Wikipedia:Reference desk/Archives/Mathematics/2024 September 20

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September 20

A^def
₌B+C

1. Since the expression in the title is a definition, we can conclude, that A exists if and only if both B and C exist as well.

2. However, we can't conclude that if B or C exist then A exists. Check: A denotes the total price, B denotes the price of the first product, and C denotes the price of the second product: if only the price $B$ of the first product exists, we still can't conclude - that the price $C$ of the second product exists - nor that the total price $A$ exists. The defintion $A{\overset {\underset {\mathrm {def} }{}}{=}}B+C$ only tells us (besides the relation $A{\overset {\underset {\mathrm {def} }{}}{=}}B+C$), that the total price $A$ exists if and only if both - the price $B$ of the first product exists - and the price $C$ of the second product exists.

3. For concluding, that if any value of the above three values A,B,C exists then all of them exist, it's sufficient to write down three definitions:

$A{\overset {\underset {\mathrm {def} }{}}{=}}B+C.$

$B{\overset {\underset {\mathrm {def} }{}}{=}}A-C.$

$C{\overset {\underset {\mathrm {def} }{}}{=}}A-B.$

4. Question: as I've indicated, it's sufficient (to write down all three defintions), but is there any shorter notation, to make sure that if any value of the above three values A,B,C exists then - all of them exist - and satisfy $A{\overset {\underset {\mathrm {def} }{}}{=}}B+C?$

HOTmag (talk) 13:26, 20 September 2024 (UTC)

Are you asking about a shorter notation? I may be looking mighty foolish, but isn't it circular and therefore meaningless to write the system of three definitions like you have above? Or rather, wouldn't writing down the first be exactly as meaningful as writing down all three, since both situations only relate A, B, C to one another? I get they're definitions and not merely statements, but I'm not really seeing the difference here. You'd have to define B, C in terms other than A to reach a more meaningful domain. Again, I could look totally foolish right now, since I've never answered a question here before. Remsense ‥ 论 13:34, 20 September 2024 (UTC)

It seems you haven't read 2#. HOTmag (talk) 13:40, 20 September 2024 (UTC)

I don't really understand it, no. You're defining A, B, C as the prices of products, but then you're writing abstract definitions of them in terms of each other. Am I missing something? Oh, did you mean to say A is the total? It makes more sense to me that way. Remsense ‥ 论 13:43, 20 September 2024 (UTC)

In any case, I still don't understand what the extra two definitions achieve: either B and C are defined outside of A or they're not, right? If we wanted to know them in terms of A, we already got that in the first definition. Remsense ‥ 论 13:45, 20 September 2024 (UTC)

I've just added an addition to 2#, to make it clear. HOTmag (talk) 13:49, 20 September 2024 (UTC)

I return to my initial question then, are you just looking for a shorter notation for this? Remsense ‥ 论 13:52, 20 September 2024 (UTC)

Yep. I've just made it clear in 4# (thanks to your question). HOTmag (talk) 13:55, 20 September 2024 (UTC)

Sorry for being slow on the uptake. I'm not sure if this needs to fit into any particular system or paradigm: anything wrong with $\forall (B,C)\in \mathbb {R} ^{+},\exists A=B+C$. Sorry if that hurts anyone to see, haven't flexed these muscles in a while Remsense ‥ 论 14:31, 20 September 2024 (UTC)

Not to bug you, but only since I'm relatively unsure of myself in this area—was this answer something like what you were looking for? Remsense ‥ 论 20:10, 20 September 2024 (UTC)

You limit the set of the Bs and the Cs to be the positive integers, but my question is general, without limiting anything. HOTmag (talk) 02:05, 22 September 2024 (UTC)

You gave the example of prices, so that's what I picked the positive reals* based on. Clearly, you can replace the set with whatever you want. Remsense ‥ 论 02:08, 22 September 2024 (UTC)

If A denotes the total price, B denotes the price of the first product, and C denotes the price of the second product, does your definition let us deduce the other definition: $C{\overset {\underset {\mathrm {def} }{}}{=}}A-B$ which I would like to deduce, bearing in mind that not all products have a price? Note: Since the latter is a defintion of C, then the existence of A and of B must be derived from the very existence of C. HOTmag (talk) 02:21, 22 September 2024 (UTC)

It seems the bit about "deducing a definition" articulates a fundamental confusion you have about what you're trying to accomplish. Definitions are stated, not deduced. Remsense ‥ 论 05:34, 22 September 2024 (UTC)

By asking whether a new definition can be deduced from an old definition, I mean whether a new claim, that was presented before as an additional definition, can be derived as consequence, from a given assumption that was presened before as an old definition.

In our case, the given assumption, was presented before as an old definition: A^def
₌B+C. The new claim, was presened before as an additional definition: $C{\overset {\underset {\mathrm {def} }{}}{=}}A-B.$ The question is, whether we can deduce the latter from the former, i.e. whether we can deduce the new claim from the given assumption. HOTmag (talk) 13:08, 22 September 2024 (UTC)

If the values we assign to B and C are logically independent from one another, then no. That's what logical independence means.Remsense ‥ 论 13:17, 22 September 2024 (UTC)

This is totally confused. The addition operation $+$ is special in that it is operand-wise strictly monotonic and therefore has, at least in the integers and reals (but not in the natural numbers) an operand-wise inverse, which we can denote using the subtraction operation $-$. In general, this is not possible.

Defining some quantity $A$ by an equation of the form $A=f(B,C)$ only makes sense if all terms in the right-hand side are defined. It is not just that this fails to define $A$ if $B$ or $C$ is not defined. It just does not make sense. And if $B$ is not defined, writing something like $B~{\overset {\underset {\mathrm {def} }{}}{=}}~\cdots$ only increases the confusion.

There is another issue in which the definedness of a defined term depends on the definedness of another term, namely when defining a function. Suppose we define a new function $f$ using existing known functions with a limited domain. For example, we may define real-valued function $f$ on the real numbers by the equation

$f(x)={\sqrt {1-x}}+{\sqrt {1+x}}.$

For $f(x)$ to be defined by this equation for some given value of $x,$ it is necessary that both ${\sqrt {1-x}}$ and ${\sqrt {1+x}}$ are defined. This is more a matter of common sense than anything else, and I see no need for some notational device to express this dependency.  --Lambiam 22:29, 20 September 2024 (UTC)

If A denotes the total price, B denotes the price of the first product, and C denotes the price of the second product, does the definition A^def
₌B+C make senee? If it does, can you deduce the definition: $C{\overset {\underset {\mathrm {def} }{}}{=}}A-B?$ HOTmag (talk) 02:13, 22 September 2024 (UTC)

You can't define something that already has a meaning. So if A, B and C all have independent meanings, as shown by the use of "denote", then none of them can be defined in terms of the other two. It may be true that A = B + C based on these meanings, but that's not a definition. In any case, you can't deduce a definition. If A has no independent meaning then you can define it to be anything, B + C, B - C, or B * C. --RDBury (talk) 05:29, 22 September 2024 (UTC)

Well, I'm presenting an analogous question (taken from a discipline very close to arithmetic), in my following thread. I hope my new question explains also my old question, but if my old question is still not clear, you can ignore it, and focus on my new question. HOTmag (talk) 13:38, 22 September 2024 (UTC)