I aim to prove some of the order properties of Bishop's Real Numbers (given on page 22 of Constructive Analysis by Bishop and Bridges.) Bishop defines a real number to be a regular sequence of rational numbers, that is, a sequence $(x_n)$ of rational numbers such that $$ |x_m - x_n| \leq m^{-1} + n^{-1}. $$ A real number is positive , or $x \in \mathbb{R}^+$, if $$ x_n > n^{-1} $$ for some $n \in \mathbb{Z}^+$. A real number is nonnegative, or $x \in \mathbb{R}^{0+}$, if $$ x_n \geq -n^{-1} $$ for all $n \in \mathbb{Z}^+$. Bishop goes on to proves a useful criteria, specifically, $x \in \mathbb{R}^+$ iff there exists a $N \in Z^+$ such that $$ x_m \geq N^{-1} $$ for $m \geq N$. Additionally, $x \in \mathbb{R}^{0+}$ iff for every $n \in \mathbb{Z}^+$ there exists $N_n \in \mathbb{Z}^+$ such that $$ x_m \geq -n^{-1} $$ for $m \geq N_n$. The order properties I wish to prove are stated in Proposition 2.9. Specifically, the statement if $x,y \in \mathbb{R}^{0+}$ then $xy \in \mathbb{R}^{0+}$. I’m missing something mechanically about this proof and am seeking insight.

It is surely also relevant to provide the definition of multiplication. Given real numbers $x \equiv (x_n)$ and $y \equiv (y_n)$ we define $xy = (x_{2kn}y_{2kn})$ where $k = max\{K_x,K_y\}$ and $K_x$ is the canonical upper bound of $x$ (i.e. the least integer greater than $|x_1| + 2$.) One can quite easily show that $xy$ is in fact a real number.

  • $\begingroup$ It doesn't seem like there's actually a question. Is the question just how to prove proposition 2.9? $\endgroup$
    – Dan Doel
    Commented Jan 7, 2022 at 8:36
  • $\begingroup$ Specifically what is stated $\endgroup$
    – ToucanIan
    Commented Jan 7, 2022 at 8:54
  • $\begingroup$ OK, so you're trying to prove that if $|x_m-x_n|\leq m^{-1}+n^{-1}$ for all $m$ and $n$, $x_n\geq-n^{-1}$ for all $n$, $|y_m-y_n|\leq m^{-1}+n^{-1}$ for all $m$ and $n$, and $y_n\geq-n^{-1}$ for all $n$, then $x_{2kn}y_{2kn}\geq-n^{-1}$ for all $n$, where $k=\max(\lceil\lvert x_1\rvert+2\rceil,\lceil\lvert y_1\rvert+2\rceil)$. I don't recall if there's a proof in the book or if this is left to the reader. If there's a proof in the book, then can you quote the part where you're stuck? Or if it's left to the reader, then can you say how you've tried to prove it and why it didn't work? $\endgroup$ Commented Jan 7, 2022 at 10:38
  • $\begingroup$ It’s left to the reader. I’ve tried using the alternative characterization of non-negative as it gives you more to work with. Thus, for every $n \in \mathbb{Z}^{+}$ we have $N_{(1,n)}$ and $N_{(2,n)}$. I’ve tried using the values to define for every $n$ an $N_n$ such that $x_{2km}y_{2km} \geq -n^{-1}$ when $m \geq N_n$. The main issues is how to string inequality’s together. Naively I tried multiplying things appropriately but the right hand side has two negatives and is problematic. $\endgroup$
    – ToucanIan
    Commented Jan 7, 2022 at 10:59

1 Answer 1


Per the construction of $k$ and $xy$ we know that $$x_{2kn}, y_{2kn} \in [-(2kn)^{-1}, k]$$ and therefore $$(xy)_n = x_{2kn}y_{2kn} \geq -(2n)^{-1} > -n^{-1}.$$

  • $\begingroup$ How are you combing the information from the first line to arrive at the inequality in the second? $\endgroup$
    – ToucanIan
    Commented Jan 7, 2022 at 17:43
  • $\begingroup$ @ToucanIan The extreme negative case is where one factor is minimal (negative) and the other is maximal (positive). The min and max are $-(2kn)^{-1}$ and $k$ respectively. $\endgroup$
    – WimC
    Commented Jan 7, 2022 at 17:46

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