Suppose that a natural number n is neither a prime nor a perfect square. Then, there exist natural numbers p and q such that p < n, q < n , p ≠ q and pq = n.
Now, note that (n - 1)! = (n - 1)(n - 2)(n - 3)...(3)(2)(1).
Since p < n and q < n, p = (n - k) and q = (n - l) where k ≠ l and k, l ∈ {∈1, 2, 3, ... , n - 1}. But we know that (n - k)(n - l) | (n - 1)!
Hence, pq | (n - 1)! and therefore n | (n - 1)!
Hence, any natural number n is either prime, a complete square or divides (n - 1)!, as required.
Suppose that a natural number n is neither a prime nor a perfect square. Then, there exist natural numbers p and q such that p < n,
q < n, p ≠ q, and pq = n.
Any natural number n is either prime, a complete square, or divides
(n - 1)! as required.
Suppose that a natural number n is neither a prime nor a perfect square. Then, there exist natural numbers p and q such that p < n, q < n, p ≠ q, and pq = n.
(n - 1)! = (n - 1)(n - 2)(n - 3)... (3)(2)(1).
Since p < n and q < n, p = (n - k) and q = (n - l) where k ≠ l and
k, l ∈ {∈ 1, 2, 3, ... , n - 1}.
But we know that (n - k)(n - l) | (n - 1)!
Hence, pq | (n - 1)! and
therefore, n | (n - 1)!
Hence, any natural number n is either prime, a complete square, or divides (n - 1)! as required.
Learn more about any natural number n, either n is prime, n is a perfect square, or n|(n-1)!
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