As the title suggest im having difficulty approaching how or where to start proving this. Only thing i can derive from the given is that $m|b - a$ is equivalent to $b - a = mk$ for some k
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3This $k$ is an integer. Show that $0 < b-a < m$ and you will have a contradiction. – player3236 Oct 16 '20 at 20:41
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$$0\leq a< b\leq m-1\implies$$
$$0\le a <b< m\implies$$
$$m>b \implies$$
$$m>b-a\implies$$
$m $ cannot divide $ b-a$.
hamam_Abdallah
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ohh i totally missed the fact that if b < m - 1 then b < m. Thank you for input – UnKnoWnZ Oct 16 '20 at 20:51
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@hamam_Abdallah: no. if $b < m - 1$ then $b \le m- 1$, so $b < m$. You can't say that UnKnoWnZ's comment is wrong. – Rob Arthan Oct 16 '20 at 22:49
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@RobArthan Your comment has not been upvoted. It means something. – hamam_Abdallah Oct 16 '20 at 23:30
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"hh i totally missed the fact that if b < m - 1 then b < m. Thank you for input" Yes, but we don't know that $b < m-1$. We only know that $b \le m-1$. But that's okay. If $b\le m-1$ then $b < m$. (That was the purpose of hamam_Abdallah's "No"-- not that you were wrong [you were correct-- but that we don't know $b < m-1$.) – fleablood Oct 17 '20 at 00:01