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I am specifically asking why the statement $a \equiv b \;(\bmod\; n)$ is equivalent to the statement $a = b + kn$, where k is some positive integer. Why is it that the difference of a and b has to be a multiple of n in order for them to have the same remainder when divided by n?

Siong Thye Goh
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Wesley
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    $k$ need not be positive right? – Siong Thye Goh Jul 11 '16 at 16:53
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    When I teach this unit, that is the very definition of $mod$ that I use. Note also that $k$ need not be positive. It could be negative or zero as well. What definition of $mod$ are you working with? – JMoravitz Jul 11 '16 at 16:54
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    Suppose they both have remainder $r$. Then $a=q_1n+r$ and $b=q_2n+r$ for some integers $q_1$ and $q_2$. Then, subtracting, we get $a-b=(q_1-q_2)n$. – André Nicolas Jul 11 '16 at 16:58
  • Yes, it is the definition. But his question "why does $n|a - b \iff Rem(a) = Rem(b)$" is a legitimate (albeit it easy) question. – fleablood Jul 11 '16 at 17:08
  • Usually $\ a\equiv b\pmod n\ $ means $\ n\mid a-b,\ $ not $\ (a\bmod n) = (b\bmod n).\ $ But they are equivalent, as is easily proved. $\ $ – Bill Dubuque Jul 11 '16 at 19:51

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$a = kn + r$ and $b = jn + r$ $\implies$ $a - b = (k - j)n$ so that's one direction done.

$a - b = mn \implies$ $a = b + mn$. So if $b = jn + r$ then $a = jn + r + mn = (j+m)n + r$. So that's the other direction done.

BTW $a \equiv b \mod n \iff n|a-b$ is usually the definition (not that they have the same remainder[*]). So the title of your post did cause me to do a double take.

[*]--- although, to be honest, that the two statements are equivalent is usually taken to be utterly obvious-- sorry, but mathematicians tend to forget that what is obvious to them might elude others.

fleablood
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Because that what being equal mod n means; it's the definition. If a and b are equal mod 5 then the difference is some multiple of 5.

A good way of thinking about modular arithmetic is with equivalence classes. Thinking the numbers 0,5,10,15,... are all equivalent to each other. As are 1,6,11,16,... it becomes clear that members of the same equivalence class are all separated by a multiple of n.

If you think of the residual $k=(0,1,2,3... n-1)$ as the starting point for each equivalence class and then each equivalence class is closed under addition or subtraction of any multiple of $n$. This definition makes clear both the equivalence of $a$ and $b$ $(mod$ $n)$, and the fact that $a-b=kn$ for some $k\in\mathbb{Z}$.

it's a hire car baby
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