Let r: Nat →Nat. Here are three conditions that r might satisfy: (1) For every i, j ∈N a t, r(i+

Let r: Nat →Nat. Here are three conditions that r might...

Key Concepts

Additive Function

An additive function is one that preserves the binary operation of addition; that is, for every pair of natural numbers i and j, the function satisfies r(i+j) = r(i) + r(j). This property is central to understanding linear mappings between arithmetic structures and is a key aspect of homomorphisms in algebra.

Recursive Definition

A recursive definition specifies the initial value, such as r(0) = 0, and provides a rule to obtain subsequent values — in this case, r(k+1) = r(k) + r(1) for each natural number k. This approach to defining functions emphasizes the construction of the entire function from its base case and recurrence relation.

Mathematical Induction

Mathematical induction is a proof technique used to establish that a property holds for all natural numbers. It involves proving the base case and then showing that if the property holds for an arbitrary natural number, it must also hold for its successor. This method is fundamental when working with recursive definitions and propositions about natural numbers.

Equivalence of Definitions

Equivalence of definitions involves proving that different formulations — such as additive, recursive, and closed-form (linear) representations of a function — are logically identical. Establishing such equivalences highlights that various approaches to defining a function lead to the same family of functions, demonstrating deep structural relationships in mathematics.

Existence Through Construction

Exhibiting a concrete example, such as defining r(n) = n or more generally r(n) = n · c for some constant c, serves to demonstrate that functions meeting these criteria do indeed exist. This constructive proof not only reinforces the theoretical equivalences but also provides tangible evidence of the concepts discussed.

Transcript

00:02 As a given here, mn belong to natural numbers, abcd belong to integers, ef belong to integers.

00:09 So statement one says that we have m equals a squared plus b square and equals c squared plus d square.

00:15 We're going to check whether we have m into n is e square plus f square.

00:19 That we need to check here.

00:20 Segment 2 is here, when you just check here, this relation is correct, is z1, z2 greater than 0.

00:27 Z1, z2 are both complex numbers.

00:29 So we start with here, statement 1.

00:30 So we have here m is a square plus b square.

00:38 That means m is given as a plus iotari in mode that will give root a plus b square and it's squared.

00:48 That will give a square plus b square.

00:49 Normally we can say n equals c plus id square.

01:00 Now i want to find out here mn for the relation to prove it's equal to e square plus f square.

01:06 So we have m n that equal so just multiply this so it's the property here we have more a b equals note a times mod b so it's coming out to be we get here a more a plus iota b times c plus id not square and simplify now inside the mode so we get ac plus iota a d plus iota bc then we get plus iota square bd.

01:42 That will give negative bd square.

01:46 So we get here and n that equals we have a c negative bd plus iota a d plus bc.

01:59 So from here, let's find out that gives a square here.

02:05 That gives a c negative bd whole square that we get here plus we have a d plus bc whole square.

02:17 Is taken as e square plus we have f square where e and f both belong to integers because every have a bcd belong to integers so enf belong to integers so we prove that mn equals e square plus f square so we have here statement one is correct next we observe here statement two that is given as mode of z1 plus z2 equals more of z1 plus mode of z2 if we have z1 z2 greater than zero so let's check this version as correct.

02:49 Let's go with that.

02:50 So we have mod of z1 plus z2.

02:58 Is that equals we have mod v1 plus more v2.

03:06 So we're going to square both sides, doing it square and the square here.

03:11 So we get more v1 square plus we get mode z2 square plus we will get v1 z2 conjugate plus zv1 conjugate 3.

03:21 That is given as two times real v1 z to conjugate because we have z1 they to conjugate plus z1 conjugate z2.

03:34 So when we add then the two complex numbers, one is complex number and its conjugate.

03:39 We always get two times the real value.

03:43 It equals we have more z1 square plus mode z2 square plus two times.

03:49 We have mode v1, v2...