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Proof that if n is an integer, ! is even if and only if n is even?

User Thinzar
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Final answer:

The statement is incorrect as n! is always even for all integers n > 1, regardless of whether n is even or odd, because the number 2 is included as a factor.

Step-by-step explanation:

The statement 'if n is even, n! is even' can be proven by considering the definition of factorial and the properties of even numbers. If n is an even integer, then n! (n factorial) is the product of all positive integers from 1 to n. Since the sequence of integers from 1 to n includes the number n itself and n is even, at least one of the factors in the product is even. As a result, the product (n!) must also be even. Conversely, if n is odd, n! will also include the factor 2 because all factorials greater than 2 include the number 2 as one of the factors, making the product even in this case as well. Therefore, the statement is not correct; n! is not even if and only if n is even because n! will always be even for all integers n > 1.

User Durul Dalkanat
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