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In evaluating elementary $\epsilon$, $\delta$ proofs of limits, one often sees the following sort of move:

$$ \left|2x - 8\right| = \left|2(x-4)\right| = 2\left|x - 4\right| \dots$$ (See e.g. here (14:00): https://www.youtube.com/watch?v=DdtEQk_DHQs&t=420s)

I suppose this makes sense if you think about going the other way, namely $\,2\!\cdot\!\left|x - 4\right| = \left|2(x - 4)\right|\,$ in this example.

On the other hand, it is a fact that $|a| \geq a$. So given that, it seems the derivation should be:

$$ |2x - 8| = |2(x-4)| = |2||x - 4| \geq 2|x - 4| \ldots $$

I'm probably missing something glaringly obvious, but naively this feels like a paradox. Where is the mistake?

$\\~\\$

P.S. Also, I'm well aware of solving absolute value equations and inequalities by cases, converting to $\pm$, etc. But is there a good source on techniques for "ridding" an expression of absolute value? Squaring both sides is one, replacing $|a|$ with $\sqrt{a^2}$ is another. (A rigorous exposition of algebraic manipulation of absolute value expressions would suffice.)

TShiong
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RTF
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    This is hard to follow. $|2|=2$, so what's the problem? Of course, if you had, say, $|ax-a|$ the best you could say was that it was equal to $|a|\times |x-1|$ since we have no information on $a$. – lulu Jul 26 '23 at 16:15
  • There's no paradox: $|2x-8| = 2|x-4|$ and $|2x-8| \geq 2|x-4|$ are both true. There isn't much reason to use the second one, though. – Matthew Leingang Jul 26 '23 at 16:24
  • @lulu I know what you mean, but see accepted answer here: https://math.stackexchange.com/questions/1750709/the-absolute-value-of-a-sum-of-two-numbers-is-less-than-or-equal-to-the-sum-of-t – RTF Jul 26 '23 at 16:29
  • I have no idea what connection you see between that answer and your example. As I say, there is no paradox here. You are free to write $|2|≥2$ if you like. It is true, just kind of useless. – lulu Jul 26 '23 at 16:30
  • @Matthew Leingang. So if $|2(x-4)| = |2||x-4| = 2|x -4|$, what prevents me from likewise inferring $|2||x-4| = 2|x - 4| = 2(x-4)$, thus removing all absolute values? Is the confusion over the difference between how $|2|$ (a constant) is evaluated vs $|x|$ (a variable)? – RTF Jul 26 '23 at 16:38
  • @lulu The connection is clearly between the statement "Note that $|a|≥a$" and my (erroneous) implicit extrapolation to the concrete case of $|2|≥ 2$. – RTF Jul 26 '23 at 16:42
  • As I say, you can write $|2|≥2$ if you want to, but why would you? We know what $|2|$ is so why not use that? If I tell you that $x=1$ you are free to write $x≥-29$, that's a true statement after all. But in doing so, you'd be discarding useful information. – lulu Jul 26 '23 at 16:46
  • @lulu My question was about why it is a legal move to go from $|2(x-4)|$ to $2|(x -4)|$. As you alluded, it wouldn't be legal if we had $|y(x-4)|$. So from a certain (perhaps very naive) point of view, it seems odd that a certain algebraic manipulation is legal when done on a constant but not on a variable. I can square a constant $2$ and I can likewise square a variable $x$, same goes for adding, subtracting, taking logs, etc. But not necessarily with moving things out of absolute values. That is a bit odd at first glance. – RTF Jul 26 '23 at 16:56
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    Nothing about this is odd. We know what $|2|$ is so we can use this information. In general, we don't know what $|y|$ is exactly (only that it is $\max (y, -y)$). If you have more information in one given situation, you can say more about it. – lulu Jul 26 '23 at 17:03
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    The rule is : $|a|=a$ if and only if $a\geq 0$. So it is legal to write $|2|=2$ and it is legal to write $|2|\cdot|x-4|=2|x-4|$. However, unless you know for sure that $x\geq 4$, you cannot write $|x-4|=x-4$ because this is false if $x-4<0$ – hdci Jul 26 '23 at 17:06
  • @lulu Nothing about it ultimately is odd, yes. However, at first glance, to a non-mathematician, I stand by what I said. (Again, emphasis on a certain naive point of view). The vast majority of elementary algebraic operations allow one to ignore the "semantics" and simply manipulate the syntax--what can legally be done to a variable can likewise be done to a constant. But here we have an exception to that. – RTF Jul 26 '23 at 17:12
  • This has nothing to do with variables and constants. It's just a matter of what information is available. Say $a,b$ are two real numbers (specific values of some very complex function, say). If we know that $a>7$ then I certainly know that $|a|=a$. If we know nothing much about $b$, then I'm stuck with $|b|=\max(b, -b)$. Again, the more information you have, the more you can say. – lulu Jul 26 '23 at 17:16
  • Even with variables. If we define the variable $x$ to be $\sqrt {y^4+100y^2+400}$ for a real variable $y$ then we know that $|x-20|=x-20$. We don't know about $|x-30|$ though. That depends on $y$, which we are not told about. – lulu Jul 26 '23 at 17:17
  • A value like $|2|$ can always be rewritten as $2$ since there's no additional information to be known other than that $2$ is...$2$. As you have argued, that is not necessarily the case with variables (sometimes there is enough info to infer $|x| = x$, and sometimes there isn't.) So in that sense the difference between values and variables is relevant, even if indirectly. – RTF Jul 26 '23 at 17:30
  • No, it's just a question of information. Say that $f(n)$ describes the parity of the $n^{th}$ digit of $\pi$, with $f(n)=-1$ if that digit is odd and $1$ if it is even. Then $f(2)=1$ (I'm counting from the decimal point here so the relevant digit is $4$) but I have no idea what $f\left(10^{100000}\right)$ is. It's a perfectly well defined number, it's just too hard for me to compute it. In that case, I can certainly write $|f(2)|=f(2)=1$ but I can't tell you anything about $|f\left(10^{100000}\right)|$. I simply lack the information I'd need to proceed further with that one. – lulu Jul 26 '23 at 17:43
  • Clarification: Should have said, I can't say whether or not $|f\left(10^{100000}\right)|=f\left(10^{100000}\right)$ or not. Of course, we know that the absolute value is $1$ because $|\pm 1|=1$. – lulu Jul 26 '23 at 17:54
  • You are going to incredible lengths just to avoid even slightly agreeing with my point, which is no longer so much about the mathematics as it is about perception. To the student solving elementary $\epsilon$, $\delta$ proofs there is a relevant distinction between variables and values to be made in terms of what can typically (forget about obscure counterexamples) pulled out of an absolute value expression--even if it's just a heuristic useful at an elementary level and ultimately the distinction breaks down (I've conceded that multiple times at this point). – RTF Jul 26 '23 at 18:03
  • After reading the discussion my understanding is that @RTF is thinking of the rules as grammatical: since a rule is $|uv| = |u||v|$ we should write $|2(x-4)|=|2||x-4|$ (taking $u=2$ and $v=x-4$), and since a rule is $|u|\geq u$ we should write $|2||x-4|\geq 2|x-4|$. But there is a difference between pure algebra and analysis. In the former, such rules are central. In the latter, we look more into the actual values. – md2perpe Jul 26 '23 at 19:01
  • @md2perpe Yes, that ended up being the glaring mistake I alluded to in the question. In my defense it is entirely obvious once it dawned on me that I was looking at things purely mechanically, sort of "on autopilot." Your comparison to grammar is good and exactly why I brought up the point about syntax vs. semantics. – RTF Jul 26 '23 at 19:22

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