Does ax = bx imply a = b in any vector space?justify

Question: Does ax = bx imply a = b in any vector space? Justify. Let v = ${(a_1, a_2):a_1, a_2 \in f}$ where f is field. Define addition of elements of v coordinate wise, and for $c \in f$ and $(a_1, a_2) \in V$ . Define $c(a_1, a_2) = (a_1, 0)$. Is vector space over F with these operations? Justify your answer by taking all condition of vector space.

Solution: In any vector space ax = bx implies that a = b false
It is only true if we take x = 0
i.e ; ax = bx \
x(a – b) = 0 either x = 0 or ( a-b) = 0
take x = 0 and any a≠b for x.

Let us suppose $(a_1, a_2), (b_1, b_2) \in V$ then
$V_{s1}: (a_1, a_2) + (b_1, b_2) = (a_1 + b_1, a_2 + b_2) = (b_1 + a_1, b_2 + a_2) = (b_1, b_2) + (a_1, a_2)$
$V_{s2}: [(a_1, a_2) + (b_1, b_2)] + (c_1, c_2) = (a_1, a_2) + [(b_1 , b_2) + (c_1, c_2)] \forall (c_1, c_2) \in f$
$V_{s3}: \exists (d_1, d_2) \in V\ s.t\ (a_1, a_2) + (d_1, d_2) = (a_1, a_2)$
$(a_1, a_2) + (0, 0) = (a_1, a_2)$
$(d_1, d_2) = (0, 0)$
$V_{s4}: \forall (a_1, a_2) \in V \exists (-a_1, -a_2) \in V$
s.t $(a_1, a_2) + (-a_1, -a_2) = (0, 0)$
$V_{s5}: \forall \in F 1.(a_1, a_2) = (a_1 ,0) = (a_1, 0) ≠ (a_1, a_2)$
Hence $v_{s5}$ does not satisfy.
$V_{s6}: \forall x, y \in F\ xy(a_1, a_2) = x(ya_1, 0)$
$V_{s7}: \forall x, y \in F$
$(x + y)(a_1, a_2)$
= $[(x + y)a_1, 0]$
= $[xa_1 + ya_1, 0]$
= $(xa_1, 0) + (ya_1, 0)$
$V_{s8}: \forall x \in f$
$x[(a_1, a_2) + (b_1, b_2)]$ =$x[a_1 + b_1, a_2 + b_2]$
=$[x(a_1 + b_1), 0]$
=$(xa_1, 0) + (xb_1, 0)$
=$x(a_1, a_2) + x(b_1, b_2)$\

$V_{s5}$ does not satisfy it is not vector space.

Question: Does ax = bx imply x = y in any vector space? Justify. Let v denote the set of ordered pairs of real numbers.if $(a_1, a_2)$ and $(b_1, b_2)$ are element of v and $c \in R$ , define $(a_1 , a_2) + (b_1 , b_2) = (a_1.b_1, a_2 + b_2)$ and $c(a_1, a_2) = (a_1, ca_2)$. Is v a vector space over R with these operations ? Justify your answer by showing all conditions of vector space.

Solution: In any vector space ax = ay implies that x = y is false.
It is only true if we take a = 0
i.e; ax = ay a(x – y) = $0 \in V$
either a = 0 , or ( x – y) = 0 if we take a = 0 and any x ≠ y for a.\

Second part
Here v denote the set of ordered pairs of real numbers. If $(a_1, a_2) and\ (b_1, b_2)$ are element of v and $c \in R$,define $(a_1 , a_2) + (b_1 , b_2) = (a_1.b_1, a_2 + b_2)$ and $c(a_1, a_2) = (a_1, c.a_2)$.
$V_{s1}: (a_1, a_2) + (b_1 , b_2) = (a_1.b_1, a_2 + b_2) = (b_1.a_1 , b_2 + a_2) = (b_1, b_2) + (a_1, a_2)$
$V_{s2}: \in [(a_1, a_2) + (b_1, b_2)] + (c_1, c_2) = (a_1.b_1, a_2 + b_2) + (c_1, c_2)$\
= $(a_1.b_1.c_1 , a_2 + b_2 + c_2) = (a_1, a_2) + (b_1.c_1 , b_2 + c_2) = (a_1, a_2) + [(b_1, b_2) + (c_1 , c_2)]$
$V_{s3}: \exists (d_1, d_2) \in V$
s.t $(a_1, a_2) + (d_1, d_2) = (a_1, a_2)$
Or,$(a_1.b_1 , a_2 + b_2 ) = (a_1 , a_2) \Rightarrow a_1.b_1 = a_1 \Rightarrow b_1 = 1$
Also,$a_2 + b_2 = a_2 \Rightarrow b_2 = 0$
Hence $(d_1, d_2) = (1, 0)$
so, $(a_1, a_2) + (d_1, d_2) ≠ (a_1 , a_2)$
$v_{s3}$ does not satisfy.
$V_{s4}: \forall (a_1 , a_2) \exists (-a_1, -a_2) \in V$ s.t $(a_1, a_2) + (-a_1 , -a_2) = [a_1.(-a_1) , b_1 + (-b_1) ] = (-a_1^2, 0)≠0$
Hence $v_{s4}$ also not satisfy.
$V_{s5}: for 1\in R 1.(a_1, a_2) = (a_1, 1.a_2) = (a_1, a_2)$
$V_{s6}: \forall x, y \in R$
$xy(a_1, a_2) = x(a_1, ya_2) = x[y(a_1, a_2)$
$V_{s7}: \forall x, y \in R$
$(x + y) [(a_1, a_2) + (b_1 , b_2)] = (x + y)(a_1, a_2) + (x + y) (b_1, b_2)$
L.H.S = $(x + y)(a_1.b_1, a_2 + b_2)$
= $[a_1.b_1 , (x + y)(a_2 + b_2)]$
=$[a_1.b_1 , (x + y) a_2 + (x + y) b_2]$
R.H.S = $[ a_1 , (x + y)a_2] + [b_1 , (x + y)b_2]$
= $[a_1.b_1 , (x + y)a_2 + (x + y) b_2]$
L.H.S = R.H.S
$V_{s8}: \forall x \in R$
$x[(a_1, a_2) + (b_1, b_2)] = x(a_1, a_2) + y(b_1, b_2)$ L.H.S
= $x[ a_1.b_1 , a_2 + b_2]$
= $[a_1.b_1 , x(a_2 + b_2)]$
= $[a_1.b_1 , xa_2 + xb_2]$
R.H.S
= $(a_1 , xa_2) + (b_1 , xb_2) = (a_1.b_1 , xa_2 + xb_2)$
L.H.S = R.H.S
Here $v_{s3}$ and $v_{s4}$ are not satisfy. Hence it is not vector space.