Proof of one of the properties of Real Coordinate Vector Spaces

[jwqwerty]

1. Homework Statement

Prove that there is an additive identity 0∈R^n: For all v∈R^n, v+0=v2. Homework Equations

Axiom of Real Numbers:
There is an additive identity 0∈R : For all a∈R, a+0=a and o+a=a

3. The Attempt at a Solution

Solution 1 (My own attempt)
: Let v=(v1, v2, v3... vn). Then by axiom of R (as stated above), for every vi (i=1, 2..., n) there exists 0∈R such that vi+0=vi. Thus, there exists 0=(0,0,...0)∈R^n.

Solution 2 (Professor's Attempt)
: Let 0=(0,0,...0)∈R^n
then v+0=(v1, v2 ... vn) + (0,0...0)
=(v1+0, v2+0, ... vn+0)
=(v1, v2 ... vn) - by axiom of R (as stated above)
=v

However, I think solution 2 has a problem that it assumed existence of 0 when the question is asking to prove its existence. Or is my thought wrong? And also, is there anything wrong with solution 1?

 

[Office_Shredder]

The professor wrote down a vector (0,0,...,0) which he then called the zero vector. He proved that this vector satisfied the required property, he didn't assume that it satisfied the property required.Your attempt ends too early in my opinion. You constructed a vector which you call 0, but at no point do you actually write down the line v+0=v which is what you need to prove happens (it only really requires one additional line to finish it)

Your confusion seems to stem from notation. 0 is often used to refer to the vector which is the additive identity. However, in both your proof and the professor's 0 is referring to a vector of all zeros - this is not a priori the additive identity. It's just a name being given to a vector, and you have to prove it satisfies the additive identity property

 

[HallsofIvy]

Your professor is using the fact that, from the definition of Rⁿ, Rⁿ contains (0, 0, ..., 0). He then proves that it is the 0 vector.

 

[jwqwerty]

The professor wrote down a vector (0,0,...,0) which he then called the zero vector. He proved that this vector satisfied the required property, he didn't assume that it satisfied the property required.

Well, but the question is asking to prove its existence. Then, I think we can't just write down a vector (0,0...,0) (zero vector) because we do not know its existene. Please correct me if I am wrong.

(it only really requires one additional line to finish it)

Also, I do not get what you mean by this.

 

[LCKurtz]

Well, but the question is asking to prove its existence. Then, I think we can't just write down a vector (0,0...,0) (zero vector) because we do not know its existence. Please correct me if I am wrong.

You are wrong. $R^n$ is the set of all n-tuples of real numbers, and that includes $\vec 0 = (0,0,0,...,0)$. You just have to prove the vector I denoted $\vec 0$ has the additive identity property.

 

[jwqwerty]

You are wrong. $R^n$ is the set of all n-tuples of real numbers, and that includes $\vec 0 = (0,0,0,...,0)$. You just have to prove the vector I denoted $\vec 0$ has the additive identity property.

Ok i understand the second solution. But why is my solution wrong? I think the first solution has no logical flaws.

 

[LCKurtz]

Solution 1 (My own attempt)
: Let v=(v1, v2, v3... vn). Then by axiom of R (as stated above), for every vi (i=1, 2..., n) there exists 0∈R such that vi+0=vi. Thus, there exists 0=(0,0,...0)∈R^n.

Ok i understand the second solution. But why is my solution wrong? I think the first solution has no logical flaws.

Your argument is incomplete, that's all. Your statement that there exists $(0,0,0...,0)\in R^n$ doesn't need proof. The fact that you named that vector 0 doesn't make it the additive identity. You have all you need but you just need to write down the steps for$$
\vec 0 + \vec v = \hbox{use the definition of vector addition and your first two sentences here to get }=\vec v$$ and its commutative cousin.