# One to one and onto.

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1. Mar 30, 2016

### TyroneTheDino

1. The problem statement, all variables and given/known data
I am supposed to prove or disporve that $f:\mathbb{R} \rightarrow \mathbb{R}$
$f(x)=\sqrt{x}$ is onto. And prove or disprove that it is one to one

2. Relevant equations

3. The attempt at a solution
I know for certain that this function is not onto given the codomain of real numbers, but I am stuck on the one-to-one definition. I believe that I am supposed to disprove it, but I am not sure.

I say disprove because for a function to be one to one all values in the domain must correspond to a value in the codomain. Since anything less than 0 is undefined, does this make it true that not everything in the domain is defined in the codomain. Or is this reasoning flawed? Do only that value that are defined in the domain matter?

2. Mar 30, 2016

### HallsofIvy

Staff Emeritus
With something that basic, focus on the definitions. A function, f, from set A to set B is said to be "onto" (more precisely "onto B") if and only if, for any y in B, there exist x in A such that f(x)= y. Here, your function is $\sqrt{x}$, A= B= R. I suggest you look at y= -1.
(And be sure that you understand that, since $f(x)= \sqrt{x}$ is a function, it is "single-valued"- that is, "$\sqrt{x}$" is NOT "$\pm\sqrt{x}$".)

For "one-to-one", a function, f, from A to B is said to be "one-to-one" if and only if two different values of x, say $x_1$ and $x_2$ cannot give the say "y": if $x_1\ne x_2$ then $f(x_1)\ne f(x_2)$. Often it is simplest to prove that by proving the "contra-positive": if $f(x_1)= f(x_2)$ then $x_1= x_2$. Here, with $f(x_1)= \sqrt{x}$, start with $\sqrt{x_1}= \sqrt{x_2}$. What can you say from that?

3. Mar 30, 2016

### SammyS

Staff Emeritus
What is the domain of your function?

4. Mar 31, 2016

### TyroneTheDino

Okay, I take from this because $\sqrt{x_1}= \sqrt{x_2}$, $\sqrt{x_1}^2= \sqrt{x_2}^2$, so x1=x2. So this function is one to one because I can prove that . Correct?

5. Mar 31, 2016

### HallsofIvy

Staff Emeritus
Yes, that is correct. Notice that if the function had been $f(x)= x^2$ then $f(x_1)= f(x_2)$ would be $x_1^2= x_2^2$ from which it follows that $x_1= \pm x_2$, not "$x_1= x_2$" so that function is not one-to-one.

6. Mar 31, 2016

### Ray Vickson

Is your $f(x) = \sqrt{x}$ even a map from $\mathbb{R}$ into $\mathbb{R}$ at all?