A lemma in the integers from calculus


by Bipolarity
Tags: calculus, integers, lemma
Bipolarity
Bipolarity is offline
#1
Nov13-12, 06:14 PM
P: 783
Suppose that M and N are natural numbers, such that N>M-1.
Prove that N≥M

The problem above is a rather minor lemma that I obtained while proving the ratio test from calculus. I was able to successfully prove the ratio test itself, but I took this lemma for granted, which I am now trying to prove.

I expected that since this lemma was rather simple, it would be easy enough to prove, but I can't seem to catch it.

Any ideas on how this might be done?

BiP
Phys.Org News Partner Science news on Phys.org
Going nuts? Turkey looks to pistachios to heat new eco-city
Space-tested fluid flow concept advances infectious disease diagnoses
SpaceX launches supplies to space station (Update)
Vargo
Vargo is offline
#2
Nov13-12, 06:41 PM
P: 350
The inequality you start with is the same as 1 > M - N. So you just need to prove that there are no integers strictly between 0 and 1. Then it follows that 0 ≥ M - N.

The positive integers do have a smallest element (Well Ordering Principle), call it s. Clearly s is greater than 0 and less than or equal to 1. You have to prove that s must be equal to 1.

That is about half the proof.
Bipolarity
Bipolarity is offline
#3
Nov13-12, 06:59 PM
P: 783
Quote Quote by Vargo View Post
The inequality you start with is the same as 1 > M - N. So you just need to prove that there are no integers strictly between 0 and 1. Then it follows that 0 ≥ M - N.

The positive integers do have a smallest element (Well Ordering Principle), call it s. Clearly s is greater than 0 and less than or equal to 1. You have to prove that s must be equal to 1.

That is about half the proof.
Thank you, I was not aware of the well-ordering principle. It completes my proof quite succinctly.

BiP

Bipolarity
Bipolarity is offline
#4
Nov13-12, 07:06 PM
P: 783

A lemma in the integers from calculus


Hey Vargo, is this reasoning correct:

s cannot be an integer if 0<s<1, or does it require further proof?

BiP
Petek
Petek is offline
#5
Nov13-12, 10:06 PM
Petek's Avatar
P: 360
Quote Quote by Bipolarity View Post
Hey Vargo, is this reasoning correct:

s cannot be an integer if 0<s<1, or does it require further proof?

BiP
Here's a proof that there are no integers between 0 and 1: Suppose that s is an integer such that 0 < s < 1. Let S = {positive integers n : 0 < n < 1}. Then S is not empty because s [itex] \in[/itex] S. By the well-ordering principle, S contains a least element m. Thus, 0 < m < 1. Multiplying by m, we get 0 < [itex]m^2[/itex] < m. Thus [itex]m^2 \in [/itex] S and [itex]m^2[/itex] < m, a contradiction. Therefore no such s exists, so there are no integers between 0 and 1.
mathsman1963
mathsman1963 is offline
#6
Nov14-12, 06:21 AM
P: 25
I may be missing something but

N > M-1 implies that N-M>-1. Since N and M are integers N-M>-1 is equivalent to N-M≥0

from which we get N≥M.
Vargo
Vargo is offline
#7
Nov14-12, 09:28 AM
P: 350
Whether it requires proof that there are no integers between 0 and 1 depends on the context. Most people would accept it without proof. If you are writing this up for a class, then you could check with your professor. In most classes, even ones that start by proving of all the basic properties of numbers, by the time you get to infinite series, then you have probably moved on past minor points such as this. So mathsman's argument would suffice.

However, since you were asking about the proof of that lemma, a careful proof would involve the Well Ordering Principle and it would follow Petek's argument.


Register to reply

Related Discussions
Let a, b be positive coprime integers. Show that if two positive integers x, y sat... Calculus & Beyond Homework 0
Fundemental lemma of the calculus of variations Calculus & Beyond Homework 3
calculus of variations fundamental lemma question - Calculus 0
Fundemental Lemma of the Calculus of Variations Calculus 2
Fundamental Lemma of Variational Calculus Calculus 5