# B Flaw in my proof of something impossible

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1. Nov 2, 2016

### Buffu

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 ?

Last edited: Nov 2, 2016
2. Nov 2, 2016

### PeroK

Leaving aside your proof, the conclusion you reach is absurd:

$g(f(x_1)) = g(f(x_2)) \implies x_1 \ne x_2$

Let $x_1 = x_2 = a$

Then $g(f(x_1)) = g(f(a)) = g(f(x_2))$

3. Nov 2, 2016

### Staff: Mentor

Those "some y1,2" don't have to have corresponding x1,2, you are reversing the logic of choosing them.

Let g be an arbitrary non-injective function R->R and f be a function {0}->{0} with (obviously) f(0)=0. See if your proof works. $g(f(x_1)) = g(f(x_2)) \implies x_1 = x_2$ is true.

4. Nov 2, 2016

### Stephen Tashi

How do you justify that implication? Is $f$ given to be 1-to-1 ?

5. Nov 2, 2016

### Buffu

Thanks for help. I am certain that I am the only one that commit such blunders.
Yes, that was a bad idea. Sorry for wasting your time by posting this silly question. I will consider twice before posting any question now on.

6. Nov 2, 2016

### Erland

No you're not. If you were, this forum would not exist :)

Don't worry. You learned something, and that's what this forum is for. You have nothing to be ashamed of. The big problem is all those people who don't understand math and don't care either.