# Proving x > 0 Implies Natural Number M for Xn > 0

• Hodgey8806
In summary, the person is asking for help in proving that if the limit of a sequence is a positive number x, then there exists a natural number M such that all terms of the sequence after the Mth term are also positive. They have attempted a solution using an epsilon-neighborhood approach and want to confirm its correctness and improve its presentation. The expert suggests a clearer and more organized way to present the argument.
Hodgey8806

## Homework Statement

I'll try to restate it more clearer (I'm sorry I'm not good with using the more mathematical text type.

Prove that if lim (Xn) = x and if x>0, then there exists a natural number M such that Xn > 0 for all n≥M.

## Homework Equations

Just to be clear, this limit definition is assumed to be to infinity as per our book for analysis.
(Xn) represents the sequence notation
And Xn will represent my term for a given n.

Scratch work pieces are that I want to choose an ε = x (the limit) at term XM
This would imply that for all n ≥ M, 0<Xn<2x

I would like to approach this using an ε-neighborhood.

## The Attempt at a Solution

Assume that the lim(Xn) = x and x > 0.
=> Given any ε>0, |Xn-x| < ε
=> -ε + x < Xn < ε + x
=> There exists a natural number M s.t. 0 < XM < 2x (This is from the scratch work)
=> For all n ≥ M, 0<Xn<2x
=> Xn > 0 for all n ≥ M

I do want to make sure this is right. I realize that it is sloppy currently and I would love to have more help in writing this a bit nicer. It makes sense to me, but it's weird to choose a term that "fits" that epsilon.

OK, since your argument is in fact mathematically correct, I will show you how to better present it. My comments will be interspersed in red.

Hodgey8806 said:

## The Attempt at a Solution

Assume that the lim(Xn) = x and x > 0.

=> [STRIKE]Given any ε>0, |Xn-x| < ε
[/STRIKE]

There is a natural number ##M## such that if ##n > M##, ##|X_n -x|<x##.

Now replace this statement:
=> -ε + x < Xn < ε + x
with: ##-x < X_n - x < x##

Now if you add x to all three sides of that inequality you can delete the crossed out steps
[STRIKE]=> There exists a natural number M s.t. 0 < XM < 2x (This is from the scratch work)
=> For all n ≥ M,[/STRIKE] 0<Xn<2x
=> Xn > 0 for all n ≥ M

Thank you very much! Sorry for the late response. I just want you to know that the help is greatly appreciated, and the flow makes it much easier to read!

## 1. How do you prove that x > 0 implies a natural number M for Xn > 0?

The proof involves using the definition of natural numbers and the properties of inequalities. We can show that if x is greater than 0, then it must be a natural number. From there, we can use mathematical induction to show that x^n will also be a natural number for any positive integer n.

## 2. What is the significance of proving x > 0 implies a natural number M for Xn > 0?

This proof is important because it establishes a connection between the properties of inequalities and the properties of natural numbers. It allows us to make conclusions about the behavior of natural numbers based on the behavior of inequalities.

## 3. Can this proof be applied to any value of x?

Yes, this proof can be applied to any value of x as long as it is greater than 0. This is because the properties of inequalities and natural numbers hold for all real numbers greater than 0.

## 4. Are there any exceptions to this proof?

No, there are no exceptions to this proof as long as the initial assumption of x > 0 is true. If x is not greater than 0, then the proof does not hold.

## 5. How can this proof be useful in real-world applications?

This proof can be useful in various fields such as physics, economics, and computer science where inequalities and natural numbers are commonly used. It allows us to make accurate predictions and draw conclusions about the behavior of certain systems or phenomena based on their initial conditions.

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