So I have this question I'm stuck at at, and it's where I have to prove that some set is not a vector space. First part of the question, I had to prove for 2 different sets with different operations that they're vector spaces, already did that. Now this following set is like this:

Given is set R with the normal addition rules. The Integers (Z) is the Scalar-field and the multiplication of integers and real numbers is defined as the scalar multiplication.

I need to prove that the above is not a vector space. So if I understood correctly, I'm given (R, +) where + is just normal addition so:

r1 + r1 := r1 + r2 for all r1, r1 ∈ R and for the multiplication, it's basically and interger * real number. z * r := z * r for all z ∈ Z and r ∈ R

I've checked all of the axioms for Vector spaces, and they're all holding out... Not sure what I'm missing...

For "+", it's associative, commutative, there is a neutral and inverse element, and group closure is there. There is a multiplicative neutral element as well, e = 1 and "*" and "+" are distributive and associative... And again, there is group closure since an integer times a real number will always give me something back in R.

Any help would be appreciated... I'm still a newbie in Linear Algebra, and I've been working on this for hours now...

  • $\begingroup$ Do the integers form a field? $\endgroup$ – Aweygan Jan 9 '17 at 1:58
  • $\begingroup$ Are the elements of your vectors in $\mathbb{Z}$ or $\mathbb{R}$? $\endgroup$ – Michael McGovern Jan 9 '17 at 2:00
  • $\begingroup$ What I wrote above is exactly what was written on the question, it was really badly formed, but I'm certain that the vectors are in R and yes the Integers form a field... I had to translate the question. Maybe I mistranslated something, but literally translated it's like this: Show that the following sets with their defined operations do not form vector space: R with the usual addition, Z as scalar field and the multiplication of integers with real numbers as scalar multiplication. $\endgroup$ – Rudy Ailabouni Jan 9 '17 at 2:04
  • $\begingroup$ The integers are not a field (they have no multiplicative inverses). $\endgroup$ – lulu Jan 9 '17 at 2:10
  • $\begingroup$ Oh right! So in order for this to be valid, R has to be an Abelian group and the Z has to form field, but in this case it doesn't? I was paying too much attention to R, and just took it for granted that Z formed a field since the question told me so... $\endgroup$ – Rudy Ailabouni Jan 9 '17 at 2:12

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