consider-the-triple-of-real-numbers-x-y-z-defined-by-the-addittion-x-y-z-x-y-z-x-x-y-y-z-z-and-scalar-multiplication-by-x-y-z-0-0-0-Show-that-all-axioms-for-a-vector-space-are-

Question Number 61840 by psyche last updated on 10/Jun/19

c o n s i d e r t h e t r i p l e o f r e a l n u m b e r s (x, y, z)

d e f i n e d b y t h e a d d i t t i o n (x, y, z) + (x^{'}, y^{'}, z^{'}) = (x + x^{'}, y + y^{'}, z + z^{'})

a n d s c a l a r m u l t i p l i c a t i o n b y α (x, y, z) = (0, 0, 0) .

S h o w t h a t a l l a x i o m s f o r a v e c t o r s p a c e a r e s a t i s f i e d e x c e p t a x i o m 8 .

Answered by arcana last updated on 10/Jun/19

if \forall (x, y, z) \in R^{3}

1 \cdot (x, y, z) = (0, 0, 0) = (x, y, z) i f f x = 0,

y = 0, z = 0

Commented by psyche last updated on 10/Jun/19

p l e a s e, c o m p l e t e t h e p r o o f

Commented by arcana last updated on 10/Jun/19

axiom 1, 2, 3, 4 define (R^{3}, “ +^{″}) is a structure of group

with sum usual in R^{3}

we need a field K because define \cdot operation .

elements in K are scalars, elements in R^{3} are vectors

if R = K (field)

\cdot : R \times R^{3} \to R^{3}

(α, a) \to α \cdot a

axiom 5

α, β \in R, (x, y, z) \in R^{3}

α + β \in R

\Rightarrow (α + β) \cdot (x, y, z) = (0, 0, 0) def . “ \cdot^{″}

β \cdot (x, y, z) = (0, 0, 0); β \cdot (x, y, z) = (0, 0, 0)

\Rightarrow (α + β) \cdot (x, y, z) = α \cdot (x, y, z) + β \cdot (x, y, z)

axiom 6

α \in R, (x, y, z), (x^{'}, y^{'}, z^{'}) \in R^{3}

α \cdot [(x, y, z) + (x^{'}, y^{'}, z^{'})] = α \cdot (x + x^{'}, y + y^{'}, z + z^{'})

(x + x^{'}, y + y^{'}, z + z^{'}) \in R^{3}

\Rightarrow α \cdot (x + x^{'}, y + y^{'}, z + z^{'}) = (0, 0, 0)

α \cdot [(x, y, z) + (x^{'}, y^{'}, z^{'})] α \cdot (x, y, z) = (0, 0, 0); α \cdot (x^{'}, y^{'}, z^{'}) = (0, 0, 0)

\Rightarrow α \cdot [(x, y, z) + (x^{'}, y^{'}, z^{'})] = α \cdot (x, y, z) + α \cdot (x^{'}, y^{'}, z^{'})

axiom 7

α, β \in R, (x, y, z) \in R^{3} . α β \in R

(α β) \cdot (x, y, z) = (0, 0, 0)

β \cdot (x, y, z) = (0, 0, 0) \Rightarrow α \cdot [β \cdot (x, y, z)] = α \cdot (0, 0, 0) = (0, 0, 0)

\Rightarrow (α β) \cdot (x, y, z) = α \cdot [β \cdot (x, y, z)]

Facebook Tweet Pin