Is the set of integers a vector space?

My friend says it's not, because he thinks vectors have to have have at least two elements, otherwise they're scalars. I say that (Z, +, *) satisfies the axioms of a vector space and that's all you need.

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1D vectors are still vectors.

Your friend should become an hero.

Vector spaces exist "over" fields; specifically, for every vector space V there is a field F such that given any vector v in V and any element c of F, c*v is in V. This is not true of the integers.

(Simpler explanation: Your friend is correct for incorrect reasons)

The integers are a module though, which is similar.

integers mod a prime though would be fine!

Z is not a field; it has non-zero elements without multiplicative inverses in Z.

A vector only needs a magnitude and direction. A scalar has only a magnitude. I'd think you could argue that any vector is made up of one dimensional components which themselves have to be vectors in order to maintain a relationship between themselves and the others.

"A vector only needs a magnitude and direction." Needs to follow parellelagram addition law too. Direction could be indicated by + or -.

>>7

Wah waaah.

>>7
A vector needs neither magnitude nor direction.
{0} is a vector space under the usual operations + and *.

>>10
Nice one (though it still has magnitude: 0).  PWND

>>11
lol - touche
0 is definitely a magnitude.

The trivial vector space is still a good reminder that the typical high school teacher definition of vectors as "magnitude and direction" is a gimmick to avoid formalism.

>>12
BUT IT DOESNT HAVE A FUCKING DIRECTION

and...?

Hint: 10 and 12 are both the same person ;)

>>11 >>12
Fail, >>10 defined no norm. {0} need not be a normed vector space, so the concept of magnitude is not intrinsically defined over its elements. >>10 is correct; it does not have magnitude.

>>1
As everyone has been saying, your friend is correct for incorrect reasons. First of all "elements" is not the term you're looking for; the number of elements is literally the total number of unique vectors in the space, so it's always either zero, one, or infinity (with some cardinality). You're thinking of the term "dimensions", and your friend is wrong when he says you need at least two.

The set of integers is not a vector space because it's not a field. The set of real numbers, however, is a field, so it is a vector space (even though it has only one dimension).

>>4
>>6
>>15

Okay, so if I'm understanding correctly: Z is not a vector space because it's not over a field, and that's because not all (in fact, only two) of its elements have multiplicative inverses. Right?

>>15
Fine, it's true that >>10 didn't define a norm.  Nevertheless, most obvious definitions of norm for vector spaces would give it a value of 0.

>>16
Your general understanding is correct.  However, "not all (in fact, only two) of its elements have multiplicative inverses" is wrong.  The only element of Z with a multiplicative inverse in Z is 1 (1/2 is the multiplicative inverse of 2, but it is definitely not in Z).

also -1, dumbass.

>>15
Fail. A norm is implicitly defined with the scalar multiplication.    If you actually need the definition of the norm, here it is: ||0||=0.  Feel better?  (Just like 17 said.)

{0} is commonly known as "the trivial vector space".  It is a subspace of every vector space, and it IS a vector space.

>>16
Consider the vector space R^2 over the field R.  Does (1,0) have a multiplicative inverse? 

Here is an outline of a proof that Z is a not a vector space over any field. Suppose that there is a field F such that Z is a vector space over F.

Giving a vector space V over a field F is the same as giving a group V and a ring homomorphism F -> End(V), where End(V) is the ring of Endomorphisms of V.  In terms of Z, we need to determine End(Z).

Z is the free group on 1 generator, so any morphism Z -> Z is determined by the image of 1.  So it is not hard to show that End(Z) = Z as a ring.

This leads us to determining ring homomorphisms F -> Z.  Every homomorphism from a field is injective (F only has one ideal, (0)), so (F,+) is a subgroup of (Z,+).  But (Z,+) is a free group, so any subgroup is also free. Hence (F,+) = (Z,+) (note F is nontrivial), and the multiplication must also agree.  Hence F is isomorphic to Z, which is a contradiction, as F is a field.

>>19
>Fail. A norm is implicitly defined with the scalar multiplication.    If you actually need the definition of the norm, here it is: ||0||=0.  Feel better?  (Just like 17 said.)

Fail. What's the norm of the vector space over 2x2 real matrices? It's not simply implied by the scalar multiplication of the field.

>>17
The "obvious" norm does not cut it when you're calling it a vector space. If you called it "the normed vector space {0}", then you could use the usual norm without having to define it.

I know I'm nitpicking, but this shit is important. I've failed lots of tests because I forgot to write stupid shit like "it is normed" just to satisfy a bunch of axioms.

>>20
Dude, that's way, way overboard. Here's a proof: "Z is not closed under scalar multiplication with R".

Here's a counterexample: You need to be able to multiply any vector (say, 2) by any real number you want (say, pi), and get back some value in the vector space. 2pi is not in Z.

>>21
He was referring specifically to the {0} vector space.

As for your claim about >>20 being overboard; your "proof" implicitly assumes that Z must be a vector space over R if it is a vector space at all, which is not the case. >>20 proved that Z cannot be a vector space over ANY field.

Also, a simpler proof than >>20 gives:

Every field either has prime characteristic, or characteristic 0. Z cannot be a vector space over a field of characteristic p (p is prime), because that gives 0 = p*1z = (1f + 1f + ... + 1f)*1z = 1f*1z + 1f*1z + ... + 1f*1z = 1z + 1z + ... + 1z = p. (Here 1f denotes the multiplicative identity of the field, and 1z denotes 1 in the integers; 1f + 1f + ... + 1f has 1f occurring p times.) So we conclude that if Z is a vector space over a field F, F must have characteristic 0. However, if F has characteristic 0, let 2f = 1f + 1f. Since F is a field and 2f is necessarily non-zero, there exists a multiplicative inverse of 2f, we will call that (1/2)f. 2f * (1/2)f * 1z = 1z, and 2f * (1/2)f * 1z = 1f*(1/2)f*1z + 1f*(1/2)f*1z = (1/2)f*1z + (1/2)f*1z. So we have an element x = (1/2)f*1z of Z satisfying x + x = 1. This is, of course, not possible.

ASSUME ITS A FIELD
THEN THE VECTOR SPACE IS A CONTRADICTION
BY THE PRINCIPLE OF MATHEMATICAL INDUCTION WE HAVE COMPLETED OUR PROOF

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>>23
Oh, elements!  That makes the proof a lot simpler.  Too much abelian categories has made me stupid. lol

>>18
oops, thx for the correction

>>20
(1,0) is an element of a vector space.  Vector spaces do not need an inverse operation (required for a field).

>>21
I'm sorry to hear that you've had lots of angry math professors.  At which educational facility did you fail said tests?

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