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I have read that Dedekind cuts allow you to define the real numbers from the rationals. For example, $\sqrt{2}$ can be defined in the following way:

  • Partition the rational numbers into two sets $A$ and $B$, such that all of the elements of $A$ are less than the elements of $B$
  • $A=\{a \in \mathbb{Q}:x<0 \text{ or } x^2 < 2$}
  • $B=\{b \in \mathbb{Q}: x > 0 \text{ and } x^2 \geq 2$}
  • Because $B$ has no lower bound, there is a 'gap' in the number line
  • We define $\sqrt{2}$ to fill this gap: $\sqrt{2}$ is the unique number such that $x^2=2, x>0$. $\sqrt{2}$ sits on the partition line that we previously drew (hence why we say $\sqrt{2}$ is that cut)

Hopefully, I understand Dedekind cuts well enough to ask this question. It seems to me that because $\pi$ is transcendental, it cannot fill a gap in the same way that $\sqrt{2}$ does. There is no polynomial equation that $\pi$ helps solve. By contrast, $\sqrt{2}$ solves the equation $x^2=2, x \geq 0$. Nevertheless, $\pi$ is a real number, and Dedekind cuts define the real numbers. So, can $\pi$ be defined using Dedekind cuts, or is more work needed? Moreover, does the usual geometric definiton of $\pi$ have a well-defined mathematical meaning if you have not constructed the real numbers from the rationals?

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We know that $$\pi=4\int_0^1\frac{dx}{1+x^2}.$$ A lower Riemann sum of this integral is $$L_n=\frac4n\sum_{k=1}^{n}\frac1{1+k^2/n^2}.$$ An upper Riemann sum of this integral is $$U_n=\frac4n\sum_{k=0}^{n-1}\frac1{1+k^2/n^2}.$$ We can define $A=\{a\in\Bbb Q:a<L_n\text{ for some }n\in\Bbb N\}$ and $B=\{b\in\Bbb Q:b>U_n\text{ for some }n\in\Bbb N\}$.

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Any sequence $c_n$ that converges to $\pi$ can be used to define such a cut, with

$$ A=\{a\in\mathbb Q\mid a\lt c_n\text{ infinitely often}\}\hphantom{\;.} $$

and

$$ B=\{b\in\mathbb Q\mid b\gt c_n\text{ infinitely often}\}\;. $$

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  • $\begingroup$ Thank you for responding. What does 'infinitely often' mean in this context? $\endgroup$ – Joe May 4 at 8:28
  • $\begingroup$ @Joe: That there are infinitely many $n\in\mathbb N$ for which $a\lt c_n$ (or $b\gt c_n$). $\endgroup$ – joriki May 4 at 8:29

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