Proving a property of vector addition

Question

Following up on this question of mine, I have constructed an addition of two vectors,. I have defined the product of a vector by a scalar and proved all usual properties of these operations, and colinearity except for the following

I have yet to prove

$\forall a \in \mathbb{R} \forall \vec{u}, \vec{v}, a(\vec{u}+\vec{v}) =a\vec{u}+a \vec{v} $

I thought about using the intercept theorem but I have yet to find an outline for the proof

Any help would be much appreciated

T. D

Answer 1

I will show you the proof of this property in elementary (planar) geometry using my own definitions.

The call any ordered pair of points $(a,b)$ a vector and denote it by $\overrightarrow{ab}$. Next we introduce a two-argument relation $\sim$ between vectors: We say that $\overrightarrow{ab}\sim\overrightarrow{cd}$ whenever

1. $a=b$ and $c=d$
2. $a\neq b,c\neq d, ab\equiv cd$ and
   1. $L(ab)=L(cd)$ and $H(ab),H(cd)$ have the same direction
   2. $L(ab)\cap L(cd)=\emptyset$ and $H(ab),H(cd)$ lie on the same side of $L(ac)$

where $L(pq)$ is a line going through $p,q$ and $H(pq)$ is a halfline with origin $p$ going through $q$.

One can prove that $\sim$ is an equivalence relation. Equivalence classes are called free vectors. Next we say that a (free) vector $\overrightarrow{w}$ is a sum of vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ whenever there exist points $p,q,r$ such that $\overrightarrow{pq}\in\overrightarrow{v}, \overrightarrow{qr}\in\overrightarrow{u},\overrightarrow{pr}\in\overrightarrow{w}$.

And given a real number $a$ we say that a (free) vector $\overrightarrow{u}$ is a product of $a$ and a vector $\overrightarrow{v}$ whenever

1. ($a=0$ or $\overrightarrow{v}=\overrightarrow{0}$) and $u=\overrightarrow{0}$
2. $a>0, \overrightarrow{v}\neq\overrightarrow{0}$ and there exist points $o\neq p,q$ such that $\overrightarrow{op}\in \overrightarrow{v},\overrightarrow{oq}\in \overrightarrow{u}$, $|oq|=a\cdot |op|$ and $p,q$ lie on the same halfline with origin $o$.
3. $a<0, \overrightarrow{v}\neq\overrightarrow{0}$ and there exist points $o\neq p,q$ such that $\overrightarrow{op}\in \overrightarrow{v},\overrightarrow{oq}\in \overrightarrow{u}$, $|oq|=-a\cdot |op|$ and $p,q$ lie on complementary halflines with origin $o$.

One can prove that for any two vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ there exists exactly one vector $\overrightarrow{w}$ which is the sum. Therefore we can denote the sum by $\overrightarrow{v}+\overrightarrow{u}$.

Similarly for any real number $a$ and vector $\overrightarrow{v}$ there exists exactly one vector which is the product. Therefore we can denote it by $a\cdot\overrightarrow{v}$.

Now we can take on the theorem you want to prove.

Lemma. For any point $o$ and a free vector $\overrightarrow{v}$ there exists exactly one point $p$ such that $\overrightarrow{op}\in\overrightarrow{v}$.

I will assume that all other axioms of vector space hold for $+$ and $\cdot$ since they can be proved independently.

Proof of the theorem. Fix free vectors $\overrightarrow{v},\overrightarrow{u}$ and a real number $a$. The theorem is clearly true if $a=0$ or $\overrightarrow{v}=\overrightarrow{0}$ or $\overrightarrow{u}=\overrightarrow{0}$. Assume otherwise. Pick a point $o$ and using the lemma find points $p,q$ such that $\overrightarrow{op}\in-\overrightarrow{v}=(-1)\cdot \overrightarrow{v},\overrightarrow{oq}\in\overrightarrow{u}$. If $o,p,q$ are collinear, it is easy to verify that $\overrightarrow{v}=b\cdot\overrightarrow{u}$ for some real number $b$ (even nonzero). Therefore $$a\cdot(\overrightarrow{v}+\overrightarrow{u})=a\cdot(b\cdot\overrightarrow{u}+\overrightarrow{u})=a\cdot(b\cdot\overrightarrow{u}+1\cdot\overrightarrow{u})=a\cdot((b+1)\cdot\overrightarrow{u})=(a(b+1))\cdot\overrightarrow{u}=(ab+a)\cdot\overrightarrow{u}=(ab)\cdot\overrightarrow{u}+a\cdot\overrightarrow{u}=a\cdot(b\cdot\overrightarrow{u})+a\cdot\overrightarrow{u}=a\cdot\overrightarrow{v}+a\cdot\overrightarrow{u} $$

Assume $o,p,q$ aren't collinear. The theorem is also trivial for $a= 1$. Asuume otherwise.

1. First case: $a>0$ find points $p',q'$ on halflines $H(op),H(oq)$ respectively such that $|op'|=a|op|,|oq'|=a|oq|$. Then $p\neq p',q\neq q'$ (because $a\neq 1$). Clearly $\overrightarrow{op'}\in a\cdot(-\overrightarrow{v})=-a\cdot\overrightarrow{v},\overrightarrow{oq'}\in a\cdot\overrightarrow{u}$ and $$[\overrightarrow{pq}]= [\overrightarrow{po}]+[\overrightarrow{oq}]=\overrightarrow{v}+\overrightarrow{u},[\overrightarrow{p'q'}]= [\overrightarrow{p'o}]+[\overrightarrow{oq'}]= a\cdot\overrightarrow{v}+a\cdot\overrightarrow{u}$$.

By the converse of intercept theorem we know that $pq\parallel p'q'$ and $|p'q'|=a|pq|$. Also halflines $H(pq), H(p'q')$ lie on the same side of $L(pp')$. On halfline $H(p'q')$ find point $q''$ such that $pq\equiv p'q''$. Clearly $\overrightarrow{pq}\sim\overrightarrow{p'q''}$ and $$a\cdot\overrightarrow{v}+a\cdot\overrightarrow{u}=[\overrightarrow{p'q'}]=a\cdot[\overrightarrow{p'q''}]=a\cdot[\overrightarrow{pq}]=a (\overrightarrow{v}+\overrightarrow{u})$$

2. Second case: $a<0$. We apply the first case for $-a>0$ and we obtain $$(-a)\overrightarrow{v}+(-a)\overrightarrow{u}=(-a)(\overrightarrow{v}+\overrightarrow{u})$$ After some computation we get what we want (all rules we apply don't require the fact we are proving).

Answer 2

It's the penultimate axiom in the table here, so don't try proving it from the other axioms.