Proof of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space

Throughout, a point of Euclidean space $\mathbb{R}^n$ is an ordered $n$-tuple of real numbers, and two points are equal exactly when their corresponding coordinates are equal; for a point $z$ we write $z_i$ for its $i$-th coordinate, the index $i$ always being a natural number with $1\le i\le n$ in the order on the natural numbers. The real numbers form a field; the eight axioms of that definition are used below without further comment, and $0$, $1$, $-t$ denote its additive identity, multiplicative identity, and the additive inverse of $t$.

Step 1 (The two operations). By the definition of the sum of points and the definition of the scalar multiple, for all points $u,v$ of $\mathbb{R}^n$ and every real number $\lambda$ the tuples $u+v$ and $\lambda v$ are again points of $\mathbb{R}^n$, with

Thus $+$ assigns a point of $\mathbb{R}^n$ to each pair of points of $\mathbb{R}^n$, and scalar multiplication assigns a point of $\mathbb{R}^n$ to each real number and each point of $\mathbb{R}^n$; these are operations of the kind required by the definition of a vector space over the field of real numbers. It remains to verify the eight conditions of that definition, and each is verified coordinatewise from the corresponding field axiom.

Step 2 (The eight conditions). Let $u,v,w$ be points of $\mathbb{R}^n$, let $\lambda,\mu$ be real numbers, and let $i$ satisfy $1\le i\le n$.

Condition 1: $((u+v)+w)_i=(u_i+v_i)+w_i=u_i+(v_i+w_i)=(u+(v+w))_i$, by associativity of addition in the field.

Condition 2: $(u+v)_i=u_i+v_i=v_i+u_i=(v+u)_i$, by commutativity of addition.

Condition 3: let $0_{\mathbb{R}^n}$ be the origin of $\mathbb{R}^n$, so that $(0_{\mathbb{R}^n})_i=0$. Then $(v+0_{\mathbb{R}^n})_i=v_i+0=v_i$, because $0$ is the additive identity. Hence $v+0_{\mathbb{R}^n}=v$ for every point $v$, and condition 3 holds with $0_{\mathbb{R}^n}$ in the role of the zero vector.

Condition 4: given $v$, each coordinate $v_i$ has an additive inverse $-v_i$ in the field, so $w=(-v_1,\dots,-v_n)$ is a point of $\mathbb{R}^n$, and $(v+w)_i=v_i+(-v_i)=0=(0_{\mathbb{R}^n})_i$. Hence $v+w=0_{\mathbb{R}^n}$.

Condition 5: $(\lambda(\mu v))_i=\lambda(\mu v_i)=(\lambda\mu)v_i=((\lambda\mu)v)_i$, by associativity of multiplication.

Condition 6: $(1v)_i=1v_i=v_i$, because $1$ is the multiplicative identity.

Condition 7: $(\lambda(u+v))_i=\lambda(u_i+v_i)=\lambda u_i+\lambda v_i=(\lambda u+\lambda v)_i$, by distributivity.

Condition 8: $((\lambda+\mu)v)_i=(\lambda+\mu)v_i=\lambda v_i+\mu v_i=(\lambda v+\mu v)_i$, again by distributivity.

Since all eight conditions hold, $\mathbb{R}^n$ with these two operations is a real vector space.

Step 3 (Claim 1: the zero vector). That $0_{\mathbb{R}^n}$ is a zero vector was shown under condition 3. Suppose $z$ is also a zero vector, that is, $v+z=v$ for every point $v$ of $\mathbb{R}^n$. Taking $v=0_{\mathbb{R}^n}$ gives $0_{\mathbb{R}^n}+z=0_{\mathbb{R}^n}$, while condition 3 applied to $v=z$ gives $z+0_{\mathbb{R}^n}=z$. By condition 2 the two left-hand sides are equal, so $z=0_{\mathbb{R}^n}$.

Step 4 (Claim 2: the additive inverse). Let $x=(x_1,\dots,x_n)$. By condition 4 the point $(-x_1,\dots,-x_n)$ satisfies $x+(-x_1,\dots,-x_n)=0_{\mathbb{R}^n}$. Conversely, if $w$ is a point of $\mathbb{R}^n$ with $x+w=0_{\mathbb{R}^n}$, then comparing $i$-th coordinates gives $x_i+w_i=0$, whence $w_i=-x_i$ by claim 1 of Additive Cancellation and Elementary Additive Identities in a Field. As this holds for every $i$, we get $w=(-x_1,\dots,-x_n)$, which proves uniqueness. Finally, by the definition of the scalar multiple and the identity $(-s)t=-(st)$ of claim 2 of Zero Products and Elementary Identities in a Field, applied with $s=1$ and $t=x_i$,

the last equality because $1$ is the multiplicative identity. Hence $(-1)x=(-x_1,\dots,-x_n)$.

Step 5 (Claim 3: agreement with the difference). Let $y=(y_1,\dots,y_n)$. By the definition of the difference of points its $i$-th coordinate is $x_i-y_i$, which abbreviates $x_i+(-y_i)$. By Step 4 the $i$-th coordinate of $-y$ is $-y_i$, so by the definition of the sum of points the $i$-th coordinate of $x+(-y)$ is $x_i+(-y_i)$ as well. The two points therefore have the same coordinates, so $x-y=x+(-y)$. $\blacksquare$