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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?

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

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