Let L1 consist of strings that have n zeros followed
by 2n ones where n is a natural number.<p>

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2a: There is no DFA that recongizes L1. We prove this using a proof by
    contradiction.  Suppose that there were such a DFA M that has k states. 
    Consider the string x = 0^{k} 1^{2k} (i.e. k zeros followed by 2k ones).
    Since x is in L1 the DFA M must end in an accepting state.  Furthermore, 
    since M has k states and k+1 states have been entered (some could be 
    repeated) when simulating it on the k zeros, by the pigeon hole principle 
    some state, say state i, has been visited at least twice when simulating M
    on the k zeros.  Consider the string x' obtained by skipping the cycle 
    from state i back to state i.  Notice that we remove some of the 0s from x
    without making any other changes.  Hence the number of 0s in x' is less 
    than two times the number of 1s in x'.  Hence x' is not in L1.  However, 
    M on input x' ends in the same state as x (which was an accepting state)
    and hence M incorrectly outputs that x' is in L1.  This contradicts that
    M correctly recongizes L1.

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