Is $(a,b)\oplus (c,d)=(a+c+1,b+d)$ and $\lambda \otimes (a,b)=(\lambda a+\lambda-1,\lambda b)$ a vectorspace?

Question

I have this question:

On a set $V=\{(a,b)|a,b\in \mathbb{R}\}$ we defined the following binary operations:

For every $a,b,c,d,\lambda\in\mathbb{R}$:

Addition: $(a,b)\oplus (c,d)=(a+c+1,b+d)$

Multiplication: $\lambda \otimes (a,b)=(\lambda a+\lambda-1,\lambda b)$

Is V a vector space?

My solution: We'll check if all of the properties of a vector space exists.

We'll start with the addition.

1. Closure: $(a,b)+(c,d)=(a+c+1,b+d)\in V$.
2. Associativity: $\left(\left(a,b\right)+\left(c,d\right)\right)+\left(e,f\right)=\left(a+c+1,b+d\right)+\left(e,f\right)=\left(a+c+e+2,b+d+f\right)$ and $\left(a,b\right)+\left(\left(c,d\right)+\left(e,f\right)\right)=\left(a,b\right)+\left(c+e+1,d+f\right)=\left(a+c+e+2,b+d+f\right)$.
3. Commutativity: $\left(a,b\right)+\left(c,d\right)=\left(a+c+1,b+d\right)$ and $ \left(c,d\right)+\left(a,b\right)=\left(a+c+1,b+d\right)$.
4. Neutral Element: $\left(a,b\right)+(-1,0)=\left(a,b\right)$
5. Negative Element: $\left(a,b\right)+\left(-\left(a,b\right)\right)=(-1,0)$

Multiplication:

1. Closure: $ \lambda\left(a,b\right)=\left(\lambda a+\lambda-1,\lambda b\right)\in V$
2. Distributivity over addition on $\mathbb{V}$: $\lambda\left(\left(a,b\right)+\left(c,d\right)\right)=\left(\lambda a+\lambda-1,\lambda b\right)+\left(\lambda c+\lambda-1,\lambda d\right)=\left(\lambda a+2\lambda+\lambda c-1,\lambda b+\lambda d\right)$ and $\lambda\left(a,b\right)+\lambda\left(c,d\right)=\left(\lambda a+\lambda-1,\lambda b\right)+\left(\lambda c+\lambda-1,\lambda d\right)=\left(\lambda a+2\lambda+\lambda c-1,\lambda b+\lambda d\right)$.
3. Distributivity over addition on $\mathbb{R}$: $\left(\lambda+\alpha\right)\left(a,b\right)=\left(a\left(\lambda+\alpha\right)+\left(\lambda+\alpha\right)-1,b\left(\lambda+\alpha\right)\right)=\left(\lambda a+\alpha a+\lambda+\alpha-1,\lambda b+\alpha b\right)$ and $\lambda(a,b)+\alpha(a,b)=(\lambda a+\lambda-1,\ \lambda b)+(\alpha a+\alpha-1,\alpha b)=(\lambda a+\alpha a+\lambda+\alpha-1,\lambda b+\alpha b)$.
4. Associativity: $\left(\lambda\alpha\right)\left(a,b\right)=(\lambda\alpha a+\lambda\alpha a-1,\alpha ab)$ and $\lambda(\alpha(a,b))=\lambda(\alpha a+\alpha-1,\alpha b)=(\lambda(\alpha a+\alpha-1),\lambda\alpha b)=(\lambda\alpha a+\alpha+\lambda-2,\lambda\alpha b)$.
5. Neutral Element: $1\ast\left(a,b\right)=\left(a,b\right)$.

I'm unsure about my calculation in 4 (multiplication), I feel like I did something wrong with the associativity there. If it was right, then it is a vector space.

Thank you for your help!

Answer 1

So we know that $(a.b) \oplus (c,d) := (a+c+1, b+d)$ and $\lambda \otimes(a,b) := (\lambda a + \lambda - 1, \lambda b) $.

So we want to look if $(\lambda\alpha)\otimes(a,b) = \lambda\otimes(\alpha(\otimes(a,b))$ for all $\lambda,\alpha\in \mathbb{R}$ and $(a,b)\in V$.

But let us notice that $$\lambda\otimes(\alpha(\otimes(a,b)) = \lambda\otimes (\alpha a + \alpha - 1, \alpha b) = (\lambda(\alpha a + \alpha - 1) + \lambda - 1, \lambda \alpha b) $$ Simplifying the righthandside of the latter gives us: $$\lambda\otimes(\alpha(\otimes(a,b)) = (\lambda\alpha a + \lambda \alpha - 1, \lambda \alpha b)$$ But this is exactly equal to $$(\lambda\alpha a + \lambda\alpha - 1, \lambda\alpha b) =(\lambda\alpha)\otimes(a,b) $$ So the fourth axiom has been satisfied.