3.4. Elementary Consequences Of The Axioms
Theorem 3.1. Uniqueness of Zero Element
In any linear space, there is only one zero element.
Proof
Axiom 5 ensures the existence of at least one zero element.
Let O1 and O2 be 2 zero elements.
Taking x  O1 , O  O2 in axiom 5, we have
(a)
O2  O1  O1
Taking x  O2 , O  O1 in axiom 5, we have
(b)
O1  O2  O2
According to the commutative law,
O1  O2  O2  O1
Substituting into (a) and (b) gives O1  O2 .
Theorem 3.2. Uniqueness of Negative Elements
In any linear space, every element has exactly one negative element.
In other words, there is exactly one y that satisfies x  y  O for any given x.
Proof
Let both y1 and y 2 be the negative of x.
Hence
y1  x  y2  y1   x  y2   y1  O  y1
  y1  x   y2  O  y2  y2
so that y1  y2 .
Theorem 3.3
For any x, y V and  ,   R , we have
(a) 0x  O
(b) O  O
(c)
   x    x      x 
(d)  x  O  either   0 or x  O
x y
(e)  x   y and   0 

(f)  x   x and x  O 
(g)   x  y     x     y    x  y
n
(h)
 x  nx
i 1
Proofs
Only proofs for items a-c will be given. Rest is left as exercise.
Proof of (a)
Let z  0x . Hence
z  z  0 x  0 x   0  0 x  0 x  z
Adding –z to both sides gives z  O .
Proof of (b)
Let z  O . Hence
z  z  O  O   O  O   O  z
Adding –z to both sides gives z  O .
Proof of (c)
Let z     x . Hence
z   x     x   x       x  0x  O
which means z is the negative of  x , i.e.,
   x    x  .
Setting z     x  gives
z   x     x    x     x  x   O  O
which means z is the negative of  x , i.e.,    x     x  .