# Thread: How to prove that a function is one-to-one?

1. ## How to prove that a function is one-to-one?

Hi everyone,

I was wondering, how can I show mathematically that f(x) = x2, defined for x <= 0 is one-to-one?

Can I do it like this:

If f(x) is one to one, then f(x1) = f(x2) if x1 = x2

so

(x1)2 = (x2)2
+-(x1) = +-(x2) but the domain is only defined for negative numbers so
-(x1) = -(x2)
therefore x1 = x2

or is this completely off?

(Sorry for the sloppy notation, I just thought it would be ok in this case, since the calculations are fairly short and simple)

2. ## Re: How to prove that a function is one-to-one?

The solution is correct, but I agree that the notation could be improved. I would write:

Suppose $f(x_1) = f(x_2)$. Then $x_1^2 = x_2^2$. Taking the square root of both sides, we find $x_1 = \pm x_2$. If $x_2<0$, then $-x_2$ is not in the domain, so $x_1 = x_2$. If $x_2 = 0$, then $x_2 = -x_2$.

The difference is $\pm x_1 = \pm x_2$ implies either $x_1 = x_2$ or $-x_1 = -x_2$, both of which mean $x_1 = x_2$. But, that is not actually what is meant. The notation I used is $x_1 = \pm x_2$ which is more general, as now it means $x_1 = x_2$ or $x_1 = -x_2$.

3. ## Re: How to prove that a function is one-to-one?

Originally Posted by Nora314
Hi everyone,

I was wondering, how can I show mathematically that f(x) = x2, defined for x <= 0 is one-to-one?

Can I do it like this:

If f(x) is one to one, then f(x1) = f(x2) if x1 = x2

so

(x1)2 = (x2)2
+-(x1) = +-(x2) but the domain is only defined for negative numbers so
-(x1) = -(x2)
therefore x1 = x2

or is this completely off?

(Sorry for the sloppy notation, I just thought it would be ok in this case, since the calculations are fairly short and simple)
to show f is injective what you need to show is that if {f(x1) == f(x2)} then {x1 == x2}

if {x1 == x2} the fact that {f(x1) == f(x2)} is trivial since the function is single valued.

4. ## Re: How to prove that a function is one-to-one?

Originally Posted by romsek
to show f is injective what you need to show is that if {f(x1) == f(x2)} then {x1 == x2}

if {x1 == x2} the fact that {f(x1) == f(x2)} is trivial since the function is single valued.
The OP meant $f(x_1) = f(x_2)$ only if $x_1 = x_2$. That was obvious from the method of proof used.

5. ## Re: How to prove that a function is one-to-one?

yep, I didn't get that far once I saw the basic error I commented on.