Yes, n^2 - n is always an integer if n is an integer. Here's why:

1. The product of any two integers is always an integer. This is a fundamental property of integers. So, n^2 (which is n*n) is an integer.

2. The difference of any two integers is always an integer. This is another fundamental property of integers. So, n^2 - n (which is an integer minus an integer) is also an integer.

Therefore, if n is an integer, then n^2 - n is always an integer.

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Yes, n^2 - n is always an integer if n is an integer. Here's why:

1. The product of any two integers is always an integer. This is a fundamental property of integers. So, n^2 (which is n*n) is an integer.

2. The difference of any two integers is always an integer. This is another fundamental property of integers. So, n^2 - n (which is an integer minus an integer) is also an integer.

Therefore, if n is an integer, then n^2 - n is always an integer.

Knowee AI · Accepted Answer

Yes, n^2 - n is always an integer if n is an integer. Here's why:

1. The product of any two integers is always an integer. This is a fundamental property of integers. So, n^2 (which is n*n) is an integer.

2. The difference of any two integers is always an integer. This is another fundamental property of integers. So, n^2 - n (which is an integer minus an integer) is also an integer.

Therefore, if n is an integer, then n^2 - n is always an integer.

Suppose n is an integer. Is n2 n always aninteger? Justify your answer.

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