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.
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
since we have the area of the front side, to get its volume we can simple get the product of the area and the length, let's firstly change the mixed fractions to improper fractions.
