punk Posted March 23, 2005 Report Share Posted March 23, 2005 Only if you restrict the domain. You mean restrict the codomain, or the range. You would be right if I were insisting that the square have an *inverse function*. Colloquially the inverse function is often just called the "inverse". It has a perfectly well definable *inverse* which is not an inverse function. Of course by allowing mappings into sets of numbers I can give the square an inverse function. Quote Link to comment Share on other sites More sharing options...

Nate T. Posted March 23, 2005 Report Share Posted March 23, 2005 If you go to Category Theory, a "mapping" is used in the sense I am using it above. It is nothing more than a set of arrows from a domain set to a codomain (range) set.Since you're using terminology out of Category Theory, I'll ask something I should have asked up front: exactly how much mathematical knowledge do you have? If I am mistaken in this definition of "mapping" then there is another term I should be using. The point is we are taking something more primitive than a function which can simply be represented by arrows from a domain set to a codomain set. Some of these "mappings" will be functions, most wont. I think the concept you're looking for is that of a relation. This generalizes the idea of a function by allowing that one point in the domain may be "related" to more than one point in the range, and that points in the domain need not be paired to any pints in the range at all. As for "inverses" versus "inverse functions," relations always have inverse relations, so that wouldn't be an issue. This is all perfectly rigorously definable. Right, but how do you overcome the algebraic closure problem I give above? Quote Link to comment Share on other sites More sharing options...

punk Posted March 26, 2005 Report Share Posted March 26, 2005 I have a BS in mathematics (I completed all the undergrad coursework recommended for pursuing grad work in math), BS and MS in physics, and I've done graduate level course work in formal logic, set theory, computation theory, yadda yadda. I'm also fairly bad at remembering the official terminology as you've noticed. But that's why god gave us reference books. I think capturing what we mean by "square root" is something we should do without worrying about closure. "Square root" is really one of our basic notions. We can make the proper steps toward closure afterward. The important point is that the "square root" is *two* numbers rather than one number. How we treat that formally is, in the end, going to depend on personal preferences. Most every formalization involves biting bullets in some places to get something the way we want it in another. Of course I admit some formalizations are better for certain types of work than others. Quote Link to comment Share on other sites More sharing options...

Nate T. Posted March 26, 2005 Report Share Posted March 26, 2005 I have a BS in mathematics (I completed all the undergrad coursework recommended for pursuing grad work in math), BS and MS in physics, and I've done graduate level course work in formal logic, set theory, computation theory, yadda yadda.That's interesting; you got a BS in math, and then pursued a PhD in physics? Was that difficult? Did you take a fair number of physics classes in your undergrad career? I think capturing what we mean by "square root" is something we should do without worrying about closure. "Square root" is really one of our basic notions. We can make the proper steps toward closure afterward. Well, if you do want to do a "multi-valued function" kind of approach, you would have to treat the sets that come out of functions like the square root function as numbers in order to make sense of equations involving radicals (and any other multivalued function like the inverse trig functions). Therefore, you'd have to extend the real number line to include every possible such multiply nested collection of sets of real numbers (in order to make sense of the arcsine of the square root of one, say). Then, you would have to find a way to denote a general member of this extension if you want to solve equations with this extension; how to do this isn't obvious, at least to me. The important point is that the "square root" is *two* numbers rather than one number. How we treat that formally is, in the end, going to depend on personal preferences. Most every formalization involves biting bullets in some places to get something the way we want it in another.It is something of a trade-off, but I do think that it's better to restrict (albeit awkwardly) the squaring function to a bijection in order to define a square root function instead of extending the real numbers in the way outlined above. After all, in the former case one can always describe the roots in terms of the one chosen by convention (this even holds true for complex roots). Of course I admit some formalizations are better for certain types of work than others. To be honest, I haven't thought out a way to extend the usual algebraic operations to relations, but I'm guessing you'd encounter messes fairly quickly. Do you have a specific kind of task in mind for which the multi-valued function approach would be better? Quote Link to comment Share on other sites More sharing options...

punk Posted March 26, 2005 Report Share Posted March 26, 2005 Thinking from physical applications where the square root appears, it isn't always the case that it is clear which of the roots has physical significance, or even that only one has physical significance. In particular, in solutions to the Dirac equation both roots are physically valid. The initial response of the physics community was to select out only one of the roots as valid. This proved to be wrong, and I would assert was largely motivated by conventions like are being advocated here. It would be preferable to preserve all the roots formally in the mathematics and let some physical considerations rule out some solutions at the end of the problem. I dont think any purely mathematical considerations should prefer one solution to another. Also misapplication of the standard of selecting out one root as representative is guilty of the "paradox" that started this thread. Quote Link to comment Share on other sites More sharing options...

Nate T. Posted March 26, 2005 Report Share Posted March 26, 2005 It would be preferable to preserve all the roots formally in the mathematics and let some physical considerations rule out some solutions at the end of the problem. I dont think any purely mathematical considerations should prefer one solution to another.Well, it's not as though the "restriction" approach eliminates the other roots from contention. It's just that after one extracts a square root from real equations like x^2 = a, one really has two equations to solve: x = +root(a) and x = -root(a). The only reason one does a "restriction" is to let root(a) be a single number, so that one can break up the possibilities into separate equations. Also misapplication of the standard of selecting out one root as representative is guilty of the "paradox" that started this thread. That's true; pedagogically, that kind of mistake does follow from a "restriction" approach. However, if they broke up their equations into two parts as above, there would be no problem. Quote Link to comment Share on other sites More sharing options...

punk Posted March 27, 2005 Report Share Posted March 27, 2005 I guess I see splitting the equations apart and treating the square root as a set as essentially the same thing. So taking y^2 = x, the convention is to write: y = +/- sqrt(x) Using sqrt(x) to be restricted to one value by convention. Which would seem to really mean: y = sqrt(x) OR -sqrt(x) So we interpret "=" in a truth-functional sense a la logic. Presumably then we have a rule that allows us to infer: y = sqrt(x) OR y=-sqrt(x) from that. My intent was just to use "{sqrt(x),-sqrt(x)}" to mean essentially what we would mean by "sqrt(x) OR -sqrt(x)". And then allow functions F(*) to work as: F({sqrt(x),-sqrt(x)} = {F(sqrt(x)),F(-sqrt(x))} And just leave out the OR which I sort of think is weird to include in arithmetic expressions since I don't tend to think of "=" as being a truth function of two arguments. Quote Link to comment Share on other sites More sharing options...

Nate T. Posted March 27, 2005 Report Share Posted March 27, 2005 My first concern to the set of solutions approach you give above is extraneous roots. Sometimes not all of the solutions arrived at via a solution method will actually satisfy the equation for various reasons-- you would have to take this into account somehow. My second concern is that you'd have to "flatten" expression involving nested radicals (or other multivalued functions). For instance, since we certainly want sqrt(sqrt(x)) = 4thrt(x), we'd have to have formally that {{sqrt( -sqrt(x)), sqrt(sqrt(x))}, {-sqrt(-sqrt(x)), -sqrt(sqrt(x))}} = {4thrt(x), -4thrt(x)}, which are sets of different levels, and wouldn't even be equal if flattened, since some of the roots on the left side are imaginary. Quote Link to comment Share on other sites More sharing options...

punk Posted March 28, 2005 Report Share Posted March 28, 2005 (edited) Lets take sqrt(x) = {a,-a}, sqrt(a) = {b,-b}, i is the usual imaginary number so: sqrt(sqrt(x)) = sqrt({a,-a}) ={sqrt(a),sqrt(-a)} ={{b,-b},{ib,-ib}} We'd expect: 4rt(x) = {b,-b,ib,-ib} So yeah, there is a problem if we want: 4rt(x) = sqrt(sqrt(x)) Maybe the better answer *is* just to interpret everything more in the mode of logic and say: sqrt(x) = a OR -a Then: sqrt(sqrt(x)) = sqrt(a OR -a) =sqrt(a) OR sqrt(-a) =(b OR - OR (ib OR -ib) =b OR -b OR ib OR -ib But now "=" is a truth function of two arguments, instead of what it is usually is taken to be. Under the normal usage of "=", we'd be inclined to say: b = b OR -b is false since "b" is certainly not the same thing as "b OR -b", but taking "=" as a truth function (of some kind) this is a true statement. I think nailing this down would get a little weird though. Edited March 28, 2005 by punk Quote Link to comment Share on other sites More sharing options...

Nate T. Posted March 28, 2005 Report Share Posted March 28, 2005 Under the normal usage of "=", we'd be inclined to say: b = b OR -b is false since "b" is certainly not the same thing as "b OR -b", but taking "=" as a truth function (of some kind) this is a true statement. Not to mention, logical connectives join propositions, not numbers. If you were to try an approach like this, it would generalize logical connectives to numbers. Quote Link to comment Share on other sites More sharing options...

Alfred Centauri Posted May 28, 2005 Report Share Posted May 28, 2005 (edited) <snip> -1=-1 sqrt(-1)=sqrt(-1) [sqrt=square root] <snip> Try it this way instead... Let sqrt(x) denote the positive root of x -1 = -1 +sqrt(-1)=+sqrt(-1) +sqrt(-1/1)=+sqrt(1/-1) +sqrt(-1)/+sqrt(1)=+sqrt(1)/+sqrt(-1) +[sqrt(-1)/sqrt(1)]=+(-1)[sqrt(1)/sqrt(-1)] (Why? Recall that 1 / i = - i) And so we cross-multiply and get: +[sqrt(-1)sqrt(-1)]=+(-1)[sqrt(1)sqrt(1)] +(-1) = +(-1) Edited May 28, 2005 by Alfred Centauri Quote Link to comment Share on other sites More sharing options...

Alfred Centauri Posted May 28, 2005 Report Share Posted May 28, 2005 I never studied imaginary numbers. I have a question about negative numbers. Are they "real"? A negative number is a positive number that I am looking at upside down. I mean, like height and depth. A depth of 1 unit is a height of -1 units. I suppose that what you are asking is along the lines of 'are negative numbers physical' in the sense of 'here is one apple, now show me negative one apple'. That might be tough but negative numbers can be given a more physical interpretation by considering antimatter. For example, under certain conditions a particle and its antiparticle can be created. Where there was no matter, now there is matter and antimatter. Think of it like this: where there was 0, now there is a +1 and a -1. (I bet I'll get some wisecracks for that one...) Quote Link to comment Share on other sites More sharing options...

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