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Let the relation $R:\mathbb{N} \rightarrow \mathbb{N}$ givenas $xRy$ iff $0 ≤x − y ≤ 1$i.e $R=\{(x,y):0 ≤x − y ≤ 1\}$ on $\mathbb{R}$. How do I find an inverse of this relation?

Is it $R^{-1}=\{(x,y):0 ≤y − x ≤ 1\}$?. Could someone help me out please?

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    To find the inverse, you literally switch $x$ and $y$. Done. :-) – 2017-02-03
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    Yea that say Wiki hehe but How can I prove it? – 2017-02-03
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    What definition of the inverse relation are you using? I ask because it ought to be an immediate consequence of the definition. – 2017-02-03
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    but then always the inverse relation exists? – 2017-02-03
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    Yes, the inverse relation always exists. As the answer of @zoli shows, every relation has an inverse relation. – 2017-02-04

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The definition of the inverse of $R$, $R^{-1}$, is that for any $x,y$ $xRy$ iff $yR^{-1}x$.

$xRy$ means that $0\leq x-y \leq 1$ that is, $xRy$ iff $x\geq y$ and $0 \leq \mid x-y\mid \leq 1$. To get the inverse we have to require $x \leq y$ and $0 \leq \mid x-y\mid \leq 1$ which is the same as that requiring $0\leq y-x \leq 1$ as the OP suggested.