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Does this statement require a proof? It seems pretty obvious to me.

"prove that there is no integer between n and n+1 where n is an integer."

Also if it does require a proof, what I need to show? Just few hints will suffice.

thanks

jitendra

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- Thread starter AlbertEinstein
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- #1

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Does this statement require a proof? It seems pretty obvious to me.

"prove that there is no integer between n and n+1 where n is an integer."

Also if it does require a proof, what I need to show? Just few hints will suffice.

thanks

jitendra

- #2

AlephZero

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Starting from Peano's axioms and defining addition using Peano's successor function, this is fairly simple.

On the other hand, Russell and Whitehead's "Principia Mathematica" takes 379 pages to get to the statement

"From this proposition it will follow, when arithmetical addition has been defined, that 1+1=2"

and proof that 1+1=2 is finally done on page 86 of Volume II - with the comment "The above proposition is occasionally useful."

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Without going into axiomatics, a simple proof could use the division algorith.

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thanks.

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What are natural numbers? What is a "negative" of a natural number? You need to build them from scratch, or axioms, as mathematicians call them. As mentioned by AlephZero, one way of defining (our long-known and beloved) numbers is by using Peano's axioms.

thanks.

- #7

StatusX

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- #8

MathematicalPhysicist

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and then prove this statement through tools of set theory.

the question is where did you get this question from?

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Einstein wasn't that much of a mathematician.

- #10

Gib Z

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Hehe heres a proof :D :

Proof By Intimidation - Trivial. Lol I love that.

But seriously now, Try proof by contradiction/example.

- #11

AlephZero

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I think the result is "obvious" to anybody who knows how to count, but has not had any formal mathematics training at all. So the way to "prove" it is to start from some axioms (e.g. Peano's) which define exactly what you mean by "counting".

If you want to use other theorems (e.g. the division algorithm) then you have to be sure you can prove the division algorithm without assuming this result, otherwise you have a circular argument.

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thanks.

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This is book description

Why does 2 times 2 = 4? What are fractions? Imaginary numbers? Why do the laws of algebra hold? And how do we prove these laws? What are the properties of the numbers on which the Differential and Integral Calculus is based? In other words, What are numbers? And why do they have the properties we attribute to them? Thanks to the genius of Dedekind, Cantor, Peano, Frege and Russell, such questions can now be given a satisfactory answer. This English translation of Landau's famous Grundlagen der Analysis-also available from the AMS-answers these important question

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by the way, why was the last post deleted that gave a proof to my question?? Was that proof okay?

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mathworld.com search peano arithmetic or axiom

- #16

HallsofIvy

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No.

Here's a modified version.

It's still not OK, but it might serve

the purpose of stimulating some criticism

(which of course is always good).

By the way, I agree with the post about

starting essentially from scratch (say Peano).

Probably the only way to give a rigorous proof.

But also probably not appropriate for someone

just leaving high school.

Preliminary assumptions:

1. N is set of Natural numbers. The set is well ordered, i.e., the set comes equipped with a well order relation. Therefore, every nonempty subset of N owns a least element.

2. 0 is in N. Clearly, it is the least element for N itself.

3. Z is symbol for integers.

First, prove there is no k in N satisfying, 0 < k < 1.

Consider it a lemma. It's purpose is to provide the contradiction sought in the following theorem.

Let A = {k | k in N and 0 < k < 1}. Assume A nonempty. Since A is a nonempty and apparently a subset of N, it contains a unique least element, say m. Now 0 < m < 1.

But then 0 < m^2 < m < 1, which is a contradiction since m is the least element of A. So A must be empty, and therefore no such k exists.

*Note*

I think you or anyone else who bothers to read this stuff will see problems with 0 < m^2 < m < 1. Anyway, I'll finish it up.

Thm.

For any n in Z, there is no k (in Z) satisfying n < k < n+1.

Assume such a k exists. Then p = k - n is in Z. But now p must satisfy 0 < p < 1, which contradicts the result given above. (Assumption here is that Z includes N.)

Incidently, I see "Foundations of Analysis", by E.Landau mentioned in a post. I'd also recommend this book.

Landau will take you from Peano, step by step, through to the real and complex numbers. But be forewarned, he'll do it in what he describes as a "merciless telegraph style".

But it's only about 125 pages and it's not an expensive purchase.

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1) For any n, n < n+1.

2) For any n, if n < m+1 then n < or = m.

It is from the second that we get the following argument:

Assume there is some integer k such that n < k < n+1.

Then by (2), k < n+1 implies that k < n or k = n. But this contradicts that n < k.

Hence, there is NO integer k such that n < k < n+1.

:P

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