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I'm just talking about (b), (c) and (d) in this question.

The way I see it, (b) is asking to prove that:

$$n \mod m = n \mod m$$which is like asking to prove that $1 = 1$.

(c) is also asking to prove that 1=1 essentially.

So how am I supposed to go about "proving" these statements b c and d. I don't know what to say other than "it just is".

  • $\begingroup$ It might be easier to figure out what to do if you think about how you would address (b) and (c) for different definitions of equivalent. For example, (b) Say $n$ and $m$ are equivalent if $n\le m$. Show that $n$ is equivalent to $n$. (Proof: Because $n\le n$ for all $n$.) Or: (c) Say $m$ and $n$ equivalent if $m = n+3k$ for some integer $k$. Prove if $m$ is equivalent to $n$, then $n$ is equivalent to $m$. Note that for the given (b), you don’t need to prove $n\mod 4 = n \mod 4$, you need to use that fact. For the given (c), use the fact that if $m\mod4=n\mod4$, then $n\mod4=n\mod4$. $\endgroup$ – Steve Kass Mar 15 '14 at 4:53
  • $\begingroup$ Yes, if one understands the words it is obvious. Conceivably the argument for d) is worth writing out. $\endgroup$ – André Nicolas Mar 15 '14 at 5:15

The relation says that $n\sim m$ iff $n\ mod\ 4\ =\ m\ mod\ 4$. So b) is trivially true. Similar observations hold for c) and d) as well. for d), if $n$ and $m$ on dividing by 4 leave remainder say $r$ and $m$ and $k$ leave remainder say $q$ then $r\ =\ q$ and hence $n\sim k$.

The best thing to understand your relation is to workout some examples, e.g., part a) will give you some clear picture as to what the relation is trying to convey.


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