By user93957


2013-09-13 15:52:03 8 Comments

Why $f(x) = \sqrt{x}$ is a function (as I found in my textbook) since for example the square root of $25$ has two different outputs ($-5,5$) and a function is defined as "A function from A to B is a rule of corre- spondence that assigns to each element in set A exactly one element in B.", so $f(x) = \sqrt{x}$ is not a function?

2 comments

@J. W. Perry 2013-09-13 16:13:43

I will assume that in this portion of your textbook it is assumed that $x \in \mathbb{R}$, and with that condition $f(x)=\sqrt{x}$ is certainly a function. Specifically $f:[0,\infty) \rightarrow [0,\infty)$. It meets the formal definition of a function (not one to many).

Your confusion is due to an inappropriate extrapolation of reasoning. Specifically this:

You know that $x^2=25 \Rightarrow x= \pm \sqrt{25}$ by the square root property. However this function in no way involves taking the square root of both sides of an equality. It is just a function $f(x)=\sqrt{x}$, and the domain is $x \geq 0$ by virtue of the fact that you are living in the real number system in this portion of your textbook.

Now $f(x)= \pm \sqrt{x}$ is certainly not a function. For example, if the question were "Let $y^2=x$. Is $y$ a function of $x$?" You would say no in this case as $y= \pm \sqrt{x}$, and you would choose $x=25$ to counter the definition of a function.

Note, your statement "the square root of $25$ has two different outputs" is false. There is only one output. However, if $y^2=25$, then $y$ has two different solutions.

@alvonellos 2014-03-27 21:13:32

This is a great answer.

@Kyle Delaney 2015-10-17 02:04:00

You said $f(x) = \sqrt{x}$ is a function if $x \in \mathbb{R}$. Doesn't that make $\mathbb{R}$ the domain? If $\mathbb{R}$ is the domain and the codomain, doesn't that mean that $f(x) = \sqrt{x}$ is not a function? My understanding is that in order to be a function, it needs to meet two requirements. Not only does it need to have only one relationship for each input value, but it needs to work for every input value. And $f(x) = \sqrt{x}$ doesn't work for every input value if the domain is $\mathbb{R}$, right?

@J. W. Perry 2015-10-20 04:00:33

@KyleDelaney Are you absolutely sure that is what I said? Read really really really closely. The domain is the first part of the mapping where I say "Specifically $f:\ldots$" If you have a new question about something you do not understand, maybe ask it. In context my English appears pretty clear here 2 years later. How to set you straight is escaping me at the moment.

@Kyle Delaney 2015-10-21 20:49:12

Boy did I strike a nerve. Wasn't trying to. Sorry.

@user93957 2013-09-13 15:52:03

For functions defined by equations, we agree on the following convention regarding the domain: Unless otherwise indicated, the domain is assumed to be the set of all real numbers that lead to unique real-number outputs. The symbol $\displaystyle \sqrt{ }$ is defined in algebra to mean the positive square root only. Thus $\displaystyle \sqrt{25} = 5$, and if you are thinking of the other root, you need to write $\displaystyle -\sqrt{25} = -5$. Consequently, the function $\displaystyle f(x) = \sqrt{x}$ do represent a function; for each input there is exactly one output.

@Daniel Fischer 2013-09-13 15:53:43

That's problematic. What if there is no order, no canonical choice, for example in $\mathbb{C}$? For real numbers $x$, $\sqrt{x}$ customarily means the non-negative square root, but there is no law, one can follow other conventions.

@imranfat 2013-09-13 16:00:41

You mean that how to take √i ??

@kaine 2013-09-13 16:04:07

Wait... why did you answer your own question?

@user93957 2013-09-13 16:04:21

@DanielFischer For functions defined by equations, we agree on the following convention regarding the domain: Unless otherwise indicated, the domain is assumed to be the set of all real numbers that lead to unique real-number outputs. So there must be a convention regarding $\sqrt{}$ and it was chosen to be a positive square root only for the set of all real numbers that lead to unique real-number outputs when dealing with functions.

@user93957 2013-09-13 16:04:53

@kaine Q/A style, share my own knowledge.

@MJD 2013-09-13 16:38:38

I am puzzled by your claim that “Unless otherwise indicated, the domain is assumed to be the set of all real numbers that lead to unique real-number outputs.”. I am not familiar with that convention, and it does not seem to make very much sense or to be consistent with what I already know. I worry that you made it up. Can you provide a source?

@user93957 2013-10-14 11:26:01

@MJD : "[...] the domain is assumed to be the set of all real numbers that lead to unique real-number outputs." That means that the domain contains all the numbers which makes of a function continuous, example the domain of this function $\displaystyle f(x) = \frac{2}{x-2}$ is all the real numbers except $2$. And if we put this $2$ into the function we will get a non-unique non-real number output. That's why we say: "[...] the domain is assumed to be the set of all real numbers that lead to unique real-number outputs."

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