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Johnny5

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Everything posted by Johnny5

  1. And while I'm at it, is the axiom of choice true or false Matt? Or is it one of those kind of statements whose truth value can vary? Regards
  2. Not for nothing, but this made no sense. Definitions of the type found in mathematics involve 'if and only if' Let me think of another example. WRONG: Definition: A is a set if either there is at least one X, such that X is an element of A, or A is equal to the empty set. RIGHT: Definition: A is a set if and only if (either there is at least one X, such that X is an element of A, or A is equal to the empty set.) Now, perhaps the definitions above lead to a contradiction which is beyond Patrick Suppes' ability to detect, but that is another issue. My point has be
  3. 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 b
  4. Coquina brought up a very good point, about the hip bone, which connects to the very reason I got involved in this thread in the first place... because almost twenty years ago, I figured out what killed the dinosaurs. I have sort of been dancing around the answer, to see what responses I'd get, but it seems this thread has died, so I will finally reveal my hidden motive for discussing the 'hip bone' issue, and the 'hopping' dinosaur issue. The theory came to me as i was reading an encyclopedia article on dinosaurs. It was an old encyclopedia, and there was an artists rendition of a bu
  5. 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 s
  6. 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.
  7. 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.
  8. 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 rotat
  9. 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]
  10. 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
  11. 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
  12. There's nothing to invent, I already know the basis of the argument. Let me see what the heck this thread was about again though. Ah yes this: Prove the following: [math] \Gamma (n+1) = n! [/math] By definition, the following statement is true: [math] \Gamma(n+1) \equiv \int_{x=0}^{x=\infty} t^n e^{-t} dt [/math] So we can focus on the integral. Here is the integration by parts formula: [math] \int u dv = uv- \int v du [/math] let dv = e^-t dt, whence it follows that v = -e^-t let u = t^n, thus du = n t^(n-1) dt Substituting we have: [math] \int_
  13. How is it false as it stands?
  14. 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,
  15. Why not? Yes' date=' i know. But x^y is being defined for as many cases as are possible in stages, because his question was what does x^y mean. No, x^0 was not being defined right there, that was a theorem that x^0 must equal 1(given that not(x=0)), as a consequence of the fact that x^m times x^n must be equal to x^{m+n}.
  16. To both of you... I just read the last two posts (each once), and was quite impressed. Let me see if i have any worthwhile comments to make. You ask me, what does x^y mean? In the case where y is a natural number, we have the following definition: Definition: [math] x^y = \prod_{k=1}^{k=y} x = x_1 \cdot x_2 \cdot x_3 ... x_y [/math] So for example, if y = 3, we have, using the definition above: [math] x^3 = \prod_{k=1}^{k=3} x = x \cdot x \cdot x [/math] In the case where y is a negative integer, and not(x=0), we have the following definition: Definition:
  17. It's not that I have a problem with conventions, there is something else going on with 0^0, that is not a "conventional" issue. Look at it this way... When x isn't equal to zero, there is a simple proof that x^0=1. And, when x=0, there is a simple proof that x*y=0, for any y. So there's a sort of blind alley with 0^0. 0=1 thing. As I say, it's not that I have a problem with conventions, but few things in mathematics are conventions. Most of the structure of mathematics is purely logical, and this is why the subject attracts good minds IMHO. I'm not sure yet what th
  18. Consider the product from k=1 to k=2 of some arbitrary function of k, f(k). [math] \prod_{k=1}^{k=2} f(k) = f(1)f(2) [/math] In the 'scalar multiplication' being considered here, both f(1)f(2) are elements of the real number system. They are not matrices, or anything else which doesn't necessarily commute. Hence... f(1)f(2) = f(2)f(1) In the case where the lower indice is greater than the upper indice we have: [math] \prod_{k=2}^{k=1} f(k) = f(2)f(1) [/math] Since the kind of multiplication being considered here is commutative we have: [math] \prod_{k=1}^{k=2} f(k
  19. Why must it not force that upon us?
  20. I don't hate it, it's just not what I'm thinking of. I'll come up with something, then post it. Basically, what I have in mind uses integration by parts in the proof, but its been so long since i formally proved it, i was hoping you knew the argument. Regards
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