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I believe there are three cases. I think I have figured out the first one.

Case 1: Let $ε$ = 1 and N > 1. Take $ N$=2. Then 1/2 < $ε$ ≡ 1/2 < 1.

Case 2: Let 0<$ε$<1. I believe that I would use the Archimedian Property, which states that, "If x is a real number, then either x is an integer or there exists an integer n, such that $n$ < $x$ < $n$ + 1. I am having trouble applying this to case 2.

And then there's the case when ε > 1, but I am confused on how to prove this as well.

Any hints would be much appreciated.

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  • $\begingroup$ Perhaps the following hint: if $\epsilon$ is very small, then $\frac{1}{\epsilon}$ is very large. Combining this with the Archimedian property... $\endgroup$
    – Matt
    Commented Feb 2, 2017 at 2:32

1 Answer 1

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HINT: Given any real number, say $x$, you can a positive integer $N$ so that $N>x$. Now what should you let $x$ be so that when you `flip both sides upside-down', you get the desired inequality?

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  • $\begingroup$ Would I let x be 1/ε ? I don't know why I can't wrap my head around this. $\endgroup$
    – Mathgirl
    Commented Feb 2, 2017 at 3:33
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    $\begingroup$ @Mathgirl Exactly! Then you have $N>1/\epsilon$ so that when you invert, you obtain $1/N< \epsilon$, which is what you wanted. You can even try it in a few cases to see how it works. If $\epsilon=5$, then $1/\epsilon =1/5$. Choose $N=1>1/5$ and inverting gives $1/N=1/1<5$. If $\epsilon=1$, then $1/\epsilon=1$ and choose $N=2$. Then $1/N=1/2<1$. If $\epsilon=1/3$, then $1/\epsilon=3$. Choose $N=5$, then $1/N=1/5<1/3$. Not that epsilon is always rational, but this gives an idea of how this works. $\endgroup$
    – mathematics2x2life
    Commented Feb 2, 2017 at 3:44
  • $\begingroup$ Thank you, that is very helpful! Does that mean I don't really even need to use the Archimedian property? $\endgroup$
    – Mathgirl
    Commented Feb 2, 2017 at 4:00
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    $\begingroup$ @Mathgirl You do use the Archimedean Property! These are concrete examples. But given any $\epsilon$, we create the number $1/\epsilon$ (we can because $\epsilon>0$ so 'flipping upside-down' doesn't cause any issues). But now we have this number $1/\epsilon$, how do we know for certain that we can choose some counting number $N$ so that $N>1/\epsilon$? Well, it is the Archimedean Property that allows us to choose such an $N$. $\endgroup$
    – mathematics2x2life
    Commented Feb 2, 2017 at 5:09

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