Johnny5 10 Posted June 13, 2005 It is not necessary to define anything in mathematics, but so long as definitions are consistent with everything which you hold to be mathematically true, there is no logical error you are making, and then you can 'do some math'. The definitions which i have presented thus far, were to answer someone elses question. They wanted to know the meaning of x^y. Now, the field axioms are being used, and this i feel goes without saying. As for the suffixes, they are not necessary at all, but they help you to understand how many times x is being multiplied by itself. In other words, there is nothing wrong with defining using the suffixes, since it is clear that [math] x_1=x_2=x_3... = x [/math] The suffixes are helpful, though not necessary. But, if you omit them, then you are assuming that the reader can infer the number of times x is being multiplied by itself, or 1/x is being multiplied by itself. By using the suffixes, you are lessening the number of deductions to be made by the external reasoning agent. 0 Share this post Link to post Share on other sites

matt grime 10 Posted June 13, 2005 I thikn you'll find that we all know what x^y means but we're trying to educate you in how mathematics works. There is no known consisitent with all uses way of defining the symbol 0^0. So? who said there was? However, x^0 has a removable singularity in a loose sense at x=0 which is the convention used in Taylor series 0 Share this post Link to post Share on other sites

matt grime 10 Posted June 13, 2005 But, if you omit them, then you are assuming that the reader can infer the number of times x is being multiplied by itself we are doing no such thing since the produvt sign tells us already or 1/x is being multiplied by itself. By using the suffixes, you are lessening the number of deductions to be made by the external reasoning agent. no, you are iontroducing a deduction that the reader will guess x_i=x for all i. there is no har in saying x.x...x n factors. and it avoids writing something that is as it stands false 0 Share this post Link to post Share on other sites

Johnny5 10 Posted June 13, 2005 no' date=' you are iontroducing a deduction that the reader will guess x_i=x for all i.[/quote'] Thats why i gave an example. You are just splitting hairs now. Some of your other comments are noteworthy, but this one is just splitting hairs. 0 Share this post Link to post Share on other sites

Johnny5 10 Posted June 13, 2005 there is no har in saying x.x...x n factors. and it avoids writing something that is as it stands false How is it false as it stands? 0 Share this post Link to post Share on other sites

Algebracus 10 Posted June 13, 2005 Johnny5: I didn't ask for the meaning of x^y because I don't know what it mean myself, but because you don't seem to have a total understanding of the expression. Your attempt of proving that 0^0 = 1 if 0! = 1: You started by stating that [MATH]e^x = \sum_{k = 1}^{\infty} x^n/n![/MATH] for all x, but do you know whether this expression is so easily written because of a convention of letting 0^0 = 1 or because of something else? (And this is another rethorical question, asked because I hope you will see the systematic approach of starting at the definitions, which are needed (to start with everything you find mathematically true is really to let "human whim" take over a bit too much), and then go upwards in a rigourous manner. A little mistake, and you will lead yourself to contradictions, possibly because the starting foundation is wrong (like in naive set theory), but more likely because you broke a rule.) 0 Share this post Link to post Share on other sites

Johnny5 10 Posted June 14, 2005 Johnny5: I didn't ask for the meaning of x^y because I don't know what it mean myself, but because you don't seem to have a total understanding of the expression. To be honest, x^y isn't hard to understand. Suppose that [math] A = x^{\frac{1}{3}} [/math] Now, cube both sides to obtain: [math] A^3 = x [/math] So pick any real number A, at random. Multiply A by itself y times. Suppose that y = 7, then we have: AAAAAAA=A^7 Set that number equal to X X = A^7 Now, take the seventh root of both sides, to obtain the number that you started with... [math] A = X^{\frac{1}{7}} [/math] So that gives you more of an idea about exponents. In order to really say that you have expressed the meaning of X^y is going to require a lot of work, but it can be done, its not impossible. You just have to handle things one case at a time. Consider something like [math] A^{\sqrt{2}} [/math] Even this is allowable, and has a unique answer. You can approximate the answer by approximating root 2 as 1.414, and you can keep making a better approximation by using more decimals. 0 Share this post Link to post Share on other sites

matt grime 10 Posted June 14, 2005 You wrote [math]\prod_{k=1}^{y}x = x_1x_2\ldots x_y[/math] that is a false statement - the LHS and the RHS are not equal. There is nothing in your post that states what x_i is at all we are forced to introduce what looks like y new variables. incidentally, x^y is defined as exp{ylogx} defined for all x strictly positive, and even negative y picking a branch of log. Branches being yet one more thing you forgot in your attempt to define powers. In anyecase, none if this has indicted that you're explaining anything about 0^0 that is a genuine "funky" problem that we are glossing over in mathematics 0 Share this post Link to post Share on other sites

Johnny5 10 Posted June 14, 2005 You wrote [math]\prod_{k=1}^{y}x = x_1x_2\ldots x_y[/math] that is a false statement - the LHS and the RHS are not equal. There is nothing in your post that states what x_i is at all we are forced to introduce what looks like y new variables. No' date=' it's not false, you are not interpreting it properly. However, i do understand your point about notation, when you say, "what looks like y new variables." More frequently than not, the subscript under the same letter indicates a new variable... hmm. Yes, that notation is of my own choosing, so i suppose i have to state that the x_i are all equal, but i did that somewhere, after you commented on it earlier in the thread. But again, it's not false based upon the meaning. Do you have any better suggestion as regards notation for that? You do understand that the LHS and RHS are supposed to be equal at this point. Probably this would do just fine: [math'] x^y \equiv \prod_{k=1}^{y} x = x_1 x_2 \ldots x_y [/math] where [math] x = x_1 = x_2 = \ldots = x_y [/math] 0 Share this post Link to post Share on other sites

Johnny5 10 Posted June 14, 2005 incidentally' date=' x^y is defined as exp{ylogx} defined for all x strictly positive, and even negative y picking a branch of log. Branches being yet one more thing you forgot in your attempt to define powers. [/quote'] I wouldn't say I 'forgot' because I wouldn't use natural log to define x^y. Primarily because historically, x^y preceded logarithms. To define it that way, would be a bit anachonistic. Nonetheless, I am not sure why you keep writing that. I saw you write that somewhere else. Maybe you could explain why anyone would want to define x^y as exp{ylog x}, at least then i could see where you are coming from. 0 Share this post Link to post Share on other sites

Johnny5 10 Posted June 14, 2005 In any case' date=' none if this has indicted that you're explaining anything about 0^0 that is a genuine "funky" problem that we are glossing over in mathematics[/quote'] That whole issue about 0^0 is secondary and deserves a thread of its own. Keep in mind, I was only answering someone's elses question about the meaning of x^y. I am going to move this to a thread of its own. 0 Share this post Link to post Share on other sites

matt grime 10 Posted June 14, 2005 you define x^y as exp(ylogx) since it is correct and negates any need to muck around with thinking what x^pi, say, as the limit of the rational approximations to pi - does that even exist? is the limit well defined? x^y is just a function, and functionally exp(ylogx) is equivalent to x^y and avoids the need to take a completion of the integral and hence rational powers to the real powers, and is generally the easiest definition. since it is obviuosly coninuous and henc, must be equal to the competion if it exists. Ie i am demonstrating that the completion is well defined by giving a function in the completed space that agrees at the rational points. you did omit branches when you claimed to definethe power 1/3 or 1/7. 0 Share this post Link to post Share on other sites

Johnny5 10 Posted June 14, 2005 you did omit branches when you claimed to definethe power 1/3 or 1/7. Yes' date=' well branches as I recall show up when you use complex variables right? Pick a complex number Z at random. Then, there are real numbers x,y such that: [math'] Z = x + iy [/math] Where i is the square root of negative one. (which has its own problems) [math] Z = R e^{i \theta} = R[cos \theta + i sin \theta] [/math] Where [math] x^2 + y^2 = R^2 [/math] [math] tan \theta = \frac{y}{x} [/math] 0 Share this post Link to post Share on other sites

matt grime 10 Posted June 14, 2005 no, what about the square root of 4? is it 2 or -2? no complex variables at all 0 Share this post Link to post Share on other sites

matt grime 10 Posted June 14, 2005 there are no problems with i, and don't pretend there are. 0 Share this post Link to post Share on other sites

Johnny5 10 Posted June 14, 2005 no, what about the square root of 4? is it 2 or -2? no complex variables at all suppose that x*x = 4. Then, either x=2 or x=-2. There are two roots to the equation. But if i see the following: [math] \sqrt {4} [/math] That single number will be 2. I don't look at that symbolism as representing two numbers. Thats why in the quadratic formula we write: [math] \pm \sqrt{B^2 - 4AC} [/math] So if faced with the following Solve for x [math] x^2 = 4 [/math] The answer is expressed as [math] \pm \sqrt{4} = \pm 2 [/math] So i still don't follow you about branches. I thought branches show up in complex variables, becase a rotation through 2pi radians results in the same complex number. 0 Share this post Link to post Share on other sites

matt grime 10 Posted June 14, 2005 when you state that sqrt(4)=2 you have chosen a branch of the square root function - the principle one. 0 Share this post Link to post Share on other sites

Johnny5 10 Posted June 14, 2005 I don't remember the definition of 'principle branch' so if it's not too hard can you say what it is quickly? And which branch of mathematics it shows up in? I googled it, and didn't find a nice answer. 0 Share this post Link to post Share on other sites

matt grime 10 Posted June 14, 2005 principle simply means "the one that makes most sense and is most widely used" we could declare the square root function to be never positive so that ^2 and ^{1/2} are mutually inverse bijections between R^- and R^+ the negative and positive reals. but generally it is better to make them inverse bijections on R^+ to itself 0 Share this post Link to post Share on other sites

Johnny5 10 Posted June 14, 2005 In what sense would you say I am using it. Also keep in mind, the terms injective, bijective, surjective, never really took hold with me. I memorized the definitions, knew them for the test, and then dumped them. I know they are 'adjectives' which operate on 'functions'... classification of different types of functions and all, but i never committed the definitions to memory. I followed you about 'principle' branch though. Not really an official definition, but better than nothing. I could swear that comes from complex variables though. 0 Share this post Link to post Share on other sites

matt grime 10 Posted June 14, 2005 eh? you declared that sqrt should be positive - that means you made a choice, something that if you were really defining x^y properly would require you to comment on. i don't really care for that attitude towards acceptable ignorance of standard terms, and it isn't my fault you didn't learn them; if you wanted sympathy, or for me to care at all, then you shouldn't make such absolute claims about things in mathematics (ie what the product symbol is) only to reveal that you know none of the basics; problems you fidn are not ones in mathematics but ones in your knowledge. there is no absolute definition of principal - note correction to spelling. the principal block contains the trivial module the principal branch of log takes +ve reals to reals, principal divisors are something else entirely, a principal ideal is generated by one element. 0 Share this post Link to post Share on other sites

Johnny5 10 Posted June 14, 2005 It wasn't my personal choice, its in the formula for the quadratic as i explained to you. As for the product symbol, i use it in a manner consistent with the field axioms, nothing more, nothing less. And you aren't going to find too many non-professional mathematicians, who will be able to explain the meanings of injection bijection surjection to you, and furthermore those definitions will boil down to a simple minded usage of binary logic anyways, point being that the definitions aren't necessary to be known, since one can communicate using just binary logic, and some set theory, and some first order logic. Example given... More formally a function f : A → B is injective if, for every x and y in its domain A, if not (x =y) then also not( f(x) = f(y) ). And now i offer the following correction to this wikipedia entry Let F denote a function from set A to set B. f : A → B The function f is injective if and only if, for every x and y in its domain A, if not (x =y) then also not( f(x) = f(y) ) When you define something, you also get the converse for free. It's not something that needs to be deduced. I guess my only point here, and it's rather minor, is that even if a person doesn't know the meaning of the term 'injective,' you can still discuss anything you want to about an injective function with that person, by using the definition. It may be more verbose, but you can still say anything you want, and furthermore you can tell them at that moment in time, that we call this kind of function injective. In my case, i remember the terms, but just not which term goes to which kind of function. And the reason for that was, I didn't like the material for some reason. I didn't like the way it was presented in any book which i had, and after introducing the definition, the author never bothered to use it again. A seemingly pointless definition. 0 Share this post Link to post Share on other sites

matt grime 10 Posted June 14, 2005 And you aren't going to find too many non-professional mathematicians' date=' who will be able to explain the meanings of injection bijection surjection to you[/quote'] Well, i'd suggest everyone doing maths as an undergrad knows those definitions in the UK as they are taught in the first semester. Moreover most people doing highschool mathematics are taugh one to one and onto here too. so they understand them even if those words aren't always used. besides you are posting in a pseudo authoratitive manner on mathematics so tough, you should know the basics. 0 Share this post Link to post Share on other sites

matt grime 10 Posted June 14, 2005 When you define something' date=' you also get the converse for free. It's not something that needs to be deduced.[/quote'] not necessarily, though it is often presumed. the author was being more correct than the average mathematician who usually presumes that when we say "we denote by FOO those things with property BAR" that it is implicit that we are saying that this uniquely characterizes them. Ie that only FOO's have the property BAR. But that is again a convention, and you hate those. 0 Share this post Link to post Share on other sites

Johnny5 10 Posted June 14, 2005 Well, i'd suggest everyone doing maths as an undergrad knows those definitions in the UK as they are taught in the first semester. Moreover most people doing highschool mathematics are taugh one to one and onto here too. so they understand them even if those words aren't always used. besides you are posting in a pseudo authoratitive manner on mathematics so tough, you should know the basics. It's just that I couldn't remember the difference between 'injective' and 'surjective' I know what a one-to-one mapping is of course, because when you say it like that, the meaning is clear. But anyways, I went away and thought about it for awhile, and will go so far as to say that the study of relations, and functions is part of set theory, and so an individual who wants to use set theory to communicate, should go to the trouble to learn which term goes to which. It's just that for some reason, I seem to have a hard time remembering which is which. I am not one who memorizes by rote. Something is lacking in the way set theory is laid out, and with an axiomatic set theory available, i doubt i would have any trouble remembering what an injective function is, or a surjective function is, or what a function is for that matter. But, as I told you once long ago, i tried to learn axiomatic set theory, from a fellow by the name of Patrick Suppes, and my conclusion was that "axiomatic set theory" is a bit off. I do not regard there as being a consistent and complete axiomatic set theory available to man. 0 Share this post Link to post Share on other sites