How to symbolize impossible discrete logarithm?

Question

The task is to find $k$ such that

$$2^k\mod 14 = 12 $$

The output is a cycle of $4$, $8$ and $2$ making this impossible. What is the correct symbol to claim task an impossibility/invalid?

$$2^k \mod 14 ≠12 $$

is the best I got.

NB: Answer is handwritten.

Whatever you want it to be in your context. IIRC old-timer C-programmers defaulted to -1. — Jyrki Lahtonen, Aug 28 '15 at 21:36
I'd say "$2^k\not\equiv12$ modulo $14$ for all $k$". Or with more symbols: $\forall k,2^k\not\equiv12\mod{14}$. — 2'5 9'2, Jun 04 '16 at 19:33

score 1 · Answer 1 · answered Aug 28 '15 at 21:37

1

$\{k \in \mathbb Z \mid 2^k \equiv 12 \mod 14 \} = \emptyset$

answered Aug 28 '15 at 21:37

Robert Israel

What does | mean here? I can't google it. – Manumit Aug 28 '15 at 21:46
@Manumit "with the property". – principal-ideal-domain Aug 28 '15 at 21:58
Use $:$ instead of $\mid$ if you prefer. – Robert Israel Aug 28 '15 at 22:04
Thank you, I understand this now. Although this may be the more correct one I find the nonexistence solution a cleaner concept. Especially as ∅, or void, seems a bit ambigious compared to non-existant. Correct me if the solution above is wrong in any way. – Manumit Aug 28 '15 at 22:15

score 1 · Accepted Answer · answered Aug 28 '15 at 21:37

1

I would write this as $$ \nexists k: (2^k \equiv 12 \mod 14) $$

answered Aug 28 '15 at 21:37

Alex Meiburg

Can you explain what that first symbol means? I can't even copy and google with correct formatting. – Manumit Aug 28 '15 at 21:46
1

The first symbol is most commonly seen as $\exists$, or "exists". For example, $\exists x: (x^2 = 4)$ means "There is some x, such that x squared is 4." This true because x could be 2. With the stroke, this is read as "not exists", and means "There is no k such that...". If you wanted to be extremely formal you could specify which domain you are drawing k from ("There is integer k...") but I expect that this will be clear from context. – Alex Meiburg Aug 28 '15 at 22:04
Is there supposed to be an intentional double spacing between 12 and mod? – Manumit Aug 28 '15 at 22:28
Ah, not really. That's just how LaTeX presents the \mod operator by default. You can write it like that, or single-spaced, or with the 2^k mod 14 = 12 notation you had earlier. – Alex Meiburg Aug 28 '15 at 22:32
Although the congruence symbol is awesome I went for ∄k: (2^k%14=12) – Manumit Aug 28 '15 at 22:36

2 Answers2