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

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

$$\alpha \odot x = \alpha^7 (x-3) + 3$$

$$x \oplus y = (\sqrt[7]{x-3} + \sqrt[7]{y-3})^7+3$$

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

this is the question. The way it looks to me x is defined as $$\sqrt[7]{x-3}$$

and just by adding you add the 7th power and the 3. I dont get what the 3 does. I dont need help with the hole question i just want help with one thing, and i hope that i will be able to do the rest then.

Most trouble i have with axiom 4.

A4 - for each $$x \in V$$ there exists an element $$-x$$ in V such that $$x \oplus (-x) = 0$$

I get this which is wrong because i get to 3 which should be 0. $$x \oplus (-x) = (\sqrt[7]{x-3} + (-\sqrt[7]{x-3}))^7+3$$

$$=0^7 +3$$

$$= 3 \neq 0$$

where do i go wrong?

• The negative of $x$ may not be $-x$ it might be another real number in this case. – Yanko Dec 9 '18 at 14:06
• Before that, you need to find the zero element, which again might not be the original zero. – Yanko Dec 9 '18 at 14:07

The confusion here is that in the statement of axiom (here called A4):

• the symbol $$0$$ refers to the additive identity of $$\oplus$$, whereas in our particular example we've already used the symbol $$0$$ for the additive identity of the usual operation +, and
• the symbol $$-$$ in $$-x$$ refers again additive structure defined by $$\oplus$$ whereas for this particular example we've already used the symbol $$-$$ to denote the usual negation operation on $$\Bbb R$$.

You might find it helpful, then, to write the relevant axiom using separate symbols as follows:

For every $$x \in V$$ there is an element $$\ominus x \in V$$ such that $$x \oplus (\ominus x) = \hat 0$$ (where $$\hat 0$$ is the additive identity of $$\oplus$$).

On the other hand, we can use the definition of $$\hat 0$$ and $$\oplus$$ to produce explicit expressions for $$\hat 0$$ additive inverse $$\ominus x$$ in terms of $$x$$ and the usual operations on $$\Bbb R$$. By definition, we need: $$x \oplus \hat 0 = x$$ for all $$x \in V$$ and $$\hat 0 = x \oplus (\ominus x) = \left(\sqrt[7]{x - 3} + \sqrt[7]{\ominus x - 3}\right)^7 + 3 .$$ Can you find expressions for $$\hat 0$$ and $$\ominus x$$?

Remark There's a more general story here, by the way. If we have any (say, for concreteness, binary) operation $$\ast: X \times X \to X$$ and a set bijection $$\phi : \hat X \to X$$, we can use the bijection to transfer the operation from $$X$$ to $$X'$$, by defining the new operation $$\hat\ast : \hat X \to \hat X, \qquad a \,\hat\ast\, b = \phi^{-1}(\phi(a) \ast \phi(b)) .$$ By construction, $$\hat\ast$$ inherits the properties of $$\ast$$.

For example, if $$(X, 0, +)$$ satisfies the above axiom, then for all $$x \in X$$ there is an element $$-x \in X$$ such that $$x + (-x) = 0$$. Then, for any bijection $$\phi: X \to \hat X$$, we have that $$\hat 0 := \phi(0) = \phi(x + (-x)) = \phi(x) \,\hat +\, \phi(-x)$$, so $$(\hat X, \hat 0, \hat+)$$ also satisies that axiom, as we can take $$\hat- y := \phi(-\phi^{-1}(y))$$. As you basically observed in your question (in which our setting is $$X = \hat X = \Bbb R$$), we can take $$\phi(x) := \sqrt[7]{x - 3}$$. Since this is a bijection and (as computing directly shows) $$a \oplus b = a \,\hat +\, b$$, we see that $$(\Bbb R, \Bbb R, \odot, \oplus)$$ satisfies the above axiom without actually computing $$\ominus x$$ as a function of $$x$$ (though it also gives us another way to do so).