I have an exercise in my Exam that says: Prove that the set V=R (doesn't specify real positive numbers, only Real) is a vector space in the Real Field given the operations:

$\ x+y = \sqrt[3]{x^3+y^3}$

$ c * x = \sqrt[3]{c*x}$

I went right ahead and tried to prove the closure under scalar multiplication, if i choose

$ c=1/2 $ and $ x = -1 $

given that doesn't specify to use Real Positive, only Real

i get

$ \sqrt[3]{-\frac 12 }$ which gives 3 roots, from which 1 is real (-0.7937) and the other 2 are complex (0.3969+0.6874i) and the 3rd same as second.

Given that there are 2 complex roots, does that mean i can't say that this is a vector space ? because it's not a real number.

Am i right ? Did i miss something ?

Thank you

  • $\begingroup$ The third root of $-\frac 12$ is not the same as the second-it has a negative real part, so is the complex conjugate. That is not germane to the problem. You should use different symbols for the operations in your "field" and the usual operations in the reals. Your definitions use the usual real operations on the right. It is common to use \oplus, giving $\oplus$ and \otimes giving $\otimes$ for the ones in your field. $\endgroup$ – Ross Millikan Jun 2 at 2:42
  • $\begingroup$ Yes indeed, it has (-0.7937) as real negative part. So do i consider (-0.7937) as real value and ignore the other 2 complex parts ? Do i go on with other axioms then ? Yes the operations are as follows : $\ x \oplus y = \sqrt[3]{x^3+y^3}$ and $ c \otimes x = \sqrt[3]{c*x}$ $\endgroup$ – AndrewM Jun 2 at 2:45
  • $\begingroup$ Yes, you should use only the real root here. I gave a little more in my answer. $\endgroup$ – Ross Millikan Jun 2 at 2:48

Everything in the problem is about the reals, so you should use the real cube root for the multiplication. Every real number has a unique real cube root. Closure is easy because it is inherited from the reals. Commutativity is also easy because of the symmetry of the definitions. Inverses, associativitiy and distributivity are the ones that take some work.

  • $\begingroup$ " Commutativity is also easy because of the symmetry of the definitions. " You mean because i can invert x and y under the cube root ? $ \sqrt[3]{x^3+y^3} = \sqrt[3]{y^3+x^3} = y \oplus x $ $\endgroup$ – AndrewM Jun 2 at 2:51
  • $\begingroup$ Yes, and $c$ and $x$ in the multiplication definition. $\endgroup$ – Ross Millikan Jun 2 at 2:54
  • $\begingroup$ I tried with associativity but things got out of hand, am i on the right path ? $ x \oplus (y \oplus z) = (x \oplus y) \oplus z \\$ => $ x \oplus\sqrt[3]{y^3 + z^3}$ => $ \sqrt[3]{x^3 + \sqrt[3]{y^3+z^3}} $ $\endgroup$ – AndrewM Jun 2 at 3:04
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    $\begingroup$ When you write it that way with implications you are assuming what you want to prove. That is why I want you to start with $x\oplus (y \oplus z)$ and use equalities to get to $(x \oplus y) \oplus z$, but you have the idea. You should not have an $\oplus$ under the radical. Everything under the radical uses the usual addition. $\endgroup$ – Ross Millikan Jun 2 at 3:16
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    $\begingroup$ The last does not render, probably because you have mismatched braces somewhere. It is better to edit these into your question or make an answer. The site will then render it or not as you type. Yes, $0$ is the additive identity and $-x$ is the additive inverse of $x$ $\endgroup$ – Ross Millikan Jun 2 at 4:46

Distribution :

$ c \otimes x = \sqrt[3]{c * x} = \\ c \otimes (x \oplus y) =\\ (c \otimes x) \oplus (c \otimes y) = \\ \sqrt[3]{c*x} \oplus \sqrt[3]{c*y} = \\ \sqrt[3]{(\sqrt[3]{c*x})^3 + (\sqrt[3]{c*y})^3} = \\ \sqrt[3]{c*x + c*y} $

$ c \otimes x = \sqrt[3]{c * x} =\\ c \otimes (x + y) = \\ \sqrt[3]{c*(x+y)} = \\ \sqrt[3]{c*x + c*y} $

$ c \otimes (x \oplus y) = (c \otimes x) \oplus (c \otimes y) $ => V Is a vector space

  • $\begingroup$ Your two approaches make a proof of distributivitiy. You have computed $c \otimes (x \oplus y)$ and $(c \otimes x) \oplus (c \otimes y)$ and gotten the same result. The proper way to present it is to start with one of these, use equal signs between the expressions, get to the end, then go to the end of the other one and go backwards to the beginning of that one. You now have the proof you seek. Implies means nothing between expressions. It means something between sentences, so you can write $a=b \implies ac=bc$ When you do a derivation you have a chain of equalities. $\endgroup$ – Ross Millikan Jun 2 at 5:00
  • $\begingroup$ I edited a bit now, how does it look ? $\endgroup$ – AndrewM Jun 2 at 5:10

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