Introduction to Advanced Mathematics by Randall R. Holmes

By Randall R. Holmes

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1). ). 1 Example n ≥ 1 we have Use induction to prove that for every integer n with 1 + 2 + 3 + ··· + n = n(n + 1) . 2 Proof We use induction. For an integer n, let P (n) denote the indicated equality. (i) The base case P (1) is the equality 1= 1(1 + 1) , 2 which holds. ). Adding n to both sides of this equation and then using simple algebra, we get (n − 1)((n − 1) + 1) +n 2 (n − 1)n + 2n = 2 n(n + 1) , = 2 1 + 2 + 3 + · · · + (n − 1) + n = so P (n) holds. By induction, P (n) holds for every integer n with n ≥ 1.

We have, 4n − 1 = 4(4n−1 − 1) + 3 = 4(3k) + 3 = 3(4k + 1), implying 3 | (4n − 1). This shows that P (n) holds. By induction, P (n) holds for every integer n with n ≥ 0.

We have x ≥ 2, so 10 − 5x ≤ 10 − 5(2) = 0. Therefore, y + 7 = (3 − 5x) + 7 = 10 − 5x ≤ 0. ” The meaning here is that n ≤ 5 being true forces 2n − 1 ≤ 9 to be true, and this is the meaning of the original if-then statement as well. This new formulation allows a second method of proof, which uses a string of implications: Statement: Proof : If n ≤ 5, then 2n−1 ≤ 9. ) We have n≤5 =⇒ 2n ≤ 10 (multiply by 2) =⇒ 2n − 1 ≤ 9 (subtract 1). Discussion: We have shown the usual method for justifying each step to the right (but the justifications are not really necessary in this case).

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