Distributive property of scalar multiplication over scalar addition I need help with a simple proof for the distributive property of scalar multiplication over scalar addition.
Help with proving this definition: $(r + s) X = rX + rY$
I have to prove the truth of the definition for a vector space. I know $X= (a_1,b_1)$. Please someone help I been stuck on this proof for three days.
Let vector $X$ in $R^2$ be represented by $X = (a_1,b_1)$ where $a_1$  and $b_1$ are real numbers--- $r$ and $s$ are real scalars
 A: You have to distinguish between the associativity,distributivity, etc laws for the underlying field (Reals) of the vector space  and the laws your are trying to prove about the vector space. Where the rubber hits the road, you are dealing with reals, so you can apply the familiar field axioms which you should know, the proof requires that your correctly transform each line correctly by applying the laws of the field and/or the axioms of the vector space.  Here's an outline:
\begin{align*}
LHS &= (r + s) X  \\\
    &= (r+s)(a_1,b_1) \\\
    &= ( (r+s)a_1, (r+s)b_1) \hspace{3.5em} \text{ (scalar multiplication of vector in vector space)} \\\
    &= ( r\cdot a_1 + s\cdot a_1, r\cdot b_1 + s\cdot b_1) \hspace{0.5em} \text{(distributive addition over multiplication of } \\\
    & \hspace{14.5em}\text{underlying field of vector space, $r,s,a_1, b_1 \in \mathbb{R}$ )} \\\
\end{align*}
\begin{align*}
RHS &= rX + sX \\\
    &= r(a_1,b_1) + s(a_1,b_1) \\\
    &= (r\cdot a_1 , r\cdot b_1) + (s\cdot a_1 , s\cdot b_1) \hspace{1em} \text{ (scalar multiplication of vector in vector space)} \\\
    &= ( r\cdot a_1 + s\cdot a_1, r.b_1 + s\cdot b_1) \hspace{1.5em} \text{(vector addition in vector space)} \\\
    &= ((r+s)a_1 , (r+s)b_1) \hspace{4.5em}\text{ (distributive addition over multiplication}\\\
& \hspace{15em}\text{of underlying field in vector space)} \\\
    &= (r+s)X  \hspace{9.5em} \text{(scalar multiplication of vector in vector space)} \\\ 
\end{align*} 
We get to the same thing so, RHS and LHS are equal, that's the proof. 
A: I believe following is what you are asking to prove : 
(r+s)X =rX+sX for X =(a,b)
By distributivity of scalar over vector and using vector space properties; we have : 
Proof :
(r+s) (a,b) = {(r+s)a, (r+s)b} = {ra+sa, rb+sb} = (ra, rb) + (sa, sb) = r (a, b) + s (a,b). 
Hope this helps.
A: Since a geometric interpretation of the condition $(r+s)X=rX+sX$ ($r,s\in\mathbb{R}$, $X\in\mathbb{R}^2$) has been asked, let me try to add it to the already existing answers. 
Geometrically speaking a vector in $\mathbb{R}^2$ (and by this I mean a Euclidean vector) is an equivalence class of directed segments. Let $A$ and $B$ be two (eventually different) points of the plane. A directed segment $\overrightarrow{AB}$ is a line segment connecting $A$ and $B$ and oriented from $A$ to $B$ (represented by an arrow from $A$ to $B$).
Two directed segments are equivalent if there exists a translation sending one to the other.
A vector is the set of all directed line segments which are equivalent to each other
: it is characterized by a length (the norm or magnitude of the vector), a direction (the line on which the segment lives) and an orientation (the direction of the arrow). Since a vector is simply a representative of an equivalence class, we may represent it as an oriented segment connecting $O=(0,0)$ to another point in the plane. With this convention, the oriented segment $\overrightarrow{AB}$ is the vector $X:=\overrightarrow{OX}$ (representing the class of $\overrightarrow{AB}$) applied to the point $A$, that is to say, starting from $A$ (or, equivalently, with tail coinciding with $A$).
Let $X$ be a vector. If $r\in\mathbb{R}$, then $rX$ is the vector that has same direction as $X$, length $|r|$ times the length of $X$ and orientation the same as $X$ if $r>0$, the opposite of $X$ if $r<0$. If $X$ and $Y$ are two vectors, then $X+Y$ is the vector obtained by the head-to-tail joining of $X$ and $Y$ (recall that $Y$ is a representative of an equivalence class of oriented segments).
Now, let $r,s\in\mathbb{R}$. For the sake of clearness, let us assume that $r>0$ and $s<0$ (the other combinations are handled analogously). On the one hand, the vector $rX+sX$ is obtained by joining the tail of $rX$ (i.e. $O$) with the head of the vector $sX$ applied to the head of $rX$. Since $sX$ has same direction of $rX$, opposite orientation and length $|s|$ times the length of $X$, the resulting vector $rX+sX$ will have the same direction as $X$, length $\big||r|-|s|\big|$ times the length of $X$ and direction which is the same as $X$ if $|r|-|s|>0$, the opposite of that of $X$ if $|r|-|s|<0$. On the other hand, the vector $(r+s)X$ has same direction as $X$, length $|r+s|=\big||r|-|s|\big|$ times the length of $X$ and direction which is the same as $X$ if $|r|-|s|>0$, the opposite of that of $X$ if $|r|-|s|<0$. Summing up, the vectors $(r+s)X$ and $rX+sX$ are the same (or, more precisely, represents the same equivalence class. One may replicate the above argument picking any oriented segment $\overrightarrow{AB}$ instead of $X:=\overrightarrow{OX}$).
The following link allows one to play around with this idea: Geogebra.
