Are These Subsets of R3 Subspaces?

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The discussion centers on determining whether certain subsets of R3, specifically those containing integers and rational numbers, qualify as subspaces. It is concluded that neither subset is a subspace because scalar multiplication by non-integers or irrationals results in vectors outside the subsets. Clarifications are made regarding the nature of scalars, emphasizing that they must come from the field of real numbers, not from R3 itself. The conversation also touches on the distinction between R3 and C3, noting that if complex scalars are used, the space would be classified as C3. Overall, the key takeaway is the importance of scalar fields in defining vector spaces.
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Homework Statement


Which of the following subsets of R3 are subspaces? The set of all vectors of the form (a,b,c) where a, b, and c are...


Homework Equations


1. integers
2. rational numbers

The Attempt at a Solution


I think neither are subspaces. IIRC, the scalar just needs to be from R3 and not, for example, an integer for 1 or a rational number for 2.

So for number 1, I can multiply the integers of vector (a,b,c) by some non-integer k, ending up with (ka,kb,kc) outside the subset, and thus not a subspace.

For number 2, I can multiply the rational numbers of vector (a,b,c) some some irrational number (say, ∏) and end up with (∏a, ∏b, ∏c), all outside the subset and thus not a subspace.

Or am I totally wrong?
 
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MoreDrinks said:
Or am I totally wrong?
No, you are totally correct. The indicated sets are not subspaces of ##\mathbb{R}^3##, for the reasons you stated.
 
MoreDrinks said:
the scalar just needs to be from R3
Correction: the scalars are elements of ##\mathbb{R}##, not ##\mathbb{R}^3##.
 
jbunniii said:
Correction: the scalars are elements of ##\mathbb{R}##, not ##\mathbb{R}^3##.

If we're dealing with complex space, can scalars be complex?

Thanks for the help!
 
MoreDrinks said:
If we're dealing with complex space, can scalars be complex?

Thanks for the help!
They can, but then it wouldn't be ##\mathbb{R}^3## anymore. It would be ##\mathbb{C}^3##.
 
jbunniii said:
They can, but then it wouldn't be ##\mathbb{R}^3## anymore. It would be ##\mathbb{C}^3##.

True, thanks! Would the correct term be that we're working in the "field" of R^3 or just R^3 space when talking about this?
 
MoreDrinks said:
True, thanks! Would the correct term be that we're working in the "field" of R^3 or just R^3 space when talking about this?
To be precise, a vector space consists of an abelian group of vectors and a field of scalars, along with some rules governing the multiplication of a vector by a scalar.

So if we want to be precise, we would say that we are working in the vector space in which the vectors are elements of ##\mathbb{R}^3## and the scalars are elements of ##\mathbb{R}##, with the usual rules of multiplication.

However, for brevity we typically say that we are working in the vector space ##\mathbb{R}^3##, and unless stated otherwise, it is understood that the scalar field is ##\mathbb{R}##.

Similarly, we may say that we are working in the vector space ##\mathbb{C}^3##, where the assumption is that unless stated otherwise, the scalar field is ##\mathbb{C}##.
 
jbunniii said:
To be precise, a vector space consists of an abelian group of vectors and a field of scalars, along with some rules governing the multiplication of a vector by a scalar.

So if we want to be precise, we would say that we are working in the vector space in which the vectors are elements of ##\mathbb{R}^3## and the scalars are elements of ##\mathbb{R}##, with the usual rules of multiplication.

However, for brevity we typically say that we are working in the vector space ##\mathbb{R}^3##, and unless stated otherwise, it is understood that the scalar field is ##\mathbb{R}##.

Similarly, we may say that we are working in the vector space ##\mathbb{C}^3##, where the assumption is that unless stated otherwise, the scalar field is ##\mathbb{C}##.

Thank you, that clears up a lot.
 

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