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Math Forum / Mathematics / Undergraduate Math / October 2008JSH: Your broken field
So yeah, it's EASY to prove I'm right with non-polynomial factorization. Posters disagreeing with me can try to attack my proof, where remember THEIR argument was that 7 is split up by functions, so they want something like: 7 = w_1(x)*w_2(x) but given 175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1) how do you force any human, or any being for that matter to multiply through in some bizarre way with functions shifting how factors multiply through by the value of x, when any idiot can just do: 7*(175x^2 - 15x + 2) = (5b_1(x) + 2)(5*7*b_2(x)+ 7)? Answer is, you can't so I'm right that the 7 multiplies the same way for ALL x, and can prove that easily enough, but what good is proof if people simply refuse to acknowledge it? You people killed a freaking math journal over denial of this result with emails that derailed some editors who did the right thing, until you broke the formal peer review system. You betrayed the mathematical field. You betrayed the entire human race. And you kept escalating that betrayal with my other research by refusing to acknowledge key facts about my prime counting function, and the partial difference equation and partial differential equation that result from it. And in refusing to properly acknowledge my recent work solving binary quadratic Diophantine equations. The point here is that you people are willfully enemies of the human species. I know many of you are living off of public funds so your behavior is parasitic in nature. You block the correct mathematical results, and block result after result after result as you are protecting your feeding grounds, and the human species be damned as like any parasites you care for yourselves first and only. So you would betray the human species itself so I hypothesize that on some level, you are not human yourselves. You are a parasitic sub-species living amongst the human species, genetically different, but hiding yourselves in certain areas where you can gain traction and survive. So you carefully pick areas where you can gain critical mass and remove the need to actually work, like "pure math", and the priesthood. So I hypothesize that your subspecies founded most human religions. You have driven the evolution of the more dominant species for some time now, but it is finally time for you to be un-masked as your parasitic nature now has your species directly at odds with my own: the human species. Genetic testing with an eye for a sub-species with the characteristics mentioned should be able to verify or end the hypothesis. I would suggest that testing mathematicians in the "pure math" field would be a good place to start in revealing the presence of the subspecies. They look human, but they are not. James Harris > So yeah, it's EASY to prove I'm right with non-polynomial > factorization. Posters disagreeing with me can try to attack my [quoted text clipped - 10 lines] > through in some bizarre way with functions shifting how factors > multiply through by the value of x, when any idiot can just do: James, did you see the example I posted, in which we let f(x) = x(x-1)/2 ? Then 2*f(x) = (x)(x-1). I believe you are arguing that, no matter what integer we plug in for x, the 2 must divide either the first factor or the second factor--and it cannot depend on what x is. This is a consequence of the "Logical Principle 2" you wrote about in your blog entry here: http://mymath.blogspot.com/2008/04/re-visiting-non-polynomial.html , and which seems to be the crux of your argument. But it is clear that your Logical Principle 2 fails in this case: f(3) = (3)(2)/2, so when x=3, it is the second factor which is divisible by 2. f(4) = (4)(3)/2, so when x=4, it is the first factor which is divisible by 2. f(5) = (5)(4)/2, so when x=5, it is the second factor which is divisible by 2. f(6) = (6)(5)/2, so when x=6, it is the first factor which is divisible by 2. The pattern is clear. Isn't this exactly the phenomenon which you have called "the tail wagging the dog"? It certainly does happen in mathematics. It is not your arithmetic which has a problem, but rather, your "Logical Principle 1" and "Logical Principle 2" are untrue. On Oct 30, 2:25 pm, anonymous.rubbert...@yahoo.com wrote: > > So yeah, it's EASY to prove I'm right with non-polynomial > > factorization. Posters disagreeing with me can try to attack my [quoted text clipped - 20 lines] > integer we plug in for x, the 2 must divide either the first factor or > the second factor--and it cannot depend on what x is. This is a That's not the same as you don't give f(x), and your deception appears to be deliberate. In my post I give 7*(175x^2 - 15x + 2) = (5b_1(x) + 2)(5*7*b_2(x)+ 7) and readers can see that the function is GIVEN and that's important which is one of the reasons I made that change to emphasize that fact as in the past I've written expression like that as P(x) = 175x^2 - 15x + 2) and 7*P(x) = (5b_1(x) + 2)(5*7*b_2(x)+ 7) which is the same thing but that gap can leave room for doubt. So why does that matter? Because your function is not true over all algebraic integers as you have f(x) = x(x-1)/2 which is NOT an algebraic integer function, as notice that at x=sqrt(2), it gives (sqrt(2)-1)/sqrt(2), which is not an algebraic integer. You cheated. In contrast, 7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7) where the a's are functions of x that are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0 is in the ring of algebraic integers for any algebraic integer x. Details matter. James Harris James Harris > On Oct 30, 2:25 pm, anonymous.rubbert...@yahoo.com wrote: > [quoted text clipped - 64 lines] > > Details matter. Here you have 7*f(x) = g(x)*h(x) and you are asking whether 7 divides g(x) or 7 divides h(x), for specific values of x. In your case f(x) = 175^2 - 15x + 2, and g(x) = 5a_1(x) + 7, and h(x) = 5a_2(x) + 7. Your 7*f(x) will be an algebraic integer for any algebraic integer x, but f(x) is not necessarily an algebraic integer for any algebraic integer x. Above, I wrote 2*f(x) = g(x)*h(x) and I am asking whether 2 divides g(x) or 2 divides h(x), for specific values of x. In my case f(x) = x(x-1)/2, and g(x) = x, and h(x) = x-1. My 2*f(x) will be an algebraic integer for any algebraic integer x, but f(x) is not necessarily an algebraic integer for any algebraic integer x. Doesn't your "Logical Principle 2," or your objection to "the tail wagging the dog," imply that 2 must either always divide g(x) or always divide h(x), no matter which particular value of x is chosen, in my case? On 31 Oct, 02:54, anonymous.rubbert...@yahoo.com wrote: > [...] > [quoted text clipped - 7 lines] > integer for any algebraic integer x, but f(x) is not necessarily an > algebraic integer for any algebraic integer x. Erm... yes it is. The algebraic integers form a ring, so any polynomial expression in algebraic integers is an algebraic integer. The problem is rather that, say, g(x)/7 may be an algebraic integer for x = 0, but not for other values of x. If I have correctly understood James' point, he believes that this somehow constitutes a "flaw" in the AIs, and also that this somehow violates the distributive property; it has been pointed out many times that the distributive property is indeed satisfied by the AIs, and that therefore his claim that the property he thinks they should have follows from the distributive property must be in error, but to no avail. The example I posted also shows that if James' "object ring" doesn't have this property he calls a "flaw" then it necessarily contains non-integer rationals, but he refuses to look at it. > On 31 Oct, 02:54, anonymous.rubbert...@yahoo.com wrote: > [quoted text clipped - 14 lines] > The problem is rather that, say, g(x)/7 may be an algebraic integer > for x = 0, but not for other values of x. Oops--I mistyped what James' f(x) is; I should have written that James' f(x) is (175x^2 - 15x + 2)/7. In this case, for some algebraic integers x, f(x) is an algebraic integer, but for others, 7*f(x) is an algebraic integer but f(x) is not. On Oct 30, 7:54 pm, anonymous.rubbert...@yahoo.com wrote: > > On Oct 30, 2:25 pm, anonymous.rubbert...@yahoo.com wrote: > [quoted text clipped - 70 lines] > > and you are asking whether 7 divides g(x) or 7 divides h(x), for Nope. Division isn't a ring operation. Details matter here, so you need to be VERY specific. In actuality I am noting how given the factorization 175x^2 - 15x + 2) = (5b_1(x) + 2)(5a_2(x)+ 1) you can get the factorization: 7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7) where the a's are functions of x that are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0. Turns out there is only one way. > specific values of x. In your case f(x) = 175^2 - 15x + 2, and g(x) = > 5a_1(x) + 7, and h(x) = 5a_2(x) + 7. Your 7*f(x) will be an algebraic [quoted text clipped - 6 lines] > > and I am asking whether 2 divides g(x) or 2 divides h(x), for specific Division is not a ring operation. > values of x. In my case f(x) = x(x-1)/2, and g(x) = x, and h(x) = x-1. > My 2*f(x) will be an algebraic integer for any algebraic integer x, [quoted text clipped - 5 lines] > always divide h(x), no matter which particular value of x is chosen, > in my case? 2. Given that the value of functions is irrelevant to the action of multiplication you can use a value of convenience when functions are involved where there is a question about how factors have distributed. But your example was 2*f(x) = x(x-1), so there is no question as you have simply that f(x) = x(x-1)/2 which works in the ring of integer, but not in general in the ring of algebraic integers. But the weirder thing here is that is nothing like my example as to start I have P(x) = 175x^2 - 15x + 2 and multiply it by 7. I could instead multiply it times 13. Can you? James Harris > On Oct 30, 7:54 pm, anonymous.rubbert...@yahoo.com wrote: > > > On Oct 30, 2:25 pm, anonymous.rubbert...@yahoo.com wrote: [quoted text clipped - 73 lines] > > Nope. Division isn't a ring operation. That is exactly why it makes sense to ask a question like " does 7 divides g(x)?" If division were a nice, well-defined operation in polynomials over the algebraic integers, like the way division works in a field, then 7 would ALWAYS divide g(x) when g(x) is nonzero-- there would always be some element A with A*7 = g(x), and this is what it means to say "7 divides g(x)." It is precisely because algebraic integers form a ring and not a field that it is meaningful to talk about divisibility there; for some algebraic integers x, there will be some y such that y*7 = x (in which case we say that 7 divides x), while for other algebraic integers x, no such y exists (in which case we say that 7 does not divide x). In the rational numbers, for instance, it's pointless to talk about divisibility; 2 divides 3, for instance, since 3/2 is a rational and 2*(3/2) = 3. But in the integers, divisibility is meaningful; there is no integer y such that 2*y = 3, so 2 does not divide 3 in the integers. > Details matter here, so you need to be VERY specific. > [quoted text clipped - 11 lines] > > Turns out there is only one way. Sure, this factorization is fine--but if I understand your argument, you are saying that, for some values of x, it is the case that 7 divides (5a_1(x) + 7) and not (5a_2(x) + 7), while for other values of x, it is the case that 7 divides (5a_2(x) + 7) and not (5a_1(x) + 7). You seem to think this violates some principle of logic, but you have just shown that it's quite true, anyway. So you have shown that a mathematical result of yours contradicts a principle which seems to you to be true--for intuitive reasons, since you do not have a proof of this principle. Doesn't this show exactly that your principle is wrong? How are you blaming other mathematicians for this, when what is contradicted is simply the principle you have assumed? I do not understand how the problem here rests with the algebraic integers and not with your "Logical Principle 2." About the rest--I think I understand what you're getting at, but at least once I misread what your f(x) is, and I would rather step out of this now and avoid confusing things further by trying to correct things I already said. > > specific values of x. In your case f(x) = 175^2 - 15x + 2, and g(x) = > > 5a_1(x) + 7, and h(x) = 5a_2(x) + 7. Your 7*f(x) will be an algebraic [quoted text clipped - 41 lines] > > - Show quoted text - > Answer is, you can't so I'm right that the 7 multiplies the same way > for ALL x, and can prove that easily enough, but what good is proof if > people simply refuse to acknowledge it? Absence of evidence is not evidence of absence. We don't know of any counter-example to Riemann's Hypothesis. But that doesn't mean we can say "HEY! No one's disproved it, so it MUST be true!" > You betrayed the entire human race. Quick, who wants to invoke Godwin's Law? > Genetic testing with an eye for a sub-species with the characteristics > mentioned should be able to verify or end the hypothesis. I thought eugenics died out around World War II. > > Answer is, you can't so I'm right that the 7 multiplies the same way > > for ALL x, and can prove that easily enough, but what good is proof if [quoted text clipped - 13 lines] > > I thought eugenics died out around World War II. But obviously not the need for it. > So yeah, it's EASY to prove I'm right with non-polynomial > factorization. Posters disagreeing with me can try to attack my [quoted text clipped - 16 lines] > for ALL x, and can prove that easily enough, but what good is proof if > people simply refuse to acknowledge it? Like you simply refuse to acknowledge this? http://groups.google.co.uk/group/sci.math/msg/b04a1231f4c0c85c In that post I apply your method to a simpler polynomial where the factorisations at x = 0 and x = 1 can be found by inspection. It shows that, despite your non-argument based on nothing but incredulity, the factors of 7 split up dependent on x in exactly the manner you say they can't. And you don't get around this by invoking "wrappers", because if what you say about them is true then 7 is a unit in the object ring. > You people killed a freaking math journal over denial of this result > with emails that derailed some editors who did the right thing, until [quoted text clipped - 49 lines] > > James Harris "But I can do what you cannot because I see you as a human being, regardless of what you said, while you clearly do not see me as one." - James Harris ( http://mathforum.org/kb/message.jspa?messageID=6450090 ) > > So yeah, it's EASY to prove I'm right with non-polynomial > > factorization. Posters disagreeing with me can try to attack my [quoted text clipped - 23 lines] > In that post I apply your method to a simpler polynomial where the > factorisations at x = 0 and x = 1 can be found by inspection. It shows The 7 still multiplies through only one way. It must. What is being multiplied cannot control how that is done. If I have 7(x+1)(x+2) = 7x^2 + 21x + 14, I can choose that as (7x+7)(x+2) = 7x^2 + 21x + 14 or (x+1)(7x+14) = 7x^2 + 21x + 14 and there is NO WAY for (x+1)(x+2) to shift how that is done based on the value of x. It's nonsensical to claim otherwise. > that, despite your non-argument based on nothing but incredulity, the > factors of 7 split up dependent on x in exactly the manner you say > they > can't. And you don't get around this by invoking "wrappers", because > if > what you say about them is true then 7 is a unit in the object ring. That's not true. It cannot be true. Consider what you and posters like you are arguing: you claim is that when multiplying a factorization by a constant number, by 7, what is being multiplied controls how that 7 is multiplied dependent on the value of x. That's ludicrous. It's NOT possible. James Harris > > > So yeah, it's EASY to prove I'm right with non-polynomial > > > factorization. Posters disagreeing with me can try to attack my [quoted text clipped - 57 lines] > > That's ludicrous. It's NOT possible. But you have just shown that it IS possible, with a specific example! > > > So yeah, it's EASY to prove I'm right with non-polynomial > > > factorization. Posters disagreeing with me can try to attack my [quoted text clipped - 25 lines] > > The 7 still multiplies through only one way. It must. And yet... it doesn't. Look at the actual example I posted and you will see that it doesn't. > What is being multiplied cannot control how that is done. > [quoted text clipped - 8 lines] > and there is NO WAY for (x+1)(x+2) to shift how that is done based on > the value of x. Not with that example, no, because in that example the factors are polynomial functions. In the application with which you start this thread, however, you factor your polynomial into non-polynomial factors. Facts which are true about polynomial factors need not be true for non-polynomial factors. > It's nonsensical to claim otherwise. > [quoted text clipped - 13 lines] > > That's ludicrous. It's NOT possible. So you say. And yet... it happens. I'm not asking you to believe me, I'm asking you to look at the actual example I posted. Here it is again: let Q(x) be the polynomial Q(x) = 25x^2 + 5x + 2 Like your original polynomial this is primitive and irreducible over the rationals. Also like your original polynomial, we can factor 7*Q(x) as 7*Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) where the a's are now the roots of a^2 - (x - 1)a + 7x^2 = 0. Like your original polynomial, at x = 0 we have that one of the a's is equal to zero, so that at x = 0 we can divide the 7 out of one term. Moreover, the other term is coprime to 7 at x = 0. So it looks the argument you used with your original polynomial, if correct, should apply to this polynomial also. Is this the case? If not, what is the difference which means that your argument no longer applies? Assuming that your argument does apply to Q(x), consider now x = 1. Here we have a = +/- sqrt(-7) so at this value of x, 7 is not a factor of either of the terms (5a_i(1) + 7). But we can find w_1(1) and w_2(1) such that w_1(1)*w_2(1) = 7 and w_i divides (5a_i(x) + 7) such that all coefficients remain in the algebraic integers, namely w_1(1) = w_2(1) = sqrt(7) Exactly the type of factorisation you claim cannot exist. Is this correct so far? Assuming it is, I believe now that your claim is that the w's must "wrappers", meaning that the factorisation 7 = w_1*w_2 is equivalent up to Object ring units to 7 = 7*1; in other words, one of the w's has 7 as a factor in the Object ring. But then the Object ring contains sqrt(7)/7 = 1/sqrt(7), and therefore also contains 1/7. You keep claiming that this can't happen. And yet here is an example where, clear as day, it happens. You can continue to claim that it can't happen, but unless you can actually find a mistake in the above example your claims stand refuted. You can argue that the example I've given can't exist, but no argument you can come up with will make that example go away. > > > > So yeah, it's EASY to prove I'm right with non-polynomial > > > > factorization. Posters disagreeing with me can try to attack my [quoted text clipped - 28 lines] > And yet... it doesn't. Look at the actual example I posted and you > will see that it doesn't. I did look at it. > > What is being multiplied cannot control how that is done. > [quoted text clipped - 48 lines] > > a^2 - (x - 1)a + 7x^2 = 0. Rick Decker posted an example like this one years ago. The simple way to understand it is to consider the factorization BEFORE you multiply by 7, so think about Q(x) = (5b_1(x) + 2)(5b_2(x) + 1) and assume that you do only have 5b_2(x) + 1 multiplied by 7, with the previous factorization so what does that mean? > Like your original polynomial, at x = 0 we have that one of the a's is > equal to zero, so that at x = 0 we can divide the 7 out of one term. [quoted text clipped - 12 lines] > w_1(1)*w_2(1) = 7 and w_i divides (5a_i(x) + 7) such that all > coefficients remain in the algebraic integers, namely Yes, you can. But how can the thing being multiplied tell what is multiplying it what to do? > w_1(1) = w_2(1) = sqrt(7) > > Exactly the type of factorisation you claim cannot exist. Nope. I simply note that starting with Q(x) = (5b_1(x) + 2)(5b_2(x) + 1) if you multiply by 7, and have sqrt(7) split between factors like: 7*Q(x) = (5*sqrt(7)*b_1(x) + 2*sqrt(7))(5*sqrt(7)*b_2(x) + sqrt(7)) then something truly remarkable happens at x=0, which is a contradiction with your original factorization! So that CANNOT have occurred so how is it possible that you have sqrt(7) as factors of both with your situation at x=1? Think about it. The answer is actually rather obvious. > Is this correct so far? Assuming it is, I believe now that your claim > is that the w's must "wrappers", meaning that the factorisation 7 = > w_1*w_2 is equivalent up to Object ring units to 7 = 7*1; in other No. In this case there is no factorization in the object ring because you're forced into the field of algebraic numbers. > words, one of the w's has 7 as a factor in the Object ring. But then > the Object ring contains sqrt(7)/7 = 1/sqrt(7), and therefore also > contains 1/7. Which isn't possible as the object ring doesn't have 1/7 in the ring. > You keep claiming that this can't happen. And yet here is an example No I don't. > where, clear as day, it happens. You can continue to claim that it > can't happen, but unless you can actually find a mistake in the above > example your claims stand refuted. You can argue that the example I've > given can't exist, but no argument you can come up with will make that > example go away. The Decker type example simply highlights a case where you can't stay in a ring which is not also a field. Naively you can say, but at x=1, you can SEE the factors, but if you're in a field that's meaningless right? But how do you know you're in a field? You noted it yourself with 1/7 being forced into the ring, so even YOU pointed out needing to be in a field. James Harris >>>>> So yeah, it's EASY to prove I'm right with non-polynomial >>>>> factorization. Posters disagreeing with me can try to attack my [quoted text clipped - 92 lines] > But how can the thing being multiplied tell what is multiplying it > what to do? This sort of language suggests that you have a peculiar, non-mathematical way of thinking about these things. A number or variable is not a sentient being. > [...] > [quoted text clipped - 26 lines] > > Q(x) = (5b_1(x) + 2)(5b_2(x) + 1) Except that that isn't "the factorization BEFORE [I] multiply by 7". If you require that the b's are algebraic integers for each x then this is a different factorisation to the one I gave above. So what? There are lots of ways to factor a polynomial into non-polynomial factors, and the one you give here is different to the one I gave. > and assume that you do only have 5b_2(x) + 1 multiplied by 7, with the > previous factorization so what does that mean? It means that you made a false assumption. You *don't* get the previous factorisation by multiplying this one by 7, as can be seen by considering x = 1: here a_2 = +/- sqrt(-7), but if you assume that 5*a_2 + 7 = 7*(5*b_2 + 1) you find that b_2 = +/- 1/sqrt(-7), which is not an algebraic integer. You are assuming what you are trying to prove, and my example shows that this is false. > > Like your original polynomial, at x = 0 we have that one of the a's is > > equal to zero, so that at x = 0 we can divide the 7 out of one term. [quoted text clipped - 17 lines] > But how can the thing being multiplied tell what is multiplying it > what to do? Anthropomorphism is not proof. The thing being multiplied doesn't "tell" the thing it is multiplying anything. Nor do things being multiplied "do" anything. What actually happens is that the product of factors I gave has 7 as a factor for all x, but the individual factors I gave have gcds with 7 which vary with x. > > w_1(1) = w_2(1) = sqrt(7) > [quoted text clipped - 12 lines] > then something truly remarkable happens at x=0, which is a > contradiction with your original factorization! No, you are again assuming what you are trying to prove. For a start, this factorisation is different to the one I gave. But more importantly, you only get a contradiction at x = 0 if you assume that factors of 7 split among the factors of Q(x) the same way for all x - but this is exactly what you are trying to prove! That your 5*b_1 + 1 might multiply by sqrt(7) to give 5*a_2 + 7 at x = 1 does not mean the same thing has to happen at x = 0. Thus, no contradiction. > So that CANNOT have occurred so how is it possible that you have > sqrt(7) as factors of both with your situation at x=1? What do you mean? You *do* have sqrt(7) as factors of both at x = 1. That's an easily verified fact. > Think about it. The answer is actually rather obvious. > [quoted text clipped - 6 lines] > In this case there is no factorization in the object ring because > you're forced into the field of algebraic numbers. But this contradicts what you wrote before. Look, here's a quote from the thread "Algebraic integer result with quadratics": *** begin quote *** 7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7) (1) [...] and years ago I came up with what I call the ring of objects: [...] In that ring I found I didn't have a problem! And can simply divide the 7 off again like before: P(x) = (5b_1(x) + 1)(5a_2(x)+ 7) or (5a_1(x) + 7)(5b_2(x)+ 1) where a_1(x) = 7*b_1(x) or a_2(x) = 7*b_2(x) *** end quote *** See? You are saying, for a polynomial exactly like mine, that given the factorisation (1) you can divide either a_1(x) or a_2(x) by 7 and remain in the ring of objects. But if that were true for my polynomial it would mean that the object ring contains 1/7. Unless there is some property by which my polynomial differs from yours, which means that your conclusion no longer applies - but every property of your polynomial that you mention in your "proof" is shared by mine. > > words, one of the w's has 7 as a factor in the Object ring. But then > > the Object ring contains sqrt(7)/7 = 1/sqrt(7), and therefore also [quoted text clipped - 14 lines] > The Decker type example simply highlights a case where you can't stay > in a ring which is not also a field. Prove it. At x = 0 and x = 1 you can divide 7 out of the product (5*a_1 + 7)(5*a_2 + 7) such that all coefficients remain in the algebraic integers. Furthermore, whenever you have challenged people to provide similar factorisations for your more complicated polynomial, they have done so. Can you show that there is some x where this is false? > Naively you can say, but at x=1, you can SEE the factors, but if > you're in a field that's meaningless right? But we're not in a field. > But how do you know you're in a field? You noted it yourself with 1/7 > being forced into the ring, so even YOU pointed out needing to be in a > field. Only if you assume that 7 divides one of the factors for all x. But this is false. The polynomial I gave provides an easy example of this. > So yeah, it's EASY to prove I'm right with non-polynomial > factorization. [snip] You misspelled IMPOSSIBLE. > James Harris Ode to James Harris Somebody said it couldn't be done, But James with a chuckle replied That "maybe it couldn't," but he would be one Who wouldn't say so till he'd tried. So James buckled right in with the trace of a grin On his face. If he worried he hid it. James started to sing and he tackled the thing And James never f.cking could do it. Somebody scoffed: "Oh, you'll never do that; At least no one has ever done it"; But James took off his coat and he took off his hat, And the first thing we knew he'd begun it. With a lift of his chin and a bit of a grin, Without any doubting or quiddit, James started to sing and he tackled the thing And James never f.cking could do it. There are thousands to tell James it cannot be done, There are thousands to prophesy failure; There are thousands to point out to James, one by one, "How hopeless that task set before you." But just buckle in with a bit of a grin, James take off your coat and go to it; Just start to sing as you tackle the thing And James, you'll never f.cking do it.  SignatureUncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/lajos.htm#a2 |
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