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Real Analysis Proof

  • Thread starter srfriggen
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  • #1
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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.
 

Answers and Replies

  • #2
jbunniii
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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
288
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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
jbunniii
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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
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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
jbunniii
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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
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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
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The bigger of the two :D
 

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