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Posts posted by John


John, you getting close to insulting me, which I (obviously) do not like
For the resord, I understand the Arhimedean perperty of the Reals perfectly, and I know exactly what is meant by a cyclic group of infinite order. So stop it.
Moreover, my comment about there being "a kargest Real number" was part of a logical argument to show a contradiction in what I assumed to be your point, that no Real number can be written as a nonterminating string of Natural numbers. If that assumption was wrong, I apologize.
Let me ask you this, then. Is it your position that "all Real numbers are finite"?
So my question was, and is, this. What do you mean by "a finite number"? Is it that no Real number can be written as a nonterminating string of Natural numbers?
Or do you mean that, viewed as a set, the "string representation" of a Real number must have finite cardinality?
Or do you mean something else that I have missed?
Calm down. I said you seem to be misunderstanding a couple of things, because your statements above were a bit off. That's not an insult, but rather a possible source of our disagreement here. Perhaps you understand everything perfectly but simply misspoke.
And yes, all real numbers are finite. That's not simply my position, but rather part of the definition and structure of the real numbers. By finite, as indicated in my previous post, I mean finite in magnitude. I'm not sure how else to explain what I mean here. The notation doesn't enter into it. Pi, for instance, is definitely irrational. Its decimal representation is "3." followed by a nonterminating and nonrepeating string of natural numbers, but obviously pi isn't infinite in magnitude.
The number in the OP, however, has no decimal point. It would be a natural number if there were only finitely many primes, but since there are infinitely many primes, as we add each new prime to the end of the number, it increases without bound. Thus, in the limit, the number becomes infinite in magnitude. No natural number (and indeed, no real number) is infinite in magnitude.
If I seem a bit curt, then please forgive me. "Infinity is just a very large number" (usually as an implied corollary to "we can divide by zero") seems to be the Math forum's version of "relativity is wrong."
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OK, I assume as you objected to trailing ellipses in a real number, this implies you think every real number can be written as a terminating string of natural numbers.
Er, no, I don't care about "trailing ellipses," and obviously any nonterminating decimal cannot be expressed as a terminating string of natural numbers. My point is that since a number consisting of a list of all primes (without a decimal point placed anywhere in the number, mind) is infinite in magnitude, that number cannot be real, since there are no infinitely large real numbers. I'm not sure why you're hung up on the ellipses, but I'm talking about the number itself, not the notation.
Since the field [math]\mathbb{R}[/math] is a total order, this in turn implies there is a largest real number. Call it [math]x[/math]. Then the Archimedean property claims that there exist some integer [math]n[/math] and some [math]y >0 \in \mathbb{R}[/math] such that [math]ny >x[/math]. Contradiction
I'm not sure what you're arguing here, but putting a total order on [math]\mathbb{R}[/math] in no way implies there is a largest real number, and if you do believe there is a largest real number, then you have some misunderstanding about how the real numbers work. The real numbers are Archimedean.
Edit: Also, you seem to have a misunderstanding about the Archimedean property. As it applies to the real numbers, the property implies that for *any* real numbers x, y such that y < x, there exists an integer n such that ny > x. That is to say, the property is a statement about *all* y < x, not just a particular y < x.
Moreover, the ingeters [math]\mathbb{Z}[/math] form (under addition) a cyclic group of infinite order, that is, given a generator = 1, then 1 + 1 + 1 +.......= 0 only if we allow this sum to go out to infinity
Well yes, the set of integers has infinite cardinality, but each natural number is itself finite in magnitude. The integers form a cyclic group because any integer can be expressed as a finite sum or difference of 1's or 1's. Adding 1 to itself as many times as we like, we'll never reach 23571113.... In any case, analytic continuation notwithstanding, the sum 1 + 1 + 1 + ... diverges, i.e. it "equals" infinity, not zero.
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Gotta love broken LaTeX. Hopefully it will be sorted soon or something.
Anyway, you are free to disagree, but the fact that no real number is infinitely large is built into the structure. It's why the real numbers have the Archimedean property. It's why the integers with addition form a cyclic group. As far as successors go, what would S(23571113...) be? What natural number n exists such that S(n) = 23571113...?
I guess, all in all, what meaningful difference (besides the symbols) is there between 23571113... and infinity?
I'm not sure what your point is regarding transfinite numbers. Perhaps if the LaTeX is fixed it will be clearer.Edit: Alright, now that the LaTeX in this thread has been restored, I still don't see your point with the transfinite numbers. I'm talking about the set of real numbers, of which the transfinite ordinals and cardinals are not elements. Rather, transfinite ordinals and cardinals describe the order types and sizes, respectively, of infinite sets.
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Hm.
Let d_{i} be the number of digits of the ith prime p_{i}, with d_{0} = 0.
[math]\sum_{i = 1}^{\infty} \left( p_{i} \times 10^{\left( \sum_{j = 0}^{i} d_{j} \right)  1} \right)[/math]
This gives us
2 * 10^{0} + 3 * 10^{1} + 5 * 10^{2} + 7 * 10^{3} + 11 * 10^{5} + 13 * 10^{7} + ... = 2.3571113...
Of course, without a good method for determining the next prime (which, as discussed at length in another thread, we don't currently have), we'd need a list of all primes in the first place to populate the terms of our sum, which would mean we might as well just concatenate the digits of all the primes in the list instead. But, eh.
Also, I'm more convinced now that the number will be irrational, since Bertrand's postulate (which I'd heard before, but apparently forgotten) guarantees that for any integer n > 1, there exists a prime p such that n < p < 2n. Thus we'll never come across a prime whose digits comprise a list of all primes below it, so my particular idea for how this number might be rational fails. I think this also rules out imatfaal's idea.1 
Alright, though you did say any formula we produce should work "in all cases," which I take to mean for general p and general x.
Having reread the thread, I noticed a few things I missed earlier. Are you essentially trying to find a general method for quickly finding the discrete logarithm? Are you looking to break elliptic curve cryptography?
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He was just noting (correctly) that [math]\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}[/math].
In any case, again, the OP's number, being infinite, is not a real number. All real numbers are finite.
It is, of course, still true that for any finite list of primes, the number formed by concatenating their digits is a natural number.0 
I'm not entirely sure what you're asking here, but a couple of things popped into my mind while reading:
1. Using the formula for new_x presented in your OP, if we have to start with x = 9, then we have (x^{2}  1)^{2} = 6400, which means the formula will never reach 10 for any p where 6400 = 0 (mod p), or for which (6400^{2}  1)^{2} = 1677721518080001 = 0 (mod p), etc.2. Though you possibly qualified it by saying "at least in the context of...", I just thought I'd note that 2 is not necessarily a generator of ℤ_{p}^{×} where p is prime. Consider, for example, p = 7.
Of course, I may be misunderstanding entirely what you're wanting to do.0 
The OP is asking about a number whose digits are a list of all the prime numbers. There are infinitely many primes, thus there are infinitely many digits in the number, thus the number is infinitely large, thus not a real number.
As for the rationality of the decimal, I don't think we can absolutely rule out the idea that, every so often, we'll arrive at a prime number whose digits are a list of all the prime numbers below it, since gaps between primes can be arbitrarily large. Someone better versed in number theory could probably say something more definite, but in my mind, while the number is almost certainly irrational, I don't know if we can prove that it is irrational.0 
The decimal is probably irrational, but yeah, I don't think we can entirely rule out the possibility of some sort of repetition.
I also wouldn't say 23571113... is a real number, though, let alone rational, since the fact that there are infinitely many primes means the number must be infinitely large. Of course, imatfaal covered this point earlier, but in light of more recent discussion, it bears repeating.
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Ah, yeah, I suppose so, in which case don't mind me. Also, the constructing all irrationals thing is wrong in any case, since something like 0.101001000100001... obviously doesn't make use of all the primes. This is what I get for trying to do math while rushing to get to work.
Of course, we can still make different numbers by moving the decimal point, but I don't think that affects much. And yeah, Endy, it's interesting to consider.
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Yeah, I can't entirely rule that sort of thing out. Since gaps between primes can be arbitrarily large, it's also possible that we'll occasionally come across a prime whose digits are some permutation of all the primes below it, which could lend itself to some repetition, though since primes can be arbitrarily large I think we'd still run into issues.
As for the number being transcendental, I have no idea, but then I think that, especially since we apparently allow repetition of primes in our number, we can probably construct at least all the irrational numbers, in which case of course some would be transcendental and others not.
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It depends on what specific restrictions there are in the construction. For instance, 29 is prime, but would something like 3.29537... be valid, since 9 is not prime? If not, are only the digits 1 (which isn't usually considered prime anyway), 2, 3, 5 and 7 allowed?
If only those five digits are allowed, then we could easily construct either a rational or an irrational number. If all the primes are required to be part of the number, then it seems clear the result would be irrational, since each prime is a unique sequence of digits and thus there would be no way to construct a repeating decimal from them (and of course the decimal would never terminate, since there are infinitely many primes).
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Not quite. In your first proof, you substituted a result from case 1 into the entirely separate case 2, which was invalid. In your second attempt, you substituted your first assumption into each case, which is valid since your assumptions apply to the entire proof.
When I first read your second proof, I also thought you'd made the same mistake, but then I read it more closely and saw you hadn't.
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[math]1^{\infty}[/math] is an indeterminate form, yes, though the reasoning presented in the original post doesn't entirely work, since, as mentioned, infinity isn't simply a number we can toss in for the purposes of arithmetic.
By the same token, since infinity isn't a number, when we see something like [math]1^{\infty}[/math], we should understand it as shorthand for the result of some limiting process. Furthermore, there are various ways to approach this limit, and various paths lead to various values. This is why it is considered indeterminate.0 
If we allow b to equal 1 and 0^b=0^1=0=0^1
Then
0^(b1)=0^0=0^(11)=(0^1)/(0^1)=0/0
Let's hear your argument against that.
I have no argument against that, as it's a perfectly valid line of reasoning. However, there are other possible interpretations, as discussed in the pages I linked earlier. As I said before, 0^{0} can be considered indeterminate, but there are other possible values, and 0^{0} = 1 is the most generally useful.
Likewise if 0^0=1 then (0^0)^1 should equal 1 since 1^(1) is equal to 1. But it does not
(0^0)^1=(0^1)/(0^1)=(1/0)/(1/0)=(1/0)*(0/1)=0/0
I'm not sure what you're getting at here, but your first step seems wrong, and since 1/0 is undefined regardless, the chain of equalities is invalid. Anyway, using the convention that 0^{0} = 1, if we have (0^{0})^{(}^{}^{1)}, then we have (0^{0})^{(}^{}^{1)} = 0^{(0)(1)} = 0^{0} = 1, as expected.
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Is then the following proof without using contrapositive but only using contradiction, correct ??
Let, a+c<b+c...............................................................................................................................................................................(1)
Let, ~(a<b).......................................................................................................................................................................................(2)
Then by using (2) and and the law of trichotomy we have: a=b or b<a
For a=b => a+c<a+c ,by substituting into (1)
For b<a => b+c<a+c => a+c<a+c,by using the transitive property of inequalities and (1)
Hence a+c<a+c ,contradiction and
Thus ~~(a<b) => a<b
This reasoning is actually correct, though if you were to present the proof (in a paper, or for an assignment, etc.), then you'd want to flesh it out a bit, and some of the wording is a bit odd.
And just for the record, your original post does include a correct proof. Specifically,
in proving the cancellation law in inequalities in real Nos we have the following proof:
Let : ~(a<b)
Let : a+c<b+c
Since ~(a<b) and using the law of trichotomy we have : a=b or b<a
for a=b => a+c=b+c.....................................................................(1)
for b<a => b+c<a+c......................................................................(2)
Thus by contrapositive we have: a+c<b+c => a<b
is sufficient to show what you want to show, though again, you'd probably want to flesh it out. It's only the substitution part that renders the proof invalid.
Of course, all of this using the contrapositive or arriving at a contradiction stuff is unnecessary, since there is a very simple direct proof, as mentioned earlier.
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This is not a conspiracy. Rather, it's simply a useful convention. 0^{0} can take multiple values (or be left undefined/indeterminate) depending on the context. The most generally useful value, however, is 1. Here are a couple of explanations why:
http://www.askamathematician.com/2010/12/qwhatdoes00zeroraisedtothezerothpowerequalwhydomathematiciansandhighschoolteachersdisagree/
http://mathforum.org/dr.math/faq/faq.0.to.0.power.html2 
My formula has not been proven wrong yet. What works to produce the same result is of (1)(1) does not equal (1) but 1 is m^22mn2(mn)^2 as stated in my forum on that link in the bottom of my post.
If we go by your original post, then we have n = 1000, m = 999. Using the modified multiplication you proposed, then, we have (9991000)(9991000) = 998001  1998000  1000000 = 1999999, which we've shown is a counterexample. So yes, unfortunately, your formula has been proven wrong.
But if you use m^22mmn^2 as suggested by you for 1,000 you will get 2004001 which is a prime (or relative prime) number. And has no factors but itself and 1.
2mn, not 2mm. Even using 2mm, though, this results in 1998001 and not 2004001. You must have made an arithmetic error somewhere.
Edit: Also, I looked at the blog post you linked. At the end, you write:
(mn)(mn)2(mn)^2=m^22mn2(mn)^2
This is not quite correct, since (mn)(mn)  2(mn)^{2} = m^{2}  2mn + n^{2}  2(mn)^{2}. That is to say, you left out an n^{2}.
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Well, if I'm understanding your meaning correctly, then there is no need to test. The fundamental theorem of arithmetic guarantees that any integer greater than 1 is either prime or the product of a unique set of primes.
There is at least one restriction on the prime factorization of each end product, though, which is that it cannot contain any of the prime factors of n (and probably of m either, though I'm not as absolutely certain).0 
Alternatively, just use n^{2} + 2mn  m^{2} to yield a positive result. Consider also that since m = n  1, this is equivalent to n^{2} + 2(n  1)n  (n  1)^{2} = n^{2} + 2n^{2}  2n  n^{2} + 2n  1 = 2n^{2}  1.
However, the conjecture does eventually fail. Without going through the large list Endy produced to look for an earlier counterexample, just let n = 1000, and we have 2n^{2}  1 = 1999999 = 17(71)(1657).0 
The law is fine, but you're misapplying it here. I'm not sure how to explain better than I have already, but I will give it a shot.
In case 1, you have that a=b. You use this to immediately say (correctly) that (a+c) = (b+c). Now, in case 2, you have that b<a. Thus, it is not the case that a=b, therefore it is not the case that (a+c) = (b+c). This is why the substitution fails: because in case 2, the statement that (a+c) = (b+c) is no longer true.0 
First i suggest you write to all those authors that they use the word "let" instead of suppose in their proofs to correct their style of writing
No need to get snippy. You were attempting a proof by contradiction, and in that sort of proof, we "assume" that certain things are true. In this case, you were assuming ~(a<b) and (a+c)<(b+c). I understood your meaning just fine, but using "let," it'd be more correct to say, in this case, "let it be the case that ~(a<b) and (a+c)<(b+c)." It's not a big deal, and "let" is fine overall, but the change does add clarity.
Second,on page104 in Suppes's book ,Introduction to Logic i find the following rule concerning substitution:
"if S is an open formula,from S and t_1 =t_2,or from S and t_2=t_1 we may deriveT,provided that T results from S by replacing 0ne or more occurrences of t_1 in S by t_2".
So in our case if we put:
S = b+c<a+c
t_1= a+c
t_2 = b+c
then the derivable formula T IS
b+c<b+c.
I understand that, but your proof involves the two separate (by the trichotomy law, as you mentioned) cases that a=b and b<a. You derived a result from the first case, which was perfectly fine, but then you substituted that result into the entirely separate second case, which is not correct.
Perhaps I can explain more clearly. In case 1, you assumed a=b and derived a result from that. In the second case, you assumed a>b, thus any results from case 1 do not apply.
Essentially, in the substitution line of your proof, you derived a contradiction from the idea that case 1 and case 2 are both true, whereas by the trichotomy law, exactly one is true.
Is this a homework exercise?
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First, about your wording: instead of "Let ~(a<b)" and "Let (a+c)<(b+c)", it's more correct to say "Assume ~(a<b) and (a+c)<(b+c)."
Second, your proof involves the two cases a=b and b<a. Substitution of the first into the second isn't a valid step, since for all real numbers a and b, a cannot simultaneously be equal to b and greater than b. That is to say, these are entirely separate cases and should not be mixed as you've done here.Third, what axioms/results are you allowed to use? If you're allowed to make use of the fact that (b<a) => (b+c)<(a+c), for instance, then assuming you can also make use of additive inverses and the associativity of addition, you can do a pretty short direct proof of this cancellation law without having to resort to an indirect proof of the contrapositive. I'd be a bit more clear and detailed here, but this seems like it might be a homework question, so I don't want to give too much away.
The contradiction arises only when you substituted "=<" with "<"
This should read a+c=<a+c, you forgot the "=" defined by (1) a+c=b+c
This is incorrect. As detailed before, the entire substitution is invalid, but even in the context of the OP's reasoning (the apparent point was to show the logical conflict between (a+c)=(b+c) and (b<a) => (b+c)<(a+c)), the equality shouldn't be carried over.
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Dune is an amazing book. The last brainrelated book I read was John Ratey's A User's Guide to the Brain. It's over a decade old at this point, but I think it's still worth reading..
For myself, I've been reading Andrzej Sapkowski's The Witcher series lately, since I've finally started playing the video games and wanted to know the backstory. Some of the translation is a bit weird, but it's been mostly enjoyable so far anyway.Also, in terms of nonfiction, I've just started William Cook's In Pursuit of the Traveling Salesman, which (as the name implies) is about computational complexity. Reviews on Amazon say it's a pretty decent book, recreational in tone but with some real substance. I'm only two chapters in so far, though, so I'll have to wait and see.
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Deep learning weight adjustment clarity sought
in Mathematics
Posted · Edited by John
If you're referring to the notation itself, it denotes a partial derivative, in this case the partial derivative of the cost function with respect to the variable w_{ij}.
Roughly, conceptually, you can think of this as referring to the rate at which the value of the function changes with respect to the variable. That is to say (since I'm not sure what mathematical training you've had), holding any other variables constant, we change w_{ij} and see how the value of the cost function varies in response.