A minimal element of a partially ordered set M is an x ∈M such that y ≦x implies y=x. Find all m

A minimal element of a partially ordered set M is an x ∈M...

Key Concepts

Partially Ordered Set

A partially ordered set (or poset) is a set together with a binary relation that is reflexive, antisymmetric, and transitive. This structure allows some, but not necessarily all, pairs of elements to be compared under the relation, which is a key aspect when analyzing order properties within the set.

Minimal Element

A minimal element in a poset is an element m such that there is no other element in the set that is strictly less than m. Formally, if any element x satisfies x ? m, then it must be that x = m. This concept is distinct from a least element, which would require that m is less than or equal to every other element in the set.

Order Relation

The order relation, often denoted as ?, establishes the framework for comparing elements in a poset. Understanding this relation is essential for defining and identifying properties like minimal elements, as it encodes the direct comparisons between elements within the set.

Transcript

00:01 Given any integer numbers l and m with l different from 0, prove that there are unique integers q and r such that m equal l times q plus r and r is greater than equal to 0 and less than the absolute value of l.

00:20 So we can recognize here the definition of the division of two integer numbers where we are dividing m over l, and for that reason, l got a bit different from zero.

00:37 And we will get a caution to a residue r.

00:42 And the key property for the residue is to be positive or zero and less than the absolute value of the divisor l.

00:54 So this is in some way definition of integer numbers.

00:59 We will have a caution and a residue, both integer numbers.

01:03 They are unique with the condition that r is great than equal to zero and less than absolute value of l.

01:12 So to prove all that, let's define the set a as the integer numbers of the form l, m minus ln, that is m minus the product of l times n, where n is an integer number, and we put the condition that the number, m minus, l n get a be greater than or equal to zero.

01:45 So it's clear that a is a subset of the integer numbers.

01:52 That is all elements in a are integer numbers.

01:56 And in fact, they are all greater than or equal to zero because all the numbers in the set get to satisfy this condition.

02:05 So we can say that the least, an upper bound to the set is zero, lower bounder below so a is a set of integer numbers bounded below by zero that is all elements in a are greater than a equal zero that's by definition of set a and because of that because we have integer numbers and bounded below we know that there is a least element in a so there exists a list element in a.

03:04 So let's say that this list element is generated by some integer number q.

03:13 So let's say that two in the integer numbers generates the list value of a.

03:33 We are not saying at all that the list elements unique.

03:39 You are only saying that because there exists the list element, let's say we generate that value that list element with some number two.

03:48 That could be maybe two different integer numbers that generate this list value.

03:56 We will prove then after that is unique, but by the time being, so far now, let's only suppose we get at least one number, q, integer number two that generates this least value of a.

04:17 So we can say that m minus l times q is the least value of a.

04:38 Then we got to have first that this number is non -negative.

04:43 That is m minus l times q is greater than equal zero.

04:47 That is because the least element of a can satisfy the condition of the element of a, that is, b greater than a 0.

05:01 And if we give a name to that number, that is we call it r, okay, then we have that, we have that simply given this name and solving for m, we have m equal lq plus r.

05:36 And of course, because r is exactly equal to m minus lq, and m minus lq is no negative, then r is no negative.

05:44 That is only a name we are given to the list element and we are calling it r.

05:51 So let's prove first that r is less than absolute value of l.

06:03 And that's the second part of this condition over the number r.

06:07 Let's see that we have already this equation here, this one here, just by giving a name r to the least element of a, and that number, because m minus lq is no negative, this number r, which is identical to m minus lq, is also the negative.

06:32 So we have this bar already of the condition over r.

06:36 We will see the other but that is r less than absolute value of l.

06:40 So we have two possibilities for the absolute value of l.

06:45 If l is positive, the absolute value is itself and it's negative is minus l.

06:50 The possibility l, different from equal series, is not possible because l is explicitly not equal to zero.

07:03 So first, possibility if l is positive, then we know that the absolute value of l is l.

07:17 And this condition, okay, so if we get something here, sorry, let's put that down here.

07:24 And first i'm going to prove that by reduction to absurd.

07:31 That is, i'm going to suppose the contrary of that.

07:36 Suppose that r is greater than or equal to the absolute value of l.

07:41 Okay.

07:43 I'm going to suppose that, and let's examine the possibilities on l...

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