When can we say that something is equal to, rather than something approaches a limit?

Question

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$

Answer 1

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.

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

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

Answer 2

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".