# Vector addition and scalar multiplication

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?

• 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 $-(4,2)=(-4,-2)\notin Q^2$. – user598858 Jul 18 at 5:19

## 1 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.

• 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. – DuncanK3 Jul 20 at 7:09
• 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. – awllower Jul 20 at 9:16
• 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? – DuncanK3 Jul 21 at 7:39
• Yes, $(2,3)+(3,4)=(6,12)$ with the provided definition. – awllower Jul 21 at 7:41