What is the logical flaw in this reasoning? Abusing $T \equiv T \vee F$.
This is not a "Haha gotchu mathematicians!" question. I'm seriously trying to learn. Please don't take my question to be insincere.
IMPORTANT EDIT: I've noticed that people are focussing too much on my examples that they're distracted by my real question which is logic based. I use various examples to explain my "logic based" confusion, I'm not confused about the examples. I do not require explanations for my examples.
This question stems from a question I asked long long long time ago and someone answered that it is incorrect to write $|x|=±x$ and I took their word for it because well... I wasn't experienced enough to ask the right questions.
Since then my instinct to deal with $|x|$ has been to use: $$ |x|= \begin{cases} x, \ x≥0 \\\\ -x, \ x <0 \end{cases}$$
because well... that is the definition of $|x|$.
However I was going through my questions yesterday, when I realised, "Wait hold on, why is the equation $|x|=±x$ incorrect? Because "$±$" means "plus OR minus", you're not insisting that $|x|$ is $x$ AND $-x$. You're only saying it is either $x$ OR $-x$.
But hold on, there's more. With that argument in mind, you can always write $\sqrt{9}=±3$ even though it's just $3$. You can even go more bonkers with this logic by writing $$\sqrt{9} = 3 \text{ or } -3 \text{ or } -193e^2$$ as long as one of them is true. You get the point, right?
You can keep adding on nonsense using the fact that $T \equiv T \vee F$ like so:
$$( \sin x = 0) \equiv (\sin x =0 \text{ or } \cos x = 0) $$ and then get $x \in \{\frac{nπ}{2} : n \in \mathbb{Z}\}$ as the solution which is absurd.
Question: Where is the logical error here?
I will focus on your problem with $T\iff T\lor F$.
Sure, under assumption that some formula $A(x)$ depending on a parameter $x$ is true (independent of the value for $x$), you may use that to deduce that $A(x)\iff A(x)\lor B(x)$ (since $T\iff T$).
However, in your examples, the formula $A(x)$ only holds true for some special values of $x$. It could thus happen that for some values of $x$, $A(x)$ fails to hold, while $B(x)$ is true. In this case, you can clearly not write $A(x)\iff A(x)\lor B(x)$ (since $F$ is not equivalent to $T$).
\begin{align}|x| &:= \begin{cases}-x &\text{ if }x<0; \\x &\text{ if }x\geq0\tag A\end{cases} \\\\ |x| &:\not=\;\pm x \\\\ |x|&=\;\pm x\tag B \\\\ \pm x&=\;|x|\tag C\end{align}
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The above statements are all true.
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$\large|x|$ is fully specified by its definition $(\text{A}).$
On the other hand, statement $(\text{B})$ merely exhibits $\large|x|$'s property.
So, $\pm x$ is less informative than $|x|.$
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\begin{align} &\forall x{\in}\mathbb R&\bigg(\,f(x)=|x|\implies f(x)=\pm x\,\bigg)\\ &\forall x{\in}\mathbb R^-&\bigg(\,f(x)=\pm x\kern.6em\not\kern-.6em\implies f(x)=|x|\,\bigg)\\ &\forall x,y{\in}\mathbb R&\bigg(|x|=|y|\iff x=\pm y\bigg) \end{align}
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Compare:
- $$\begin{align}\lvert2x\rvert&=x-1\\\iff\bigg(x<0 \,\text{ and }-2x=x-1\bigg) \:&\text{ or }\: \bigg(x\geq0 \,\text{ and }\, 2x=x-1\bigg)\\\iff\bigg(x<0 \text{ and } x=\frac13\bigg) \:&\text{ or }\: \bigg(x\geq0 \text{ and } x=-1\bigg)\\\iff x&\in\emptyset\end{align}$$
- $$\begin{align}\lvert2x\rvert&=x-1\\\iff\pm2x=x-1 \,&\text{ and }\, x-1\geq0\\\iff x&\in\emptyset\end{align}$$
- $$\begin{align}\lvert2x\rvert&=x-1\\\implies\pm2x&=x-1\\\iff x&\in\left\{-1,\frac13\right\}\end{align}$$ (Here, both solutions being extraneous is due to starting with an inconsistent equation.)
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Notice that, unlike statement $(\text{B}),$ statement $(\text{C})$ never gets conflated with definition $(\text{A}).$ This is because displaying $|x|$'s properties/possibilities before the subject itself makes the statement feel less definitive.
$$f(x)=|x| \;\:\text{ for } x\in\mathbb R\tag1$$ can be translated as $$\forall x{\in}\mathbb R\;\bigg(\big(x<0\to C(x)\big) \land \big(x\not<0\to K(x)\big)\bigg),$$
whereas $$f(x)=\pm x \;\:\text{ for } x\in\mathbb R\tag2$$ can be translated as $$\forall x{\in}\mathbb R\;\bigg(K(x)\lor C(x)\bigg).$$
Statement $(1)$ fully describes the function (in which each input has a certain output) $$f(x)=|x|,$$ whereas statement $(2)$ conveys only partial information about its behaviour.
The term $\pm$ (or sometimes $\mp$), like any notation, carries a lot of meaning and connotations, because we see certain notations in certain places. To use $\pm$ is to say that the choice of either plus or minus should make sense, and perhaps depending on further context one is preferred over the other. To say $|x|$ is to unambiguously say that the quantity is positive and equal in absolute value to $x$.
$|x|\neq\pm x$ because the left hand side is a uniquely determined positive value, whereas $\pm$ is an ambiguous statement suggesting both states of plus or minus are valid unless further context to the question says otherwise. If you are doing algebraic manipulation, and you write $\pm x$ instead of $|x|$, you will find yourself in a nightmare of superimposed states, where you must deal with both cases of plus and minus and as such it is an inferior notation to $|x|$. There is neither reason nor motivation to write $|x|$ as $\pm x$, and $|x|$ is always only one value, is always unique; $\pm x$ is not.
You say one can go bonkers with this logic, saying $\sqrt{9}=3\vee -193e^2$. There is a good quote from somewhere, I don't remember exactly where, saying that a good notation frees the mind to focus on the problem at hand. $\sqrt{9}=3\vee -193e^2$ is not a good use of the $\vee$ notation, and is indeed bonkers as you say. I don't think it is so much a logical error but more a semantic error: logical conjunctions like "or" generally signal that either state is possible, and in any further working or proof we must account for all the states. You can chain "or"s and other conjunctions, and the point of doing this is to logically determine one or more solutions to whatever problem you're facing - introducing absurdities serves absolutely no purpose. It is not correct to write $3\vee-193e^2$ because to take mathematical notation to such a highly pedantic level is to undermine the purpose of notation in the first place, and it will trip you up to write like that if you ever write a proof or work on a harder problem, because littering the working with absurdities and ambiguous notation is not how we do maths.