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I know that it is true that $(a+b) \mod \ c=a \ mod \ c+ b \ mod \ c.$

I would like to know: Is it true that $|a| \ mod \ c \le |a+b| \mod \ c + |b| \mod \ c$

BaoLing
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    It is false, in general, that $(a+b)\bmod c=a\bmod c+b\bmod c$ to begin with. Indeed, for $c=2$ and $a=b=1$, the left-hand side is $0$, whereas the right-hand side is $2$. What is true is that $(a+b)\bmod c=(a\bmod c+b\bmod c)\bmod c$. – egreg Jun 29 '17 at 19:50
  • @BaoLing See e.g. here for a proof of the identity in the prior comment. – Bill Dubuque Jun 29 '17 at 20:16

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No. Let $a = 2$, $c = 3$, and $b = 1$. Then $|a| \mod c \equiv 2$, but $|a + b| + |b| \equiv 0 + 1 \equiv 1 \mod c$.

Duncan Ramage
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