Question

Question: Can you please break this proof down to a level where a high school student can understand it? Thank you.

Example:

Prove that the product of n successive integers is always divisible by n!

If the product of n successive integers is always divisible by n!, then we would only need to prove it is true for positive integers. If one of the integers in the product is zero, it will always be true. If the integers are negative, n! will divide by their absolute value.

Proof by contradiction:

If there is a number of n successive positive integers whose product is not divisible by n!, then we can choose the smallest and call it N. N must be greater than 2 because the product of any two successive integers is always even. Therefore, there must be an integer, m, such that (m+1)(m+2)...(m+N) is not divisible by N! Of these numbers, m, let M be the smallest. M must be positive since N! is divisible by N!. So, we are supposing that (M +1)(M+2) …(M+N) is not divisible by N!

(M+1)(M+2) …(M+N-1)(M+N) = M[(M+1)(M+2)...(M+N-1)]+N[(M+1)(M+2)...(M+N-1)]

Using our choice of M, n! divides into M[(M+1)(M+2)...(M+N-1)]. Using our choice of N, (N-1)! divides (M+1)(M+2)...(M+N-1) and therefore N! divides N[(M+1)(M+2)...(M+N-1)].

In combination, N! Divides the right side of the last equation. This contradiction establishes the result.

Answer 1