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I encountered this strange notation in an online set theory course. I've emailed my instructor, but he's on vacation so I was wondering if any of you knew what it meant. Apparently it equals 8.

Edit: Here's a complete question involving the notation again--I didn't give enough context the first time

Andrés E. Caicedo
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push33n
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  • Perhaps the notation was explained at an earlier point in the course. – Ethan Bolker Jan 05 '18 at 16:32
  • Why did you just show us the notation? Why didn't you show us the entire question? The more context we have to more likely we will be able to answer. – fleablood Jan 05 '18 at 16:38
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    It is actually a right side subscript of the $+$. – orole Jan 05 '18 at 16:49
  • Seeing the second link I believe if it is the first appearance of the symbol $+_{....}$ then possibly knowing that symbol was believed to have been a prerequisite for the course or maybe that was introduced in the notations section. – Aditya Guha Roy Jan 05 '18 at 16:55
  • After seeing your first link, I though that it's probably addition modulo 12. Your second link confirmed this since it's talking about $\mathbf{Z}_{12}.$ Thus, the symbol refers to addition modulo 12. – md2perpe Jan 05 '18 at 20:03

2 Answers2

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My guess is that $a+_nb$ is the remainder of the division of $a+b$ by $n$.

José Carlos Santos
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  • that works for the first case I posted but it doesnt work for one of the other cases. it was my first though also! – push33n Jan 05 '18 at 16:42
  • actually, never mind. you're right! thank you – push33n Jan 05 '18 at 16:44
  • @push33n" What do you mean by "doesn't work"? Which "case" is it that you think this doesn't make sense in? – hmakholm left over Monica Jan 05 '18 at 16:46
  • Did you know that from before ? – Aditya Guha Roy Jan 05 '18 at 16:55
  • The example is badly written in that it looks like it is claiming $2 +{12} 2 = 2$ and that $a +{12} a = 2$ always. Which is obviously not true. But if you read the final paragraph $a +_{12} a = 2$ only if $a = 1$ or $a = 7$ so $a = 1,7$ are the "solutions" and the rest were supposed to be false. – fleablood Jan 05 '18 at 17:11
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I think Jose Carlos Santos has the correct answer.

If so, to go into a little more detail:

1) then $12$ is a subscript of the PLUS sign (and to the right) and not of the $10$.

2) Just like $+$ or $-$ or $\times$ or $\div$, the operation $+_{12}$ is a binary operation so that $a \ +_{12} \ b$, takes two numbers, $a$ and $b$, does... something... to them, and the result is a third number $c$. In this case, what $+_{12}$ does is adds the number and takes the remainder.

$10 +_{12} 10 = $ the remainder of $10 + 10$ divided by $12 = $ the remainder of $20$ when divided by $12 = 8$.

This is called modular arithmetic and it's applications and usefulness should be clear for systems that are period rather than infinite. It's very useful for discussing number theory and divisiblity of integers.

For example: the familiar "rule of three" that if you add up the digits of a number, and you add up the digits of the results of that, and you add up the digits of that, then the number is divisible by $3$ if you end up with $3,6,$ or $9$ and not otherwise; becomes easy to state and prove. You don't have to repeat "and then add those digits, and then add those" because $+_3$ is about remainders and the result will be a single digit. So the statement is "A number is divisible by $3$ if and only if the $+_3$ of its digits is $0$". And the proof is: $10_3 = 9 +_3 1 = 1$ so $10 \times_3 k = k_3$. So if if $n = d_0 + 10*d_1 + 10^2*d_2 + ..... + 10^n*d_n$ then $n_3 = d_0 +_3 + 10 \times_3 d_1 + 10^2\times_3 d_2 + ..... + 10^n\times_3 d_n = d_0 +_3 d_1 +_2 d_2 +_3 .... +_3 d_n$

fleablood
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