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Use the pumping lemma to show that the following languages are not regular languages L = {an bm | n = 2m}

alaa raed
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Choose a string a^2p b^p. The pumping lemma says we can write this as w = uvx such that |uv| <= p, |v| < 0 and for all natural numbers n, u(v^n)x is also a string in the language. Because |uv| <= p, the substring uv of w consists entirely of instances of the symbol a. Pumping up or down by choosing a value for n other than one guarantees that the number of a's in the resulting string will change, while the number of b's stays the same. Since the number of a's is twice the number of b's only when n = 1, this is a contradiction. Therefore, the language cannot be regular.

Patrick87
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L={anbm|n=2m} Assume that L is regular Language Let the pumping length be p L={a2mbm} Since |s|=3m > m (total string length) take a string s S= aaaaa...aabbb....bbb (a2mbm times taken) Then, u = am-1 ; v= a ; w= ambm. && |uv|<=m Now If i=2 then S= am-1 a2 ambm = a2m-1bm Since here we are getting an extra a in the string S which is no belong to the given language a2mbm our assumption is wrong Therefore it is not a regular Language

Netaji
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