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I have two relations:

$R_1:=\{(x,y)| x^2=y\} \subseteq \mathbb{N} \times \mathbb{R}$

$R_2:=\{(y,x)| x^2=y\} \subseteq \mathbb{R} \times \mathbb{N}$

$R_1$ is a total function

$R_2$ is a partial function

I´m trying really hard to understand but just struggle to get it. Could someone correct my understanding of it. So for $R_1$:

I take a $x$ from $\mathbb{N}$, for example $4$ $\subseteq$ $\mathbb{N}$ the corresponding $y=2$. That seems to apply. Generally,$ y=x^2 \subseteq \mathbb{N}$ and thus $\mathbb{N}\subseteq \mathbb{R}$. For $R_2$ I mean I just dont see the difference, the $y$ are still coming from $\mathbb{R}$ and $x$ from $\mathbb{N}$. So taking a negative $y$ still results in positive $x$. Could someone explain me that?

Mamba
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1 Answers1

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It's not true that $4\subseteq \mathbb{N}$, you must mean $4\in \mathbb{N}$. For $R_2$, notice that $(y,x)\in R_2$ as long as $y$ can be seen as the square of $x$, with $x$ being a natural number, this is not always possible! (I will clarify that). This is not the same as $R_1$, where $(x,y)\in R_1$ as long as $x$ can be seen as the square root of the real number $y$ (and this is always achievable).

In some way, $R_1$ and $R_2$ contain the "same" elements, and when I say that I mean that $(x,y)\in R_1$ if and only if $(y,x)\in R_2$, but the main difference is that for all $x\in \mathbb{N}$ there is a $y\in \mathbb{R}$ such that $(x,y)\in R_1$ while not for all $y\in \mathbb{R}$ there is a $x\in \mathbb{N}$ such that $(y,x)\in R_1$.

Daniel
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