If L1- an- n- 0 and L2- bn- n- 0 - consider I- L1- L2 is a regular languageII- L1- L2 - anbn- n- 0 Which one of the following is CORRECT-

L_1={ a^n| n≥ 0 } L_2={ b^n| n≥ 0 } I. nbsp; L_1.L_2={ a^m b^n| m, n≥ 0 } = a^*b^* is regular II. L_1.L_2≠{ a^nb^n| n≥ 0 } ⇒ II is not correct. ∴ Only I ...

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If L1= an| n≥...

Question

If $L_{1} = {a^{n} ∣ n \geq 0}$ and $L_{2} = {b^{n} ∣ n \geq 0}$ , consider

I. $L_{1} \cdot L_{2}$ is a regular language

II. $L_{1} \cdot L_{2} = {a^{n} b^{n} ∣ n \geq 0}$

Which one of the following is CORRECT?

Neither I not II

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Only II

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Both I and II

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Only I

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Solution

The correct option is D Only I
$L_{1} = {a^{n} ∣ n \geq 0}$

$L_{2} = {b^{n} ∣ n \geq 0}$

I. $L_{1} . L_{2} = {a^{m} b^{n} ∣ m, n \geq 0} = a^{} b^{}$ is regular

II. $L_{1} . L_{2} \neq {a^{n} b^{n} ∣ n \geq 0}$

$\Rightarrow I I$ is not correct.

$∴$ Only I is correct.

Suggest Corrections

Similar questions

Q. Consider two languages,

L_{1} a n d L_{2}

Which of the following correctly represents

L_{1} \cdot L_{2}

, where

" \cdot "

is the concatenation operation?

L_{1} = {a^{n} | n \geq 0}

and

L_{2} = {b^{n} | n \geq 0}

Q. Consider the two languages

L_{1} and L 2

where

L_{1}

is regular and

L_{2}

D C F L .

Then the language

L_{3} = (L_{1} \cap L_{2}^{*})^{'}

is/are true about

L_{3}

Q. Let L =

L_{1} \cap L_{2}

, where

L_{1} a n d L_{2}

are languages as defined below:

L_{1}

a^{m} b^{m} c a^{n} b^{n} | m, n \geq 0

}

L_{2}

a^{i} b^{j} c^{k} | i, j, k \geq 0

}
Then L is

Q. Given the following two languages:

$L_{1} = {a^{n} b^{n} | n \geq 0, n \neq 100}$

$L_{2} = {w \in {a, b, c} * | n_{a} (w) = n_{b} (w) = n_{c} (w)}$

Which of the following options is correct?

Q. If

L_{1} \leq L_{2}

and

L_{2}

is undecidable then

L_{1}

is:

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If L1={an∣n≥0} and L2={bn∣n≥0}, consider I. L1⋅L2 is a regular language II. L1⋅L2={anbn∣n≥0} Which one of the following is CORRECT?

The correct option is D Only IL1={an∣n≥0} L2={bn∣n≥0} I. L1.L2={ambn∣m,n≥0}=a∗b∗ is regular II. L1.L2≠{anbn∣n≥0} ⇒II is not correct. ∴ Only I is correct.

If $L_{1} = {a^{n} ∣ n \geq 0}$ and $L_{2} = {b^{n} ∣ n \geq 0}$ , consider

I. $L_{1} \cdot L_{2}$ is a regular language

II. $L_{1} \cdot L_{2} = {a^{n} b^{n} ∣ n \geq 0}$

Which one of the following is CORRECT?

The correct option is D Only I
$L_{1} = {a^{n} ∣ n \geq 0}$

$L_{2} = {b^{n} ∣ n \geq 0}$

I. $L_{1} . L_{2} = {a^{m} b^{n} ∣ m, n \geq 0} = a^{} b^{}$ is regular

II. $L_{1} . L_{2} \neq {a^{n} b^{n} ∣ n \geq 0}$

$\Rightarrow I I$ is not correct.

$∴$ Only I is correct.