# Prove 1/n < a < n: Real Analysis Homework

• srfriggen
In summary, the mathematician is trying to find an integer that satisfies both inequalities, but is not able to do so.
srfriggen

## Homework Statement

Prove that if a>0, there exists n in N such that 1/n < a < n.

## The Attempt at a Solution

I am starting with a>0 and trying to manipulate, algebraically, to get n > a > 1/n.

From a > 0 I can add 1 to both sides to obtain, a+1 > 1. Then I can choose some n such that:

n > a+1 > 1.

From there, divide by n to achieve the desired result for one part of the inequality:

1> (a+1)/n > 1/n. We know from above that n > 1, so I can write:

n > (a+1)/n > 1/n.

From here I'm not sure what to do.

Try proving the two inequalities separately:

There exists an integer n in N such that a < n

There exists an integer m in N such that 1/m < a

Then think about how to combine these results.

jbunniii said:
Try proving the two inequalities separately:

There exists an integer n in N such that a < n

There exists an integer m in N such that 1/m < a

Then think about how to combine these results.

1. By the Archimedean Property, a < n for some n in N.

2. Again, by the Archimedean Property, 1 < am, for some m in N. So, 1/m < a.

Now I have: 1/m < a < n.

I don't see how I can justify that m=n.

OK, there exists an integer n in N such that a < n.

If you make n larger, is the inequality still true?

Similarly, there exists an integer m in N such that 1/m < a.

Does this remain true if you make m larger?

jbunniii said:
OK, there exists an integer n in N such that a < n.

If you make n larger, is the inequality still true?

Similarly, there exists an integer m in N such that 1/m < a.

Does this remain true if you make m larger?

yes, the inequalities both hold.

So could I just say, "We can make n and m large to arbitrary values such that m=n." Would that finish the proof?

srfriggen said:
yes, the inequalities both hold.

So could I just say, "We can make n and m large to arbitrary values such that m=n." Would that finish the proof?

Well, the question asks for a specific integer that satisfies both inequalities.

So if n satisfies the first inequality, and m satisfies the second one, what integer will satisfy both?

jbunniii said:
Well, the question asks for a specific integer that satisfies both inequalities.

So if n satisfies the first inequality, and m satisfies the second one, what integer will satisfy both?

so nm?

The bigger of the two :D

## 1. What is the purpose of proving 1/n < a < n?

The purpose of proving 1/n < a < n is to demonstrate the boundedness of the sequence a and establish its limit as n approaches infinity. This is important in real analysis as it helps to understand the behavior of the sequence and its convergence or divergence.

## 2. How do you prove 1/n < a < n using real analysis?

To prove 1/n < a < n, we first need to show that the sequence a is bounded by 1/n and n. This can be done by using the Archimedean property, which states that for any real number x, there exists a natural number n such that n > x. We can then use the squeeze theorem to prove that the limit of a as n approaches infinity is also bounded by 1/n and n.

## 3. Why is it important to use real analysis in proving 1/n < a < n?

Real analysis is essential in proving 1/n < a < n because it provides a rigorous mathematical framework for analyzing the behavior of real-valued functions and sequences. By using real analysis, we can prove the theorem with precision and rigor, ensuring that our results are accurate and reliable.

## 4. Can you provide an example of a sequence that satisfies 1/n < a < n?

One example of a sequence that satisfies 1/n < a < n is the sequence a_n = 1/n. This sequence is bounded by 1/n and n and its limit as n approaches infinity is 0, satisfying the conditions of the theorem.

## 5. How does proving 1/n < a < n relate to the concept of convergence in real analysis?

Proving 1/n < a < n is closely related to the concept of convergence in real analysis. By showing that the sequence a is bounded by 1/n and n and that its limit as n approaches infinity exists, we can conclude that the sequence converges. This is because a sequence is said to converge if it approaches a definite limit as the number of terms increases, which is precisely what we are proving in this theorem.

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