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30 May, 04:05

What is wrong with the following proof that for every integer n, there is an integer k such that n < k < n + 2? Suppose n is an arbitrary integer. Therefore k = n + 1.

a) Nothing.

b) The "therefore" phrase is illogical. The fact that n is an integer does not force us to define k as n + 1.

c) The proof writer mentally assumed the conclusion. He wrote "suppose n is an arbitrary integer", but was really thinking "suppose n is an arbitrary integer, and suppose that for this n, there exists an integer k that satisfies n < k < n+2." Under those assumptions, it follows indeed that k must be n + 1, which justifies the word "therefore": but of course assuming the conclusion destroyed the validity of the proof.

d) Since the theorem required the condition that k satisfies n < k < n + 2, a complete proof should at least mention that the chosen k satisfies that inequality, even if there is nothing to prove algebraically.

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  1. 30 May, 04:28
    0
    c) The proof writer mentally assumed the conclusion. He wrote "suppose n is an arbitrary integer", but was really thinking "suppose n is an arbitrary integer, and suppose that for this n, there exists an integer k that satisfies n < k < n+2." Under those assumptions, it follows indeed that k must be n + 1, which justifies the word "therefore": but of course assuming the conclusion destroyed the validity of the proof.

    Step-by-step explanation:

    when we claim something as a hypothesis we can only conclude with therefore at the end of the proof. so assuming the conclusion nulify the proof from the beginning
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