Use mathematical induction to prove that each statement is true for every positive integer n. 2 is a factor of n² - n.

Verified step by step guidance

1. First, identify the statement to prove using mathematical induction: For every positive integer \(n\), \(2\) is a factor of \(n^{2} - n\). This means we want to show that \(n^{2} - n\) is divisible by \(2\) for all \(n \geq 1\).
2. Start with the base case: check if the statement holds for \(n = 1\). Substitute \(n = 1\) into the expression \(n^{2} - n\) and verify if the result is divisible by \(2\).
3. Assume the statement is true for some positive integer \(k\), i.e., assume \(2\) divides \(k^{2} - k\). This is called the induction hypothesis.
4. Using the induction hypothesis, prove the statement for \(n = k + 1\). Substitute \(k + 1\) into the expression to get \((k + 1)^{2} - (k + 1)\) and simplify it.
5. Show that the expression for \(n = k + 1\) can be written as a sum of terms, one of which is divisible by \(2\) by the induction hypothesis, and the other terms are also divisible by \(2\). This will complete the induction step and prove the statement for all positive integers \(n\).

Key Concepts

Mathematical Induction

Mathematical induction is a proof technique used to show that a statement holds for all positive integers. It involves two steps: proving the base case (usually n=1) is true, and then proving that if the statement holds for an arbitrary integer k, it also holds for k+1.

Divisibility and Factors

Understanding divisibility means recognizing when one integer is a factor of another. In this problem, showing that 2 is a factor of n² - n means proving that n² - n is always even, i.e., divisible by 2, for every positive integer n.

Algebraic Manipulation of Expressions

Algebraic manipulation involves rewriting expressions to reveal properties like factors or divisibility. For n² - n, factoring it as n(n - 1) helps identify that the product of two consecutive integers is always even, which is key to the proof.