If $Q$ is the set of positive real numbers.
$Q^2 = \{(x,y)\mid x, y \in Q\} $
can be shown with operations of vector addition and scalar multiplication using the formulas
$(x_1, y_1) + (x_2, y_2) = (x_1x_2, y_1y_2)$ and $ c(x, y) = (x^c, y^c)$ where $c$, a real number, is a vector space.
Find the following vectors in $Q^2$ : the negative of $(4, 2)$, the vector $c(x,y)$ where $c= 1/3$ and $(x, y) = (9, 15)$ and the zero vector.
Now I assume the question is asking to show that vector addition and scalar multiplication work for all three of the things that need to be found. I can see this works for the zero vector if we let the components of $x$ and $y$ equal to $0$ then both scalar multiplication and addition would produce the zero vector.
I know the negative of $(4,2)$ is $ -(4, 2)$ and $ \dfrac13(12, 18) = (4, 6)$ but I can't see how both the formulas for vector addition and scalar multiplication work for them. Am I missing something?
Best Answer
Rough hints:
The zero vector is a vector $(x_0, y_0)$ such that for all $(x,y)\in Q^2$, we have $(x_0,y_0)+(x,y)=(x,y)$. So this means $x_0x=x$ and $y_0y=y$. What do you think these $x_0, y_0$ would be?
Now the negative of a vector $(x,y)$ is a vector $(x',y')$ such that $(x,y)+(x',y')=(x_0,y_0)$. This implies $xx'=x_0$ and $yy'=y_0$. Given the values of $x_0$ and $y_0$ from the previous paragraph, what can you conclude about $x'$ and $y'$.
And the definition of scalar multiplication is clear enough: $\frac13(9,15)=(9^{\frac13},15^{\frac13})$.
If you are stuck somewhere, feel free to ask for more details.
Hope this helps.