Prove that for every $\epsilon\gt 0$ there exists a natural number $\bar{n}$ such that for all natural numbers $n\geq\bar{n}$, $1/n\lt\epsilon$.

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Lesson Parent: 
Upper and Lower Bounds
Problem: 

Prove that for every $\epsilon\gt 0$ there exists a natural number $\bar{n}$ such that for all natural numbers $n\geq\bar{n}$, $1/n\lt\epsilon$.

Answer: 

It is true that for every $\epsilon\gt 0$ there exists a natural number $\bar{n}$ such that for all natural numbers $n\geq\bar{n}$, $1/n\lt\epsilon$.

SAMPLE SOLUTION

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Solution: 

Recall that the natural numbers do not have an upper bound. Therefore, for every <Sign in to see all the formulas> there exists an <Sign in to see all the formulas> such that <Sign in to see all the formulas>. Since <Sign in to see all the formulas> and <Sign in to see all the formulas>, <Sign in to see all the formulas>.

Now if a natural number $n$ is greater than <Sign in to see all the formulas> and because <Sign in to see all the formulas>, then we know that <Sign in to see all the formulas>. Therefore, <Sign in to see all the formulas>. Since $n$ is arbitrary and greater than <Sign in to see all the formulas> our proof is complete.

Another way of saying this is that between $0$ and any real number <Sign in to see all the formulas>, there exists a rational number $1/n$ such that <Sign in to see all the formulas>. This is a smaller version of the proof that for every two real numbers, there is a rational number between them. We say that rational numbers are dense in the set of real numbers. We will put a link to this more general proof in later.