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From a book, I have got proof as follows, that otherwise I would have solved using just logic that any consecutive $3$ values have one and only one value in equivalence class $\equiv 0\pmod 3$, and have difficulty in understanding. I have divided the given proof into two sequential parts, and hope that the first part's understanding will be enough:

Part (i) : Since any integer can be written in the form $3n \text{ or } 3n\pm 1$, the difference of two integers of the same form is a multiple of $3$, and therefore not less than $3$.

Part (ii) : But the difference of any two of three consecutive integers is less than $3$, so that the three consecutive integers are respectively of the above three forms, among which one is of the form $3n$, that is, a multiple of $3$.

In Part (i), the problem lies in logic, as for given form of two integers as : $3n, 3n + 1$, the difference is $-1$ and is less than $3$, as the value of $n$ has to be the same.

jiten
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1 Answers1

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Here "difference" means "absolute difference". Let $x$ and $y$ be two distinct integers of same form; then $x = 3n + r$ and $y = 3m + r$, for some integers $n$ and $m$, and $r = 0, 1$ or $2$.

Then $x - y = 3(n - m$), and $n- m$ cannot be $0$ as the integers are distinct. Hence the absolute difference between $x$ an $y$ will never be less than $3$.

G Tony Jacobs
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user512074
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