# Proof of the 1-1 Function

1. Aug 5, 2008

### roam

g(x) = $$\frac{2\left|x\right|}{\sqrt{x^2 +1}}$$

Without sketching any graphs, show that g is NOT one to one.

3. The attempt at a solution

Well, I know that in this situation it is required to show that if $$f(x_{1}) = f(x_{2})$$, then $$x_{1} = x_{2}$$

$$f(x_{1}) = f(x_{2})$$ => $$\frac{2\left|x_{1}\right|}{\sqrt{x_{1}^2 +1}}$$ = $$\frac{2\left|x_{2}\right|}{\sqrt{x_{2}^2 +1}}$$

This is interesting! but I'm not quite sure what I should do next...

$$(2\left|x_{1}\right|) (\sqrt{x_{2}^2 +1})= (2\left|x_{2}\right|) (\sqrt{x_{1}^2 +1})$$ ?

Does anybody know how to complete this proof?

Last edited: Aug 5, 2008
2. Aug 5, 2008

### uman

Careful. You seem to be trying to prove that it IS one-to-one, and you have the right method of doing that.

But you're being asked to prove that it IS NOT one-to-one! For this you only need to give a counterexample.

Can you think of any x and y such that f(x) = f(y) but that x is not equal to y? Think about it. If not I will give you a hint.

3. Aug 5, 2008

### Varnick

I'll give a hint, hopefully this won't be too obvious. Both x terms are not "Normal" x terms, consider the difference that makes. (If this is not a rigorous enough proof, sorry).

V

4. Aug 5, 2008

### HallsofIvy

Staff Emeritus
The best way to prove something is NOT true is to find a counter example. Look at Varnick's (rather cryptic) hint.

5. Aug 5, 2008

### spideyunlimit

See, g(x) is an odd function..
g(x) = g(-x)
[because of the mod and square]

I think you got it now!

6. Aug 5, 2008

### snipez90

Well g(x) = g(-x) implies that g(x) is even, which certainly means that g(x) is not one-to-one.

7. Aug 5, 2008

### theneedtoknow

there's another test you can do to prove that a function is not 1 to 1
take the first derivative, and if the first derivative changes sign anywhere, the function is not one to one (if it doesn't change sign, it means the function always increases or always decreases which makes it 1 to 1) This approach is much more useful in other cases where its not so obvious that the sign of x doesn't matter for the value of the function.

8. Aug 5, 2008

### roam

I think giving a counter-example is the easiest way to prove it's not one to one!

We want to show that the function is even and it satisfies f(x) = f(-x)
Therefore the function can't be one to one

Can we try the counter-example f(2) vs f(-2)?

$$\frac{2\left|2\right|}{\sqrt{2^2 +1}}$$ = $$\frac{2\left|-2\right|}{\sqrt{-2^2 +1}}$$

Then we show they're not equal. Is it sufficient?

9. Aug 5, 2008

### rock.freak667

I think this would also be a good way to show it. 2 values of x are mapped onto one corresponding value of f(x), thus it's not 1-1.

10. Aug 5, 2008

### HallsofIvy

Staff Emeritus
Well, except for the minor detail that the are equal! That's why this function is NOT "one-to-one".

11. Aug 5, 2008

### roam

Yes, so what do you think is the best notation to use?

12. Aug 5, 2008

### Redbelly98

Staff Emeritus
Just evaluate the two expressions you gave in post #8, to show that they are the same value.

And be careful when evaluating (-2)^2, which is different than -(2^2).

13. Aug 6, 2008

### spideyunlimit

Sorry, my bad, I meant to say even function.

14. Aug 6, 2008

### roam

$$\frac{2\left|2\right|}{\sqrt{2^2 +1}}$$ = $$\frac{2\left|-2\right|}{\sqrt{-2^2 +1}}$$

$$\frac{2 . \left|2\right|}{3}$$ = $$\frac{2 . \left|-2\right|}{3}$$

$$\frac{4}{3}$$ = $$\frac{-4}{3}$$

Is that right?

(Hmm, there shouldn't be an "equal" sign... lol! but I don't know what else to use. But we wanted to prove that they are even, $$f(x) = f(-x)$$)

Yes, I noticed that! no problem!

15. Aug 6, 2008

### HallsofIvy

Staff Emeritus
No, that's not right at all! |-2|= 2 so there SHOULD be an equal sign. And that's the whole point!
$$\frac{2 . \left|2\right|}{3} = \frac{2 . \left|-2\right|}{3}$$
$$\frac{4}{3}= \frac{4}{3}$$

16. Aug 6, 2008

### Redbelly98

Staff Emeritus
Also, instead of 4/3 it should really be

$$\frac{4}{\sqrt{2^2+1}}$$

$$= \frac{4}{\sqrt{4+1}}$$

$$= \frac{4}{\sqrt{5}}$$

But, as has been said several times, the whole point is that the two expressions are equal.

Last edited: Aug 6, 2008