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A lemma in the integers from calculus 
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#1
Nov1312, 06:14 PM

P: 783

Suppose that M and N are natural numbers, such that N>M1.
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 


#2
Nov1312, 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. 


#3
Nov1312, 06:59 PM

P: 783

BiP 


#4
Nov1312, 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 


#5
Nov1312, 10:06 PM

P: 361




#6
Nov1412, 06:21 AM

P: 25

I may be missing something but
N > M1 implies that NM>1. Since N and M are integers NM>1 is equivalent to NM≥0 from which we get N≥M. 


#7
Nov1412, 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. 


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