1++x=x$ so, $\displaystyle 1$ is the zero vector of $\displaystyle (V,++)$ .

Re: prove that a set V is a vector space

Quote:

Originally Posted by

**agentmulder** How can V as defined above be a vector space? The zero vector is missing from set V...

Have i misunderstood the question?

(Wondering)

I think the zero vector in this case is $\displaystyle 1$ because $\displaystyle x++1=x$ and $\displaystyle a**1=1$.

Re: prove that a set V is a vector space

Quote:

Originally Posted by

**alexmahone** I think the zero vector in this case is $\displaystyle 1$ because $\displaystyle x++1=x$ and $\displaystyle a**1=1$.

suppose x is 2...then 2++1=2?

oh wait...i see now... x++y=xy so 2++1=(2)(1)=2

Re: prove that a set V is a vector space

let me ask you if my interpretation of a**x = a^x is correct...

suppose 1 plays the role of the 0 vector then a**1 is the zero vector since scalar multiplication on the 0 vector gives the zero vector.

now, 1^a = 1 for real number scaler a and real number 1

since 1 is the 0 vector...it checks out

is this correct?

but somehow the vectors are being mixed up with the scalers... or am i thinking too much?

(Itwasntme)