Let a b c and d be real numbers such that a SolutionThe axio

Solution

The axiom of choice is equivalent to the statement that | c | |d | or | d| | c| for every c,d
Let f(x) = c( xb/ ab ) + d( xa/ ba ).
Then f : [a, b] [c, d] one-to-one and onto f : (a, b) (c, d) one-to-one and onto
So that
If | a| < | b |
then | A | = | B | (Cantor–Bernstein–Schroeder theorem)
The axiom of choice is equivalent to the statement that | a | |b | or | b | | a| for every a.b
| c | < | d |
So that
(axb)x(cxd)--------1
Lets say axb=x
Xx(cxd)=(x.d).c-(x.c.d)-----2
Then
x.d=(axb).d
=a.(bxd)
x.c=(axb).c
=a.(bxc)
Su statute in 2 we get
L.H.S=
{a.(bxd)}.c-{a.(bxc)}.d=constant