# Thread: prove function is one-to-one

1. ## prove function is one-to-one

I am asked to prove that the following function is one-to-one.

$f(x)=\sqrt(x), x \geq 0$

I understand that the proof template for this is:

Suppose f(x) = f(y) ... then x = y.

So, is this sufficient?

Suppose $f(x) = f(y)$. Then $\sqrt(x) = \sqrt(y)$. Square both sides, and we have that $x = y$. Therefore, $f(x)$ is one-to-one.

2. Originally Posted by centenial
I am asked to prove that the following function is one-to-one.
$f(x)=\sqrt(x), x \geq 0$
Suppose f(x) = f(y) ... then x = y.
So, is this sufficient?
Suppose $f(x) = f(y)$. Then $\sqrt(x) = \sqrt(y)$. Square both sides, and we have that $x = y$. Therefore, $f(x)$ is one-to-one.
Well that works. Does it not?
BTW: In LaTeX $$\sqrt{(x)} = \sqrt{(y)}$$ gives $\sqrt{(x)} = \sqrt{(y)}$.
Note the extra braces.

3. Thanks Plato. It just seemed so trivial, I wasn't sure if I was missing something. I've been making an effort to do all my homework in latex - I thought that square root looked funny. thanks for the tip!

4. Originally Posted by centenial
I am asked to prove that the following function is one-to-one.

$f(x)=\sqrt(x), x \geq 0$<---This is not a function.
I understand that the proof template for this is:

Suppose f(x) = f(y) ... then x = y.

So, is this sufficient?

Suppose $f(x) = f(y)$. Then $\sqrt(x) = \sqrt(y)$. Square both sides, and we have that $x = y$. Therefore, $f(x)$ is one-to-one.
You do not have a function. By definition, every element in the domain of a function must have only one image in the codomain. A square root of a number has two values, the positive and negative.

5. Originally Posted by novice
You do not have a function. By definition, every element in the domain of a function must have only one image in the codomain. A square root of a number has two values, the positive and negative.
No, that's not correct. The Function square root always gives you a positive value. $\sqrt{4}=2\neq -2=-\sqrt{4}$. Another thing, is to solve a cuadratic equation $x^2=4$ that effectively has 2 solutions.

6. Originally Posted by felper
No, that's not correct. The Function square root always gives you a positive value. $\sqrt{4}=2\neq -2=-\sqrt{4}$. Another thing, is to solve a cuadratic equation $x^2=4$ that effectively has 2 solutions.
That's incorrect. The correct function should be $f(x)= |\sqrt{x}|$, for $x >0$.

$f(x)= \sqrt{x}$, for $x >0$ is not a function.

By definition, the square root of a positive number has a positive and a negative values, and the square root of a negative number has the real and the imaginary parts.

7. Originally Posted by novice
That's incorrect. The correct function should be $f(x)= |\sqrt{x}|$, for $x >0$.

$f(x)= \sqrt{x}$, for $x >0$ is not a function.
By definition, the square root of a positive number has a positive and a negative values, and the square root of a negative number has the real and the imaginary parts.
novice you are wrong, sadly wrong on this point.
$f(x)=\sqrt{x}$ is function with domain $[0,\infty )$.

You make a common mistake made by many novices.
Here it is $\sqrt{4}=2$ and it is absolutely true that $\sqrt{4} ~{\color{blue}\not=}-2$.
It is true that $4$ has two square roots. They are $\sqrt{4}~\&~-\sqrt{4}$.

8. Originally Posted by Plato
novice you are wrong, sadly wrong on this point.
$f(x)=\sqrt{x}$ is function with domain $[0,\infty )$.

You make a common mistake made by many novices.
Glad to hear from you. I thought you were ignoring us.
Originally Posted by Plato
Here it is $\sqrt{4}=2$ and it is absolutely true that $\sqrt{4} ~{\color{blue}\not=}-2$..
That's what I learned in highschool. I thought it was an over simplification.