First, these were homework questions. However, having already submitted them, asking this doesn't break any code.

This is a new definition of vector addition in $\mathbb{R}^3$:

$$(x_1, y_1, z_1) +(x_2, y_2, z_2)=(x_1 + x_2 + 6, y_1 + y_2 + 6, z_1 + z_2 + 6),$$

scalar multiplication is the same:

$$c(x, y, z)= (cx, cy, cz).$$

My goal is to know if $\mathbb{R}^3$ is a vector space or not.

There are 10 definitions of a vector space, I am struggling with additive identity and additive inverse: the additive identity gives me the 0 vector which in this case is $(-6,-6,-6)$ since: for $U+0=U$, I need the zero vector to be $(-6,-6,-6)$.

Now that I have defined the 0 vector I can add the inverse to get 0:

$$ \begin{align} (x_1, y_1, z_1)+(-(x_1, y_1, z_1)) & = (-6,-6,-6), \\ (x_1 - x_1 + 6, y_1 - y_1 + 6, z_1 - z_1 + 6) & = (-6,-6,-6), \end{align} $$

so I have $(6,6,6) =(-6,-6,-6)$ which is false. Yet according to the solutions this should be a vector space.

My other issue is a different question in which addition and scalar multiplication are redefined like so:

$$ \begin{align} x + y & = xy, \quad \text{Addition} \\ cx & = x^c, \quad \text{Scalar multiplication} \end{align} $$

$V$ is the set of all positive real numbers and again I was expected determine if it is a vector space.

Additive identity:

This time 0 will be 1 so that $x \cdot 1 = x$.

Additive inverse:

$x+(-x) = 1$ which is translated to $x(-x) = 1$,

so I am left with $-x^2 =1$.

My final issue with this problem is also the associative property which states that ($c$ and $d$ and $z$ are real numbers)

$$c(dx) = cd(x),$$

if I define $z = c \cdot d$, this can be translated with the new rules to:

$$x^{d^{c}} = x^{z},$$

which is false, although the solutions specify it should be a vector space.

So my hope is that somebody can point out what I am missing.

Thank you very much,


1 Answer 1


First part: You have to identify that additive inverse of $(x,y,z)$ is $(-x-12,-y-12,-z-12)$.

2nd part: The additive inverse should actually be $\dfrac{1}{x}$.




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