CSE 311: Section 3

Task 1 – This is Really Mod

Let nn and mm be positive integers.

Consider the following claim: for any integers aa and bb, if amba \equiv_m b, then anba \equiv_n b.

  1. Write a formal proof that the claim holds, given that n|mn\ |\ m.

    Hint: the claim to prove is ab((amb)(anb))\forall a\, \forall b\, ((a \equiv_m b) \to (a \equiv_n b)). The fact we are given is that n|mn\ |\ m.

  2. Translate your formal proof to an English proof.

Task 2 – Extended Euclidean Algorithm Practice

  1. Find the multiplicative inverse of yy of 7 mod 33. That is, find yy such that 7y1(mod 33)7y \equiv 1 (\text{mod } 33), You should use the extended Euclidean Algorithm. Your answer should be in the range 0y<330 \leq y < 33.

  2. Now solve 7z27z \equiv 2 (mod 33) for all of its integers solutions zz.

Task 3 – Induction with Equality

For all nn \in \mathbb{N}, prove by induction that i=0ni2=16n(n+1)(2n+1)\displaystyle \sum_{i=0}^n i^2 = \frac{1}{6}n(n+1)(2n+1).

Task 4 – Strong Induction

Consider the function a(n)a(n) defined for n1n \geq 1 recursively as follows. a(1)=1a(1) = 1 a(2)=3a(2) = 3 a(n)=2a(n1)a(n2) for n3a(n) = 2a(n-1) - a(n-2) \text{ for } n \geq 3 Use strong induction to prove that a(n)=2n1a(n) = 2n - 1 for all integers n1n \geq 1.