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.