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I have a relation $f$ defined as

$$ f=\{(x, y) \mid y=\sqrt{x} \;\text { and }\; y, x \in \Bbb R\} $$

Now this relation is a function. But how? A function is a relation whose every element in domain have only one image in codomain. Now if I have $4$. Then \begin{align} \sqrt{4}=\pm 2 \end{align}

So I have two ordered pairs whose second element is square root of first element: \begin{align} (4,2) \quad \text { and } \quad(4,-2) \end{align}

Now you can observe that 4 has two images. Therefore relation $f$ must not be a function. But it is! How this is possible?

Bernard
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    @AlexanderGeldhof but the function in this question is defined as $f(x)=√x$ . So if we don't follow the positive sign convention of square root then the argument of question is legit. –  Dec 29 '19 at 11:44

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By convention, $\sqrt{x}$ means the positive square root of $x$. So $\sqrt{4}=2$.

Also note that in order for $f$ to have codomain $\mathbb{R}$, you need to restrict the domain of $f$ to the non-negative reals.

A. Goodier
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    But how do I know when to use only positive value of √x or not. In solving quadratic equation one must consider both possiblity of answers but here we are considering only principle answer. –  Dec 29 '19 at 11:18
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    @Orientalem When solving for example $z^2=4$, we have $z=\pm\sqrt{4}=\pm 2$. We have roots $\sqrt{4}=2$ and $-\sqrt{4}=-2$. By itself, $\sqrt{4}$ is just $2$. – A. Goodier Dec 29 '19 at 11:21
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    Ohh! Thanks mvpq. So if I define a relation $f$: \begin{align} f= \ (x, y) | y=\pm \sqrt{x} \text { and } y, x \in R} \end{align} then this relation $f$ isn't a function 'cause of more then one image. –  Dec 29 '19 at 11:32
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    Yes, that's right – A. Goodier Dec 29 '19 at 11:34
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With the usual positivity convention there is a difference between

$$R_1 = \{(x,y) \mid x= y^2 \}$$ and

$$R_2 = \{(x,y) \mid \sqrt{x}= y \}$$

$(4,-2),(4,2) \in R_1$ but of these two only $(4,2) \in R_2$.

$R_1$ is not a function and $R_2$ is.

Henno Brandsma
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