Question

Prove that (n² - 4n + 5) is positive for all integers ((n ∈ ℤ)).

Answer 1

To prove that (n² - 4n + 5) is positive for all integers, we complete the square to show it as the sum of a non-negative square and a positive number. As this sum will always be positive, it confirms that the expression in question is indeed positive for all integer values of n.

Step-by-step explanation:

To prove that the expression (n² - 4n + 5) is positive for all integers, we can use the method of completing the square. Completing the square involves transforming the quadratic equation into a perfect square trinomial and a constant. The expression is already in standard quadratic form, where the coefficient of the n² term is 1.

Let's complete the square for the expression:

1. Divide the coefficient of the n term by 2, which is -4/2 = -2.
2. Square this result, which gives us (-2)² = 4.
3. Add and subtract this square within the expression. This gives us (n² - 4n + 4 + 1). Notice that we've added 4 and subtracted 4 to keep the expression equivalent to the original.
4. We can rewrite this as ((n - 2)² + 1), which is the sum of a square and a positive number.

A square of any real number is always non-negative, and the sum of a non-negative number and a positive number is always positive. Therefore, (n - 2)² is non-negative, and when adding 1, the result will be strictly positive. Since this holds true for all values of n, it proves the original expression is positive for all integers n.