Proof of ℝⁿ Is a Real Vector Space

Routine facts about negatives in a commutative ring are used without comment (Toolkit for AL2). By Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (iv) and Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (v), the sum, the scalar multiple, the negative and the origin of $\mathbb{R}^{n}$ are the pointwise ones on $\mathbb{R}^{I}$ with $I=[n]$. So the addition is a binary operation on $\mathbb{R}^{n}$ and the scalar multiplication a function from $\mathbb{R}\times\mathbb{R}^{n}$ to $\mathbb{R}^{n}$ (Pointwise Operations on Real-Valued Maps (i)), and the rules of Rules for Pointwise Operations on Real-Valued Maps: a Commutative Ring, Scalar Multiples and Evaluation apply to them. Let $x,y\in\mathbb{R}^{n}$ and $\lambda,\mu\in\mathbb{R}$ be arbitrary.

We begin with the conditions on the addition. By Rules for Pointwise Operations on Real-Valued Maps: a Commutative Ring, Scalar Multiples and Evaluation (i), $\mathbb{R}^{n}$ is a commutative ring under the pointwise operations, with zero $0$, and the negative of $x$ in this ring is $-x$. In particular, by Commutative Rings (i), the addition is associative and commutative and has $0$ as a right identity, which gives Vector Space over a Field (i) and Vector Space over a Field (ii), and $x+0=x$ for every $x$, so the origin is a vector $z$ as required in Vector Space over a Field (iii). It is the only one: if $z\in\mathbb{R}^{n}$ satisfies $x+z=x$ for every $x\in\mathbb{R}^{n}$, then

$$z=z+0=0+z=0.$$

Hence the negatives condition Vector Space over a Field (iv) has only to be checked for $z=0$, and there $y=-x$ serves, because $x+(-x)=0$ holds in the ring.

The conditions on the scalar multiplication are the scalar laws of Rules for Pointwise Operations on Real-Valued Maps: a Commutative Ring, Scalar Multiples and Evaluation (iii):

$$\lambda(\mu x)=(\lambda\mu)x,\qquad1x=x,\qquad\lambda(x+y)=\lambda x+\lambda y,\qquad(\lambda+\mu)x=\lambda x+\mu x,$$

which are, in this order, Vector Space over a Field (v), Vector Space over a Field (vi), Vector Space over a Field (vii) and Vector Space over a Field (viii).

All eight conditions of Vector Space over a Field (ix) therefore hold, and $\mathbb{R}^{n}$ is a vector space over $\mathbb{R}$ with its operations and constants, that is, a real vector space (Vector Space over a Field (x)). We saw that the origin is its only vector $z$ with $x+z=x$ for every $x$, so it is the zero vector (The Zero Vector, Negatives and Differences in a Vector Space (i)); and since $x+(-x)=0$, the negative of $x$ is $-x$ (The Zero Vector, Negatives and Differences in a Vector Space (ii)). This proves (i).