<aside> 💡 If $v_1,v_2,v_3 \in V$ such that $v_1+v_3=v_2+v_3,$ then $v_1=v_2$
</aside>
$v_1+v_3=v_2+v_3$ →(1) We know from properties of vector spaces,$\exists v_3^{'}$ such that $v_3+v_3^{'} = 0$ So, (1) becomes: $v_1+v_3+v_3^{'}=v_2+v_3+v_3^{'}$ $\implies v_1+(v_3+v_3^{'})=v_2+(v_3+v_3^{'})$, $\implies v_1 + 0 =v_2+0 \implies v_1=v_2$
Suppose there exists $w\in V$ such that $v+w=v$ $\forall ~~v\in V$→(1). Then $v+w=v+0$ ($\because$ we know that $v+0=v)$ →(2) Then from (2) and the cancellation law, $w=0$ Hence the vector 0 is unique.
Similar proof as above.
In any vector space $V$, the following statements are true:
We know that $(0+0)v=0v+0v$ (from properties) →(1) Also $(0+0)v=0v$ ($\because$ $(0+0) = 0$) →(2) From (1) and (2): $0v+0v=0v$ $\implies$$0v+0v=0v+0$ ($\because$Any vector $v+0=v$ (from properties))→(3) So, From (3) and cancellation law of vector addition, $0v=0$
