Answer:
Let
x1 = (a, b) and x2 = (a?, b?)
f(x1)
= f(a, b) = (b, a)
f(x2)
= f(a?, b?) = (b? a?)
Now
f(x1) = f(x2)
(b,
a) = (b?, a?)
b = b? and a
= a?
x1 =
(a, b) = (a?, b?) = x2
f(x1)
= f(x2)
is one-one.
Let
y = (a?, b?)
(co-domain of
f)
Also
let x = (a, b)
be
its pre-image of f
(b, a) = (a?,
b?)
b = a?, a =
b?
x = (a, b) =
(b?, a?)
f(x) = f(b?,
a?) = (a?, b?) = y
Corresponding
to every ordered pair (a?, b?) there exists (a, b) such that
f(a,
b) = (a?, b?)
is onto.
Hence
f is bijective.
You need to login to perform this action.
You will be redirected in
3 sec