Show that $(\mathbb{R}, \mathbb{R}, \oplus , \odot )$ is a vector space

Question

Show that $(\mathbb{R}, \mathbb{R}, \oplus , \odot )$ is a vector space if $\oplus$ and $\odot$ are defined by

for all vectors $x, y \in \mathbb{R}$ and scalars $\alpha \in \mathbb{R}$.

I have to show this, but I don't even know where to start

Answer 1

This is a typical case of transport of structure via bijections. In general, assume you have an arbitrary field $K$, a $K$-vector space $V$ and a permutation $f$ of $V$ (i.e. a bijection from $V$ to itself); via this permutation you can define two new operations, an internal one on $V$ given by $$\oplus: V \times V \to V, x\oplus y=f(f^{-1}(x)+f^{-1}(y))$$ and an external one $$\bullet: K \times V \to V, \lambda \bullet x=f(\lambda f^{-1}(x))$$

Then it is an easy verification that the new structure $(V, \oplus, \bullet)$ is also a $K$-vector space such that $f$ implements an isomorphism between the original structure on $V$ and the new one; this new structure is said to be transported from the old one via $f$.

In your particular case the original structure is that of $\mathbb{R}$ as a left vector space over itself and the permutation considered is $$f: \mathbb{R} \to \mathbb{R}, f(x)=\sqrt[5]{x+7}$$