# A lemma in the integers from calculus

by Bipolarity
Tags: calculus, integers, lemma
 P: 774 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
 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.
P: 774
 Quote by Vargo 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

P: 774

## 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
P: 358
 Quote by Bipolarity Hey Vargo, is this reasoning correct: s cannot be an integer if 0
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 $\in$ S. By the well-ordering principle, S contains a least element m. Thus, 0 < m < 1. Multiplying by m, we get 0 < $m^2$ < m. Thus $m^2 \in$ S and $m^2$ < m, a contradiction. Therefore no such s exists, so there are no integers between 0 and 1.
 P: 17 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.
 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.

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