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Good afternoon. I am relatively new to modular arithmetic. I have been asked to show that if $ a\equiv b,\ b\equiv c \pmod n$, then $a\equiv c \pmod n$. My idea was to write that $a\equiv b \pmod n$ is the same as writing $a \bmod n = b\bmod n$. Similarly, if $b\equiv c\pmod n$ then $b\bmod n = c \bmod n$. From here, I combined both equations to say $a\bmod n = c\bmod n$, which is the same $a\equiv c\pmod n$, hence true. The only question I have is, can my professor mark this wrong because of saying $a\equiv b\pmod n$ means the same as $a\bmod n = b\bmod n$ or is this statement correct. If it is not, how should I express it?

Bill Dubuque
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Sasha Casas
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    Welcome to Mathematics Stack Exchange. Note that $a\equiv b\pmod n\iff n\mid a-b$; cf. this question – J. W. Tanner Jun 24 '21 at 21:37
  • You need to be clear about what you mean by $a \bmod n$. In general, $a \bmod n$ refers to $[a]_n$ which is the set of all integers that are congruent to $a$ modulo $n$. While it is true that $a \equiv b \pmod{n} \iff [a]_n=[b]_n$, one should be clear that the latter is a set equality. Your result can be obtained if you use the definition as suggested by @J.W.Tanner – Anurag A Jun 24 '21 at 21:40
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    Yes, you can prove transitivity of congruence that way - using the equivalence with remainders. However, to check your proof we need to see the full proof. Also, did you already prove said equivalence with remainders? If not then you would need to prove it as a Lemma supporting your proof (lacking such might be the point of your prof's critique). – Bill Dubuque Jun 24 '21 at 22:01
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    I get the feeling you have the same slight issue that I did when I started learning modular arithmetic as a computer scientist, which is that you want to use "mod" like the modulo operator we use in computer science. As Anurag points out, this isn't really correct notation. Tanner's suggestion likely leads to the quickest proof – Stephen Donovan Jun 24 '21 at 22:03
  • @Stephen There is nothing wrong with using the $!\bmod!$ operation - which is used in mathematics as well as computer science. See the link in my prior comment for further discussion on the relation between mod as a normal-form operation vs. equivalence relation (congruence). Update the above link is not what I intended - it should be this answer. – Bill Dubuque Jun 24 '21 at 22:06
  • I edited your question to use correct MathJax notation. If this is not what you meant then please edit it further. – Bill Dubuque Jun 24 '21 at 22:21
  • @AnuragA What you wrote is generally not true. Rather $,a\bmod n,$ usually denotes the standard mod normal-form operation. Occasionally $,a\bmod n,$ or $\ a\pmod n,$ is used for the entire equivalence class $,a+n\Bbb Z,,$ but that is far less common in elementary contexts - see the answer linked in my prior comment. – Bill Dubuque Jun 24 '21 at 22:26
  • In any case, as in the links, there is no need to use remainders, simply note that the multiples $n\Bbb Z$ of $n$ are closed under addition, so if $,a-b,,b-c\in\Bbb n\Bbb Z$ then so too is their sum $,a-c.\ $ The question will likely soon be closed as a dupe unless you clarify it to show that it is not. – Bill Dubuque Jun 24 '21 at 23:37

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