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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?

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 $$hatast : hat X to hat X, qquad a ,hatast, b = phi^{-1}(phi(a) ast phi(b)) .$$ By construction, $hatast$ 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).