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

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

• What's the definition of a vector space? – mizh Dec 16 '19 at 11:42
• For example, one of the properties of a vector space is that $(x \oplus y) \oplus z = x \oplus (y \oplus z)$. Using these definitions, can you verify that that is true? – Mees de Vries Dec 16 '19 at 12:03

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 $$f: \mathbb{R} \to \mathbb{R}, f(x)=\sqrt[5]{x+7}$$