Hello I´m new to formal language and searching the solution for the following task:
Language: $L = \{0^{2i+1}|i\in\mathbb{N}_0\}$
Alphabet: $\Sigma = \{0\}$
I'm searching the resultion (sic) for: $\Sigma^+\setminus L$.
Hello I´m new to formal language and searching the solution for the following task:
Language: $L = \{0^{2i+1}|i\in\mathbb{N}_0\}$
Alphabet: $\Sigma = \{0\}$
I'm searching the resultion (sic) for: $\Sigma^+\setminus L$.
It seems like you're not certain on the terminology, so I'll try to explain the notation further.
The possible characters in the alphabet ($\Sigma$) is just zero. By definition, $\Sigma^+$ is all possible strings of alphabet characters with length greater than zero. So elements of $\Sigma^+$ are $\{0,00,000,\ldots\}$.
Now, in formal languages, $a^b$ means $\underbrace{a\ldots a}_{b \text{ times}}$. So $L$ consists of all strings that look like $\underbrace{0\ldots 0}_{2i+1 \text{ times}}$ for $i \in \mathbb{N}_0$.
From here, try writing out what strings in $L$ look like in the same way that I wrote out strings in $\Sigma^+$. Then to find $\Sigma^+ \setminus L$, look at what strings are in $\Sigma^+$ but are not in $L$.
Hint 1: Write out some words in $\Sigma^+ \setminus L$ and try to find a pattern.
Hint 2: Write out the condition that a word be in $L$. Then negate it.
This task (question) is a little bit prickly:
0$^{2i+1}$ means every odd natural number. So, the language is made up of 0$^1$ and 0$^3$ and 0$^5$... to infinite. However, zero exponentiate with a natural number is always zero.
ok, but our alphabet $\Sigma$ has only one entry, the zero. And the alphabet $\Sigma^+$ \ L (\ means without) is our alphabet without zero (-> 0$^{2i+1}$ = 0, ever), so it's empty, because $\Sigma$ has only 0.
That's what I would give to an answer, but I'm not sure. Might be it's right, may be it's rubbish