- #1
Buffu
- 849
- 146
Given :- $$g(f(x_1)) = g(f(x_2)) \implies x_1 = x_2$$
Question :- Check whether ##g(x)## is injective or not.
Now this is of-course false; counter examples are easy to provide. But I proved that ##g(x)## must be one-one even after knowing the fact it must not.
Here is the proof :-
Let ##g(x)## be many-one
let ##f(x_1) = y_1## and ##f(x_2) = y_2##
Now,
##g(y_1) = g(y_2) \implies y_1 \ne y_2## for some ##y_1, y_2##
##y_1 \ne y_2 \implies f(x_1) \ne f(x_2) \implies x_1 \ne x_2##
Putting all together, we get,
##g(f(x_1)) = g(f(x_2)) \implies x_1 \ne x_2##
Which is contradictory to the given statement. So by proof of contradiction, we conclude that ##g(x)## is one-one under the given condition.
Which of my statement(s) are false ?
Question :- Check whether ##g(x)## is injective or not.
Now this is of-course false; counter examples are easy to provide. But I proved that ##g(x)## must be one-one even after knowing the fact it must not.
Here is the proof :-
Let ##g(x)## be many-one
let ##f(x_1) = y_1## and ##f(x_2) = y_2##
Now,
##g(y_1) = g(y_2) \implies y_1 \ne y_2## for some ##y_1, y_2##
##y_1 \ne y_2 \implies f(x_1) \ne f(x_2) \implies x_1 \ne x_2##
Putting all together, we get,
##g(f(x_1)) = g(f(x_2)) \implies x_1 \ne x_2##
Which is contradictory to the given statement. So by proof of contradiction, we conclude that ##g(x)## is one-one under the given condition.
Which of my statement(s) are false ?
Last edited: