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

In summary, the mathematician is trying to find an integer that satisfies both inequalities, but is not able to do so.
  • #1
srfriggen
306
5

Homework Statement



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

Homework Equations





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.
 
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  • #2
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.
 
  • #3
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.
 
  • #4
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?
 
  • #5
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?
 
  • #6
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?
 
  • #7
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?
 
  • #8
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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