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As an example, if we have a binary term $x$, like this

$x = 0.d_1 d_2 d_3 \dots$

Where $d_1 = 1$ if $x < \frac{\pi}{ 10}$ else $d_1 = 0$

$d_2 = 1$ if $x < \frac{\pi}{ 10}$ else $d_2 = 0$

$d_3 = 1$ if $x < \frac{\pi}{ 10}$ else $d_3 = 0$

$\dots$

$d_n = 1$ if $x < \frac{\pi}{ 10}$ else $d_n = 0$

$\dots$

Then we will probably agree that we approach our limit or $\lim\limits x \to \frac{\pi}{ 10}$

but will $x = \frac{\pi}{ 10}$ or will $x$ always be $x < \frac{\pi}{ 10}$

Note: $ \frac{\pi}{ 10} = 0.314159 \dots$ or binary $ > 0.01010000011011001 \dots$

Ivan Hieno
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    The definition of $x$ is circular: $x$ depends on $d_i$ and $d_i$ depends on $x$. – Cuspy Code Oct 14 '18 at 12:56
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    Perhaps you intended to use the finite truncations of $x$ in your setup. As it is you have defined the bit sequence $d_i$ in two (inconsistent) ways. – hardmath Oct 14 '18 at 13:03
  • @Cuspy Code the $d_i$ depends on $ \frac{\pi}{ 10}$, if $d_1$ is 1/2 then $x$ will be greater than the target $ \frac{\pi}{ 10}$, even if all the other digits are 0, so it must be 0. And $d_2$ must be 1/4 because if all the other values were 0, would still be less than our target. And so on, each $d_i$ can only have one value, thus it is not circular. – Ivan Hieno Oct 14 '18 at 15:02
  • @hardmath I don't understand why you say "inconsistent", as there is only one way our bit sequence can be. Or so it seems to me. – Ivan Hieno Oct 14 '18 at 15:15
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    Suppose that $x$ and $\frac{\pi}{10}$ are given numbers. The condition $x \lt \frac{\pi}{10}$ is then either always true or always false, but you've used this same condition to "define" the bit sequence $d_i$ in the second way. But now either all $d_i=1$ or all $d_i=0$, depending on whether that condition involving $x$ and $\frac{\pi}{10}$ happened to be true or false. I suggested that maybe you intended to use the truncations of (the binary expansion of) $x$ in some fashion, and thus avoid defining all the $d_i$'s to be the same bit. – hardmath Oct 14 '18 at 15:26
  • @hardmath I can use that method to convert base 10 to base 2 with a spreadsheet. Just set every row to 0 then set the first cell to 1/2. Then if the sum of all rows is less than the target, then our binary number will begin with 0.1 else 0.0, then set the next cell to 1/4 and if the sum of all rows is less than our target, then that bit is 1 else 0. The idea is simple, there just may exist no math notation to represent the idea, but that is irrelevant to the question being asked. I just want to know do we "approach" or "equal" our target? – Ivan Hieno Oct 14 '18 at 16:36
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    I'm pretty sure I understand the idea for building the "base 2" representation of a number "with a spreadsheet" that you allude to, but the language of your Question repeatedly tests the condition $x \lt \frac{\pi}{10}$ to determine bit $d_i$. This is not the test "if the sum of all rows is less than our target". – hardmath Oct 14 '18 at 16:49

2 Answers2

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No, we wouldn't agree.

In general, given the binary digits $d_1, d_2, \ldots$, $ x = .d_1d_2d_3 \ldots $ is a single number, not something depending on $n$. It's actually $$\sum_{k=1}^\infty d_k 2^{-k} = \lim_{n \to \infty} \sum_{k=1}^n d_k 2^{-k}$$ so the limit as $n \to \infty$ is "built-in".

Robert Israel
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All expansions below are binary, not decimal.

You're treating $x$ as both a sequence and a real at the same time; as a result, the definition of $x$ you've written is indeed circular. Put another way:

It doesn't make sense to check "Is $x<{\pi\over 10}$?" without having fully defined $x$ first, so you can't refer to $x$ while defining $x$.

This doesn't mean that your idea is meaningless, just that you haven't written what you intend. So before I can answer your question, I need to clear up your definition of $x$.

What you're really trying to do is define a sequence of approximations to the real $x$ you want to build. Informally, to get from $x_n$ to $x_{n+1}$ we "tack on" the biggest binary digit to the end of $x_n$ which keeps us below ${\pi\over 10}$. Formally, we define $x_n$ recursively (or inductively) as follows:

  • $x_0=0$.

  • Having defined $x_n$, we let $x_{n+1}=x_n+2^{-n-1}$ if that's $<{\pi\over 10}$ and $x_{n+1}=x_n$ otherwise.

For example:

  • $x_0+{2^{-1}}={1\over 2}>{\pi\over 10},$ so $x_1=0.0$ (which is of course a silly way to write $0$, but it's clearer in this context I think).

  • $x_1+2^{-2}={1\over 4}<{\pi\over 10}$, so $x_2=0.01$.

  • $x_2+2^{-3}={3\over 8}>{\pi\over 10}$, so $x_3=0.010$.

And so forth. We then define $x$ to be the limit of the $x_i$s: $$x=\lim_{n\rightarrow\infty}x_n.$$

Now in everything we write from now on, we need to be careful to distinguish between the specific (and now well-defined!) real number $x$ and the sequence $(x_n)_{n\in\mathbb{N}}$ of real numbers we've built (and which we've used to define $x$). For example, when you write $$\lim x,$$ what you really want to write is $$\lim_{n\rightarrow\infty} x_n,$$ since you take the limit of a sequence rather than a single number.

Having repaired the language, we can now answer your question:

$x={\pi\over 10}$, even though for each $n$ we have $x_n<{\pi\over 10}$.

Noah Schweber
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