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- Apr 13, 2013

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- Thread starter evinda
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- #1

- Apr 13, 2013

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- Jan 30, 2012

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What would be the alternative: that the difference between one prime number and theso we see that the difference of one prime number from the next one is greater or equal to one...

In fact, it is given by the lemma; see point 2 here. More importantly, if the lemma does not state something explicitly, it does not follow that it is false.So,if $p+(|vy|+1)|vy|$ was the next prime after $p+|vy|^{2}$,$|vy|$ should be greater than one,something that is not given from the pumping lemma.

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- Apr 13, 2013

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Ok,I understand.....What would be the alternative: that the difference between one prime number and thenextone is zero?

In fact, it is given by the lemma; see point 2 here. More importantly, if the lemma does not state something explicitly, it does not follow that it is false.

- Jan 30, 2012

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- Apr 13, 2013

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Could I show it maybe like that?

If we add $i|vy|$ at the length of $s$,it must still belong in $L$.

So $p+i|vy|$ must be a prime number for each $i \geq 0$. For $i=p$ we have $p+p|vy|=p(1+|vy|)$. $p$ is a prime greater than 1 and $|vy|>1$, so $ p(1+|vy|)$ is not a prime since it is written as a product of two factors greater than 1.

- Jan 30, 2012

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What is $s$ and what is $p$?I have to show that the language $L=\{a^{k},\text{ k is a prime }\}$ is not context-free..I thought that I could show this,using the pumping lemma.I took the word $s^{p}$

I don't understand the phrase "add ... at the length of ...".and said that if we add $i|vy|$ at the length of $s$

Why is $p$ prime? In fact, whether $p$ is prime is irrelevant for the rest of the proof.So $p+i|vy|$ must be a prime number for each $i \geq 0$. For $i=p$ we have $p+p|vy|=p(1+|vy|)$. $p$ is a prime greater than 1

The variant of the lemma from Wikipedia only says $|vy|\ge1$.and $|vy|>1$

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- Apr 13, 2013

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Nice..thank you!!!I think this works. Very well! But I have several remarks about the presentation.

What is $s$ and what is $p$?

I don't understand the phrase "add ... at the length of ...".

Why is $p$ prime? In fact, whether $p$ is prime is irrelevant for the rest of the proof.

The variant of the lemma from Wikipedia only says $|vy|\ge1$.