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?

  • $\begingroup$ What is the field of scalars? Since $Q$ is set of positive reals, so with the normal addition operation $Q^2$ is a not a group, because $-(4,2)=(-4,-2)\notin Q^2$. $\endgroup$ – user598858 Jul 18 at 5:19

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.

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  • $\begingroup$ I could see the scalar multiplication clearly, but I thought both had to be shown and I couldn't see how $1/3 (9, 15)$ works with vector addition. Because with the zero vector you can show the addition but you could also show that $(0^c, 0^c) = (0,0)$ Couldn't you? The vector addition is also confusing because I thought it normally worked like an addition, e. g. $(2, 3) + (3, 4) = (5, 7)$ but the formula would indicate the answer as $(6, 12)$ which is why it looks like $x0$ and $y0 = 0$ initially, but when you have $x0x = x$ it makes it seem like it equals 1. Maybe I'm, just confusing myself. $\endgroup$ – DuncanK3 Jul 20 at 7:09
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    $\begingroup$ You are probably confusing yourself. The addition and scalar multiplication here have nothing to do with the usual addition and scalar multiplication. In particular, the zero element $0$ of the reals does not have to appear in the zero element here. And yes, $x_0=y_0=1$. And $\frac13(9,15)+\frac13(9,15)+\frac13(9,5)=({(9^{\frac13})}^3,{(15^{\frac13})}^3)$. I don't know if this clarifies your confusion though. $\endgroup$ – awllower Jul 20 at 9:16
  • $\begingroup$ Hmm, so I only need to show it works for the formulas provided and not the usual way. So if I was to use the example I made of (2, 3) + (3,4) the answer would be (6, 12)? Is that correct? Or would I still need to show how scalar multiplication works with that example? $\endgroup$ – DuncanK3 Jul 21 at 7:39
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    $\begingroup$ Yes, $(2,3)+(3,4)=(6,12)$ with the provided definition. $\endgroup$ – awllower Jul 21 at 7:41

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