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Thread: prove function is one-to-one

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

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

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

    I understand that the proof template for this is:

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

    So, is this sufficient?

    Suppose $\displaystyle f(x) = f(y)$. Then $\displaystyle \sqrt(x) = \sqrt(y)$. Square both sides, and we have that $\displaystyle x = y$. Therefore, $\displaystyle f(x)$ is one-to-one.
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  2. #2
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    Quote Originally Posted by centenial View Post
    I am asked to prove that the following function is one-to-one.
    $\displaystyle f(x)=\sqrt(x), x \geq 0$
    Suppose f(x) = f(y) ... then x = y.
    So, is this sufficient?
    Suppose $\displaystyle f(x) = f(y)$. Then $\displaystyle \sqrt(x) = \sqrt(y)$. Square both sides, and we have that $\displaystyle x = y$. Therefore, $\displaystyle f(x)$ is one-to-one.
    Well that works. Does it not?
    BTW: In LaTeX [tex]\sqrt{(x)} = \sqrt{(y)}[/tex] gives $\displaystyle \sqrt{(x)} = \sqrt{(y)}$.
    Note the extra braces.
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  3. #3
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    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!
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  4. #4
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    Quote Originally Posted by centenial View Post
    I am asked to prove that the following function is one-to-one.

    $\displaystyle f(x)=\sqrt(x), x \geq 0$<---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 $\displaystyle f(x) = f(y)$. Then $\displaystyle \sqrt(x) = \sqrt(y)$. Square both sides, and we have that $\displaystyle x = y$. Therefore, $\displaystyle f(x)$ 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.
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  5. #5
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    Quote Originally Posted by novice View Post
    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. $\displaystyle \sqrt{4}=2\neq -2=-\sqrt{4}$. Another thing, is to solve a cuadratic equation $\displaystyle x^2=4$ that effectively has 2 solutions.
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  6. #6
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    Quote Originally Posted by felper View Post
    No, that's not correct. The Function square root always gives you a positive value. $\displaystyle \sqrt{4}=2\neq -2=-\sqrt{4}$. Another thing, is to solve a cuadratic equation $\displaystyle x^2=4$ that effectively has 2 solutions.
    That's incorrect. The correct function should be $\displaystyle f(x)= |\sqrt{x}|$, for $\displaystyle x >0$.

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

    $\displaystyle f(x)= \sqrt{x}$, for $\displaystyle x >0$ 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.
    $\displaystyle f(x)=\sqrt{x}$ is function with domain $\displaystyle [0,\infty )$.

    You make a common mistake made by many novices.
    Here it is $\displaystyle \sqrt{4}=2$ and it is absolutely true that $\displaystyle \sqrt{4} ~{\color{blue}\not=}-2$.
    It is true that $\displaystyle 4$ has two square roots. They are $\displaystyle \sqrt{4}~\&~-\sqrt{4} $.
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  8. #8
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    Quote Originally Posted by Plato View Post
    novice you are wrong, sadly wrong on this point.
    $\displaystyle f(x)=\sqrt{x}$ is function with domain $\displaystyle [0,\infty )$.

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