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John L.
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Proof that a Prove the language of numbers divisible by 3 is not regular using pumping lemma

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Raphael
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babou
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I have a language $L$ that I think is not regular:

$L = \{w\in \{0,1,...,9\}^* \enspace | \enspace w \enspace \text{is a decimal representation of a number divisible by 3}\}$$L = \{w\in \{0,1,...,9\}^* \; | \enspace w \enspace \text{is a decimal representation of a number divisible by 3}\}$

I'm using pumping lemma in my proof. My main main idea is the following. I choose $w = 3$, according to the pumping lemma $w = xyz$ with $xy^iz$ where $i \in \mathbb{N}_0$. But for $i=0$ we have $\varepsilon$ which is not in the language ($\varepsilon$ is not divisible by 3). Therefore the language is not regular.

Is my idea right?

I have a language $L$ that I think is not regular:

$L = \{w\in \{0,1,...,9\}^* \enspace | \enspace w \enspace \text{is a decimal representation of a number divisible by 3}\}$

I'm using pumping lemma in my proof. My main main idea is the following. I choose $w = 3$, according to the pumping lemma $w = xyz$ with $xy^iz$ where $i \in \mathbb{N}_0$. But for $i=0$ we have $\varepsilon$ which is not in the language ($\varepsilon$ is not divisible by 3). Therefore the language is not regular.

Is my idea right?

I have a language $L$ that I think is not regular:

$L = \{w\in \{0,1,...,9\}^* \; | \enspace w \enspace \text{is a decimal representation of a number divisible by 3}\}$

I'm using pumping lemma in my proof. My main main idea is the following. I choose $w = 3$, according to the pumping lemma $w = xyz$ with $xy^iz$ where $i \in \mathbb{N}_0$. But for $i=0$ we have $\varepsilon$ which is not in the language ($\varepsilon$ is not divisible by 3). Therefore the language is not regular.

Is my idea right?

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Buddy
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