proving that $a^6 in L$ (tricky) (pumping lemma)
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i've encountered a quite tricky question that i don't understand, and would appreciate your help with:
it is said that by the pumping lemma, for a certain L there exists(guarenteed) n=2. also known that $aa in L$. prove that $a^6 in L$
so from what i understand, because it is guranteed that for some L, because of the pumping lemma, n=2, it means that for a word $w in L$, w can be represented as such: w=abc, so that $|ab| leq n$, $|b| geq 1$ and for every $i in N$: $ab^ic in L$. so if $aa in L$, then the boundary(first condition) is n=2 as given, but does it mean that it can only one letter, so that $a^6 in L$? basically, i think that i should be using the third condition(for every $i in N$ $ab^ic in L$, but how can i ommit a and c so i could show that i=6, i.e $a^6 in L$
tried to explain the question and my way of thinking the best i could. how can i show that $a^6 in L$?
computer-science automata
edited Dec 2 at 15:29
saulspatz
13.5k21327
asked Dec 2 at 14:48
mathnoobie
93
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mathnoobie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
0
down vote
favorite
i've encountered a quite tricky question that i don't understand, and would appreciate your help with:
it is said that by the pumping lemma, for a certain L there exists(guarenteed) n=2. also known that $aa in L$. prove that $a^6 in L$
so from what i understand, because it is guranteed that for some L, because of the pumping lemma, n=2, it means that for a word $w in L$, w can be represented as such: w=abc, so that $|ab| leq n$, $|b| geq 1$ and for every $i in N$: $ab^ic in L$. so if $aa in L$, then the boundary(first condition) is n=2 as given, but does it mean that it can only one letter, so that $a^6 in L$? basically, i think that i should be using the third condition(for every $i in N$ $ab^ic in L$, but how can i ommit a and c so i could show that i=6, i.e $a^6 in L$
tried to explain the question and my way of thinking the best i could. how can i show that $a^6 in L$?
computer-science automata
edited Dec 2 at 15:29
saulspatz
13.5k21327
asked Dec 2 at 14:48
mathnoobie
93
New contributor
mathnoobie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Think of what $b$ can be in your situation for the word $aa$. Remember that if $n=2$ is the pumping constant, the word $aa$ satisfies the initial conditions to be pumped.
– NL1992
Dec 2 at 15:30
i might be very confused, but b should be a? really confused and would appreciate your help with it
– mathnoobie
Dec 2 at 20:24
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
i've encountered a quite tricky question that i don't understand, and would appreciate your help with:
it is said that by the pumping lemma, for a certain L there exists(guarenteed) n=2. also known that $aa in L$. prove that $a^6 in L$
so from what i understand, because it is guranteed that for some L, because of the pumping lemma, n=2, it means that for a word $w in L$, w can be represented as such: w=abc, so that $|ab| leq n$, $|b| geq 1$ and for every $i in N$: $ab^ic in L$. so if $aa in L$, then the boundary(first condition) is n=2 as given, but does it mean that it can only one letter, so that $a^6 in L$? basically, i think that i should be using the third condition(for every $i in N$ $ab^ic in L$, but how can i ommit a and c so i could show that i=6, i.e $a^6 in L$
tried to explain the question and my way of thinking the best i could. how can i show that $a^6 in L$?
computer-science automata
edited Dec 2 at 15:29
saulspatz
13.5k21327
asked Dec 2 at 14:48
mathnoobie
93
New contributor
mathnoobie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
i've encountered a quite tricky question that i don't understand, and would appreciate your help with:
it is said that by the pumping lemma, for a certain L there exists(guarenteed) n=2. also known that $aa in L$. prove that $a^6 in L$
so from what i understand, because it is guranteed that for some L, because of the pumping lemma, n=2, it means that for a word $w in L$, w can be represented as such: w=abc, so that $|ab| leq n$, $|b| geq 1$ and for every $i in N$: $ab^ic in L$. so if $aa in L$, then the boundary(first condition) is n=2 as given, but does it mean that it can only one letter, so that $a^6 in L$? basically, i think that i should be using the third condition(for every $i in N$ $ab^ic in L$, but how can i ommit a and c so i could show that i=6, i.e $a^6 in L$
tried to explain the question and my way of thinking the best i could. how can i show that $a^6 in L$?
computer-science automata
computer-science automata
edited Dec 2 at 15:29
saulspatz
13.5k21327
asked Dec 2 at 14:48
mathnoobie
93
New contributor
mathnoobie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 2 at 15:29
saulspatz
13.5k21327
asked Dec 2 at 14:48
mathnoobie
93
New contributor
mathnoobie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 2 at 15:29
saulspatz
13.5k21327
edited Dec 2 at 15:29
saulspatz
13.5k21327
edited Dec 2 at 15:29
saulspatz
13.5k21327
13.5k21327
asked Dec 2 at 14:48
mathnoobie
93
New contributor
mathnoobie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 2 at 14:48
mathnoobie
93
asked Dec 2 at 14:48
mathnoobie
93
93
New contributor
mathnoobie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
mathnoobie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
mathnoobie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Think of what $b$ can be in your situation for the word $aa$. Remember that if $n=2$ is the pumping constant, the word $aa$ satisfies the initial conditions to be pumped.
– NL1992
Dec 2 at 15:30
i might be very confused, but b should be a? really confused and would appreciate your help with it
– mathnoobie
Dec 2 at 20:24
add a comment |
1
Think of what $b$ can be in your situation for the word $aa$. Remember that if $n=2$ is the pumping constant, the word $aa$ satisfies the initial conditions to be pumped.
– NL1992
Dec 2 at 15:30
i might be very confused, but b should be a? really confused and would appreciate your help with it
– mathnoobie
Dec 2 at 20:24
1
1
Think of what $b$ can be in your situation for the word $aa$. Remember that if $n=2$ is the pumping constant, the word $aa$ satisfies the initial conditions to be pumped.
– NL1992
Dec 2 at 15:30
Think of what $b$ can be in your situation for the word $aa$. Remember that if $n=2$ is the pumping constant, the word $aa$ satisfies the initial conditions to be pumped.
– NL1992
Dec 2 at 15:30
i might be very confused, but b should be a? really confused and would appreciate your help with it
– mathnoobie
Dec 2 at 20:24
i might be very confused, but b should be a? really confused and would appreciate your help with it
– mathnoobie
Dec 2 at 20:24
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Ok, let's fix some easier notation:
We say that a language $L$ satisfies the pumping lemma for $p>0$ if for any word $win L$ with $|w|geq p$ we have that $w=xyz$ with the properties:
$$1. pgeq |xy|, |y|>0
$$
$$2. forall iin Bbb N: xy^izin L
$$
So in our case, $y$ (which is being 'pumped') would either be $a$ or $aa$.
answered Dec 2 at 21:09
NL1992
718
New contributor
NL1992 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
so basically after asserting that y is either a or aa, is it sufficient to say that because of this and the properties of the pumping lemma, $a^6 in L$?
– mathnoobie
Dec 2 at 21:34
Yes, and you can divide to cases and prove it directly if you want to.
– NL1992
Dec 2 at 21:38
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Ok, let's fix some easier notation:
We say that a language $L$ satisfies the pumping lemma for $p>0$ if for any word $win L$ with $|w|geq p$ we have that $w=xyz$ with the properties:
$$1. pgeq |xy|, |y|>0
$$
$$2. forall iin Bbb N: xy^izin L
$$
So in our case, $y$ (which is being 'pumped') would either be $a$ or $aa$.
answered Dec 2 at 21:09
NL1992
718
New contributor
NL1992 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
so basically after asserting that y is either a or aa, is it sufficient to say that because of this and the properties of the pumping lemma, $a^6 in L$?
– mathnoobie
Dec 2 at 21:34
Yes, and you can divide to cases and prove it directly if you want to.
– NL1992
Dec 2 at 21:38
add a comment |
up vote
0
down vote
Ok, let's fix some easier notation:
We say that a language $L$ satisfies the pumping lemma for $p>0$ if for any word $win L$ with $|w|geq p$ we have that $w=xyz$ with the properties:
$$1. pgeq |xy|, |y|>0
$$
$$2. forall iin Bbb N: xy^izin L
$$
So in our case, $y$ (which is being 'pumped') would either be $a$ or $aa$.
answered Dec 2 at 21:09
NL1992
718
New contributor
NL1992 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
so basically after asserting that y is either a or aa, is it sufficient to say that because of this and the properties of the pumping lemma, $a^6 in L$?
– mathnoobie
Dec 2 at 21:34
Yes, and you can divide to cases and prove it directly if you want to.
– NL1992
Dec 2 at 21:38
add a comment |
up vote
0
down vote
up vote
0
down vote
Ok, let's fix some easier notation:
We say that a language $L$ satisfies the pumping lemma for $p>0$ if for any word $win L$ with $|w|geq p$ we have that $w=xyz$ with the properties:
$$1. pgeq |xy|, |y|>0
$$
$$2. forall iin Bbb N: xy^izin L
$$
So in our case, $y$ (which is being 'pumped') would either be $a$ or $aa$.
answered Dec 2 at 21:09
NL1992
718
New contributor
NL1992 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ok, let's fix some easier notation:
We say that a language $L$ satisfies the pumping lemma for $p>0$ if for any word $win L$ with $|w|geq p$ we have that $w=xyz$ with the properties:
$$1. pgeq |xy|, |y|>0
$$
$$2. forall iin Bbb N: xy^izin L
$$
So in our case, $y$ (which is being 'pumped') would either be $a$ or $aa$.
answered Dec 2 at 21:09
NL1992
718
New contributor
NL1992 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 2 at 21:09
NL1992
718
New contributor
NL1992 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 2 at 21:09
NL1992
718
answered Dec 2 at 21:09
NL1992
718
718
New contributor
NL1992 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
NL1992 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
NL1992 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
so basically after asserting that y is either a or aa, is it sufficient to say that because of this and the properties of the pumping lemma, $a^6 in L$?
– mathnoobie
Dec 2 at 21:34
Yes, and you can divide to cases and prove it directly if you want to.
– NL1992
Dec 2 at 21:38
add a comment |
so basically after asserting that y is either a or aa, is it sufficient to say that because of this and the properties of the pumping lemma, $a^6 in L$?
– mathnoobie
Dec 2 at 21:34
Yes, and you can divide to cases and prove it directly if you want to.
– NL1992
Dec 2 at 21:38
so basically after asserting that y is either a or aa, is it sufficient to say that because of this and the properties of the pumping lemma, $a^6 in L$?
– mathnoobie
Dec 2 at 21:34
so basically after asserting that y is either a or aa, is it sufficient to say that because of this and the properties of the pumping lemma, $a^6 in L$?
– mathnoobie
Dec 2 at 21:34
Yes, and you can divide to cases and prove it directly if you want to.
– NL1992
Dec 2 at 21:38
Yes, and you can divide to cases and prove it directly if you want to.
– NL1992
Dec 2 at 21:38
add a comment |
mathnoobie is a new contributor. Be nice, and check out our Code of Conduct.
mathnoobie is a new contributor. Be nice, and check out our Code of Conduct.
mathnoobie is a new contributor. Be nice, and check out our Code of Conduct.
mathnoobie is a new contributor. Be nice, and check out our Code of Conduct.
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1
Think of what $b$ can be in your situation for the word $aa$. Remember that if $n=2$ is the pumping constant, the word $aa$ satisfies the initial conditions to be pumped.
– NL1992
Dec 2 at 15:30
i might be very confused, but b should be a? really confused and would appreciate your help with it
– mathnoobie
Dec 2 at 20:24