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Posts posted by John


I'd start by considering the geometric interpretations of vector addition and subtraction. There are a couple of ways to go about getting your answer, depending on what exactly you notice about the resulting shapes.
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While I don't have time to thoroughly read the paper right now, I did look through it briefly. It doesn't seem to have much to do with my post, so no, it won't do.
You'll have to find someone just a tad less mathematically mature than I am if you're looking to drown someone in equations.
More seriously, if you do find something that addresses the point (read: shows that predicate logic is isomorphic to probability theory), though, I would like to read it (probably after finals this week).
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Cox's "theorem" shows how Bayesian logic can serve as an extension of propositional logic. The first link in the third paragraph of ydoaPs' OP explains this nicely. Perhaps this is what kristalris is referring to.
However, propositional logic isn't sufficient for all of mathematics. Firstorder logic at least is required for much of the foundation of mathematics, and higher orders are also sometimes used when mathematics runs up against the limits of firstorder. I haven't been able to find anything to show that Bayesian logic is also an extension of predicate logic, but perhaps kristalris knows of something along those lines, and I'd be interested in seeing it if so. Until then, the claim that all of mathematics is reducible to Bayes seems a bit off.
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... sounds like here is just closer to 0 than he thinks. The idea of it being "not 0" inherently makes it finite, therefore x/ can not be infinite because it's dividing by a finite (although undefinable) number.
It's not a matter of what I think. The hyperreal number system is used as part of nonstandard analysis. It's simply an example of an area where "infinitely small" makes sense, whereas it doesn't for the real numbers.
The problem of it not being able to be defined because it's undefined is no longer a problem... because it's defined to simply communicate the idea that an infinite number of points that do not take up space can fit into a finite space (or infinite points on a line segment).
[...]
So, there ya go. All problems addressed with the sole exception of "it's not in my math book."
If anyone has any difference objections, I'd love to see them... but all objections that have been presented, so far, are debunked above.
Division by zero is left undefined because definitions lead to logical absurdity. "Not being able to be defined because it's undefined" is nonsense, and "it's not in my math book" doesn't matter and isn't an argument anyone's made.
And you haven't debunked anything.
John has wasted a great deal of effort
Maybe.
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If we're defining the operations in those terms, then the following is more accurate:
Multiplication tells us the result of summing some quantity some number of times. For example, 7*3 is the result of adding three sevens (or seven threes) together, and this equals 21.
Division tells us how many times we need to subtract a quantity from another quantity to reach zero. For example, 7/3 is 2 1/3 (or, if you like, 2.333...) because after subtracting two threes from seven, we must then subtract one (which is a third of three) to reach zero.
Now consider 7/0. This operation should tell us how many zeroes we must subtract from seven to reach zero. Now, whether we subtract three zeroes, a trillion zeroes, or a googolplex of zeroes, the result will still be seven. In fact, any real number of zeroes subtracted from seven will simply result in seven. Thus, there is no real number solution to 7/0. We might say the result must exceed any real number, i.e. 7/0 must be infinity. But as we've seen before, this leads to problems too.
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Therefore " Divide 7 by 0" really means "Divide 7 by nothing"
And isn't that same as Don't divide 7 by anything  ie leave it unchanged.
So where's the difficulty? If you divide a number by 0, you don't divide it at all. The number stays unchanged.
Doesn't that make sense?
An immediate consequence of this definition is the construction of the rational numbers breaks. This is because part of said construction is defining an equivalence relation such that given integers a, b, c and d, with b and d not equal to zero, the two ordered pairs (a,b) and (c,d) are related iff ad = bc, i.e. two rational numbers a/b and c/d are equivalent (read: equal) if and only if ad = bc. But, relaxing the restriction on b and d and using your definition here, we have
[math]\frac{7}{0} = 7 = \frac{7}{1}[/math]
which is problematic since [math]7 \times 1 \neq 7 \times 0[/math]. We also have that for two distinct integers m and n, m = n, which is a contradictory result. This is because [math]m \times 0 = n \times 0[/math], so [math]\frac{m}{0} = \frac{n}{0} \implies m = n[/math].
Of course, we could just find some other way to construct the rationals, but certain other problems arise.
For instance, given [math]\frac{7}{0} = 7[/math], we have
[math]\begin{array}{rcl}7 \times \frac{1}{0} & = & 7 \times \frac{1}{1} \\ \frac{1}{7} \times 7 \times \frac{1}{0} & = & \frac{1}{7} \times 7 \times \frac{1}{1} \\ \frac{1}{0} & = & \frac{1}{1},\end{array}[/math]
i.e. 1/0 is just another way of writing 1. This begs the question of why we need the definition in the first place, since 1/1 works perfectly well. But even ignoring that, it should be clear that if a/b = c/d, then b/a = d/c, so since we have 1/0 = 1/1, we also have 0/1 = 1/1, which is certainly not true.
Edit: Another problem with [math]\frac{7}{0} = 7[/math] is that we could just as easily say the following:
[math]\begin{array}{rcl}7 \times \frac{1}{0} & = & 7 \times \frac{1}{1} \\ 7 \times \frac{1}{0} \times \frac{1}{7} & = & 7 \times \frac{1}{1} \times \frac{1}{7} \\ 7 \times \frac{1}{0} & = & 7 \times \frac{1}{7} \\ 7 & = & 1,\end{array}[/math]
which is another contradiction.
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The properties of the real numbers under addition and multiplication are such that it is perfectly valid to multiply both sides of an equation by a value. The positioning of n doesn't matter, as multiplication of real numbers is commutative. My line of reasoning is correct if we're treating division by zero as a valid operation subject to the usual rules of arithmetic (aside from, of course, the one that says division by zero is undefined) and infinity as a quantity that can be placed in an equation. There is nothing dishonest about it.
And that's the whole point: simply defining division by zero to be infinity leads to a logical contradiction. Defining it to be some other value will also lead to a contradiction. And thus, whatever use you might find for it, at the end of the day, such a definition would yield inconsistent mathematics, which means (if it were taken to be correct) it could be used to prove anything at all. While Gödel showed us that we cannot use mathematical reasoning to prove mathematics is consistent (unless it isn't), we do strive to avoid inconsistencies where we can.
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I very well may be misunderstanding something, but:
this was posted earlier (I'm having issues getting the formula in a quote box:
fake quote box
. Then we have . We can then multiply both sides of the equation by , yielding . Simplifying, we arrive at , which by the transitivity of equality yields . But this is a contradiction, since we assumed
/fake quote box
How are you going from to ? You can drop the 1 on the left side because n x 1=n... but if , you can't just drop the n on the right side of the equation.
n * 0 * infinity = (n * 0) * infinity = 0 * infinity
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So, perhaps, it would be appropriate that the idea of checking [math]\frac {N}{0}= \infty[/math] with [math]\frac {\infty}{0}=N[/math] simply isn't a rule that holds true any more than checking [math]N \times 0 = 0 [/math] with [math]\frac {0}{0}=N[/math].
I'm assuming your second equation is supposed to be [math]N = 0 \times \infty[/math], but regardless, this is true. I'm not sure what your point is with the rest. The property that [math]\forall n \in \mathbb{R}, n \times 0 = 0[/math] is a result of the definition of multiplication for real numbers. While we can use various examples to illustrate why the definition makes sense, we don't need to check it.
It's just Okie dokie that one simply isn't a rule that holds true... yet because the other equally doesn't hold true, we just give up on the whole thing?
No. I mentioned in a previous post that while dividing by zero leads to illogical conclusions when dealing with the real numbers, there are other contexts in which division by zero is definedbut in those contexts, other properties of numbers and division that we take for granted are also slightly altered.
If points on a line don't float your boat, go to the three dimensional version: A Box has a finite area. How many infinitely small objects can fit inside that box? Area A/Area B=number of objects that fit in the box. X/0=[math] \infty[/math]
How is that not dividing by 0?
When dealing with the real numbers, "infinitely small" doesn't make sense. There is a number system called the "hyperreal numbers," in which infinitesimals (these are quantities smaller than any nonzero number, but not equal to zero) and infinity exist. In this system, if we let [math]\epsilon[/math] be an infinitesimal, then [math]\frac{1}{\epsilon} = \infty[/math] and [math]\frac{1}{\infty} = \epsilon[/math], but division by zero is still undefined.
(and no, I'm not ignoring your point about number lines. Comparing infinitely small points with the infinite number of numbers between numbers is a perfectly valid illustration. IMO, a better illustration of dividing by 0 being something other than infinity actually came from a video game I was playing called Heroes of Newerth. A character's ability targets a certain area and does, say, 500 damage divided equally among the number of targets he hits. One target takes full damage. if two targets are in the area, each takes half... but if he hits zero targets, how much damage did he do? 0, obviously. In this case X/0=0. Or, less nerdy: If I make a fruit cake and divide it equally among the people who want a slice... and 0 people want any... how much cake does each person get? 1/0=0 in this situation. However, this is part of the language of math. In any language, sometimes the same thing can have different meanings with a different context. From what I see, there are two perfectly legitimate ways to illustrate X/0 and two completely legitimate answers. The fact that there are two answers depending on the context of the problem doesn't mean the problem is invalid.
The examples here don't really hold water. In the video game, if there are no enemies around, then the attack clearly misses. But as it happens, we can just as easily say the full amount of damage was applied to the zero enemies, i.e. [math]\frac{500}{0} = 500[/math], which doesn't make much sense. The same applies to the cake example. You could say you gave zero cake, the entire cake, or any portion of the cake, to zero people.
The fact that division by zero can yield multiple answers is a problem, though not necessarily a fatal flaw. The bigger issue is the fact that any definition either leads to contradictions, if the usual rules of arithmetic are applied, or doesn't seem to be any more useful than just leaving it undefined in the first place, if the usual rules of arithmetic don't apply.
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K, John. Let's extend that.
[math]A \times B = C [/math] therefore [math]\frac {C}{B}=A[/math]
So, if we can't divide by 0 on the basis of [math]1 = 0 \times \infty[/math] being wrong... then why can we multiply by 0?
So, if [math] 1 \times 0 = 0[/math] then [math]\frac {0}{1}=0[/math], which we agree is legitimate. But also [math]\frac {0}{0}=1[/math]... which obviously doesn't work out. Yet [math]A \times B = C [/math] therefore [math]\frac {C}{B}=A[/math]
This rule only holds for nonzero B. To elaborate, the full series of steps is as follows:
[math]\begin{array}{rcl} A \times B & = & C \\ A \times B \times \frac{1}{B} & = & C \times \frac{1}{B} \\ A \times 1 & = & C \times \frac{1}{B} \\ A & = & \frac{C}{B}\end{array}[/math]
If B = 0, then this breaks down, since 1/B is undefined.
So can we not Multiply by 0's anymore? Obviously we can, even though sometimes it doesn't check out.
Multiplication of any real number by zero always works out.
Being difficult to work backwards doesn't mean it's impossible to work forwards. Specifically, how is the above example of dividing Length A by Length B to get the number of Blengthed segments not a legitimate division? We know there are infinite points in a given line segment because each point doesn't have a length... thus Length A/0=[math]\infty[/math]. Is that not dividing by 0? Or do you believe there are a finite number of lengthless points on a line segment?
I don't think a line is defined as "an infinite collection of points" in geometry. Rather, it's the path of a point moving in one direction, if it's even defined based on a point at allit may be taken as a primitive object. It is true that in analytic geometry, we might define a line as the collection of points with coordinates satisfying a given linear equation, but even this definition isn't based on division by zero. Your last question was answered above, when I explained one way in which we might show that a line segment contains infinitely many points.
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Infinity is not a number, and cannot be put into standard arithmetic expressions as you're trying to do here. Doing so leads to logical absurdity. For example, suppose [math]\frac{1}{0} = \infty[/math]. Then we have [math]1 = 0 \times \infty[/math]. We can then multiply both sides of the equation by [math]n \neq 1[/math], yielding [math]n \times 1 = n \times 0 \times \infty[/math]. Simplifying, we arrive at [math]n = 0 \times \infty[/math], which by the transitivity of equality yields [math]n = 1[/math]. But this is a contradiction, since we assumed [math]n \neq 1[/math]. Of course, we could say that while 1/0 is defined to be infinity, we can't manipulate that equation as I've done here. But then I'd ask what the point is of making the definition in the first place.
There are some contexts in which division by zero is defined, but these don't really conform to the usual notions we have about numbers. Some examples you might enjoy are the real projective line and wheel theory. Note that these structures have some odd properties (for instance, in the article about the real projective line, the author notes that ordering doesn't really work as we'd expect, so we end up with 4 > 3 and 3 > 4 both being true).
As for the line segment example, one option is this: if we consider the length of the segment to be an interval on the real number line, then we can form a bijection from that interval to the real numbers, thus showing us that there are infinitely many points on the line segment (in fact, there are as many points on the line segment as there are real numbersain't infinity grand?).0 
Divisibility by 7 can be checked by doubling the last digit, subtracting that from the number formed by the rest of the digits, and seeing if the result is divisible by 7. For example, given 315, you'd double 5 to get 10, then take 3110 to arrive at 21, which is divisible by 7. Thus, 315 is divisible by 7. The method can be applied recursively.
A number is divisible by 8 if its last three digits are divisible by 8. I guess to check the last three digits, see whether it's even. If so, divide by 2, then see if the resulting number is divisible by 4.2 
[math]x = \frac{at^2}{2}[/math]
[math]t^2 = \frac{20\cdot 2}{10}[/math]
[math]t = 4s[/math]
Unless I'm misreading something, this is not correct. Remember the full equation for x is [math]x = x_{0} + v_{0}t + \frac{1}{2}at^{2}[/math]. Taking the car's initial position to be 0, and the edge of the cliff to be at 20 m, we thus have [math]20 = \frac{65000}{3600}t + \frac{10}{2}t^{2}[/math]. Ignoring the negative root, this gives us [math]t = \frac{8}{9} \approx 0.889\textnormal{ s}[/math], which makes sense, given that 65 km/hr is already a little over 18 m/s.
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Can someone proove btw that every symmetrical number (although, that's what I suspect) is dividable by 11?
(1881, 3649229463, 635536, ...)
This is true if the number has an even number of digits. For odd numbers of digits, it's not necessarily the case, e.g. in the case of palindromic primes.
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Considering how long certain discussions have been allowed to continue on these forums, I think it's safe to say the moderators err on the side of tolerance more than suppression of strange ideas.
The thread in question, and the websites linked therein, don't provide much in the way of evidence. Indeed, the websites are so silly I'd almost think the entire thing is one big troll designed to see just how crazy an idea people are willing to believe. But in a world where Scientology is a thing, one can never be sure.
Also, while logical consistency is great (not that this applies to the Time Cube idea), it doesn't trump experimental evidence in describing the universe. I'm actually not sure the argument is a matter of evidence so much as a semantic dispute. If the thread is reopened, then perhaps we can explore things a bit. However, it's understandable if the thread remains closed.1 
The equation is
[math]\lim_{k \to \infty} \frac{F_{k+1}}{F_{k}} = \varphi[/math]
which, since the limit of a quotient is equal to the quotient of the limits of the numerator and denominator, ends up giving us
[math]\lim_{k \to \infty} F_{k+1} = \varphi \lim_{k \to \infty} F_{k}[/math].
This, of course, ends up with infinity on both sides of the equation.
What you must keep in mind when dealing with this limit is the index k never actually reaches infinity. It simply increases without bound. We can (theoretically) take arbitrarily large pairs of consecutive Fibonacci numbers and find their ratio, and the answer will be within some neighborhood of phi, but phi itself will still be irrational.
As for multiplying an irrational number by a rational number, let n be an irrational number. Multiplying it by 0, of course, will give us 0. So assume multiplying it by some nonzero rational number a/b will result in a rational number c/d. Then what we have is [math]\frac{a}{b}n = \frac{c}{d}[/math]. But this means [math]n = \frac{b}{a}\frac{c}{d}[/math]. Since [math]\frac{bc}{ad}[/math] is rational, n is therefore a rational number, which contradicts our assumption that n is irrational.1 
Are you sure you aren't misreading the answer given in your book?
Solving it one way, I get
AX + B = X
B = X  AX
B = (I  A)X
(I  A)^{1}B = X
which is your answer.
Solving it another way, I get
AX + B = X
AX = X  B
AX  X = B
(A  I)X = B
X = (A  I)^{1}(B)
which is closer to the book's answer.
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1000 cm^{3} = 0.001 m^{3} is correct. If you have that equation and wish to know the conversion factor for 10 cm^{3}, then you'll want to divide both sides of the equation by 100, i.e. 1000 cm^{3} / 100 = 0.001 m^{3} / 100, which will give you 10 cm^{3} = 0.00001 m^{3}, as desired.
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A circle is twodimensional. In the Euclidean plane, it is the set of all (x, y) such that x^{2} + y^{2} = r^{2}, where r is the radius of the circle.
As for the 2D reality being perceived as 3D concept, you might enjoy reading about the holographic principle. That article may be a bit too technical, but I'm sure Googling around would yield some more accessible introductions to the concept. It's not precisely what you're talking about, but it's similar enough in spirit.0 
Well, 10 cm^{3} is indeed 0.00001 m^{3}. Note that 0.00001 is one hundredthousandth. Since we've shown that 1 cm^{3} is one millionth of 1 m^{3}, it follows that 10 cm^{3} must be one hundredthousandth of 1 m^{3}.
Without knowing what exactly you plugged in, I don't know what problem you're encountering (if any) or how to solve it.
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I'm assuming you're talking about the real numbers and Euclidean geometry.
A real number by itself has no spatial extent, thus we might call it dimensionless or zerodimensional, i.e. it is simply a point.
If we take the entire set of real numbers, then we have an infinitely long line on which every point is associated with a real number, and this line is onedimensional.
We can then take all ordered pairs (x, y) where x and y are real numbers (this is the Cartesian product [math]\mathbb{R} \times \mathbb{R}[/math]), and the result is the familiar Euclidean plane, which is twodimensional.
If we then add a third real coordinate z, and take all the resulting ordered triples (x, y, z) (which, similarly to before, is the Cartesian product [math]\mathbb{R} \times \mathbb{R} \times \mathbb{R}[/math]), then we have Euclidean space, which is threedimensional.
We can go even higher than three, though, by adding more coordinates, taking the set of all ntuples (a_{1}, a_{2}, ..., a_{n}) for any natural number n, resulting in Euclidean nspace, which is ndimensional. We often denote this by [math]\mathbb{R}^{n}[/math], e.g. the plane is [math]\mathbb{R}^{2}[/math] and the number line is just [math]\mathbb{R}[/math]. Sometimes [math]\mathbb{E}^{n}[/math] is used to emphasize that we're dealing with a Euclidean space.
Given all this, we might define the dimensionality of a space as the number of coordinates required to specify a point in that space.I'm not sure what you mean when you say shapes are 3D. For instance, polygons are shapes, but they are twodimensional. Polyhedra are also shapes, and they are indeed threedimensional. We also have higherdimensional shapes, e.g. the tesseract in four dimensions. The tesseract is just one example of a class of ndimensional objects known as hypercubes, which, as the name implies, are ndimensional analogues of the familiar square or cube.
As for reality being 4D, I guess with the introduction of Minkowski spacetime that's valid, with the understanding that the time dimension isn't the same as the three space dimensions. The physicists could provide more details about that.
This post is a bit ramblish, as I need sleep, but hopefully it's helped some.
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This was worked out from equating them as follows: 10 cm^3 = 0.1 m^3 ... then, (10)^3 cm^3 = (0.1)^3 m^3 ... which gives 1000 cm^3 = 0.001 m^3, and then using those values in the equation (the process stated).
A minor nitpick here: The first equation should be 10 cm = 0.1 m. The rest of your reasoning then follows.
The problem however is that while this works, it doesn't seem to work out the same via the process of simply going:
(10 cm * 10 cm * 10 cm) / (100 cm * 100 cm * 100 cm) = ..
And since its a volume (or in cubic units), why exactly should accounting for the volume by means of, (x * y * z) / (x * y * z) to put it that way, not work the same ???
You have cm^{3} in both the numerator and denominator, and so they will cancel out. However, what you will end up with is (10 cm)^{3} / (100 cm)^{3} = 0.001, which makes sense since 100^{3} cm^{3 }= 1000000 cm^{3} = 1 m^{3}. This is why unit conversion involves multiplying ratios of different units together to arrive at a final sensible unit for the result. In this case, to match your previous reasoning in expressing (10 cm)^{3} = 1000 cm^{3} as some number of m^{3}, you'd want:
[math](10 \textnormal{cm})^{3} \times \frac{(1 \textnormal{m})^{3}}{(100 \textnormal{cm})^{3}} = 1000\textnormal{cm}^{3} \times \frac{1 \textnormal{m}^3}{1000000 \textnormal{cm}^{3}} = \frac{1000 \textnormal{cm}^{3}}{1000000 \textnormal{cm}^{3}} \times 1 \textnormal{m}^{3} = 0.001 \textnormal{m}^3[/math]
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Ha, I see it now. I would likely never have guessed that. Good job, md65536
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Ohhhh, so I did step 3 backwards? Don't worry about responding if the answer is yes. And  thankyou  in advance.
Well, I see you said not to worry about responding, but I thought I should clarify.
There's nothing wrong with starting with the desired conclusion and working backwards, so long as you make sure your steps are reversible. In your final written proof, you'll need to remember to do this reversal, so that the inductive step proceeds from the assumption to the conclusion, rather than from the conclusion to the assumption.
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The Bayesian Machine
in General Philosophy
Posted · Edited by John
You're misunderstanding how mathematics works. In mathematics, we state axioms, which are taken to be true, and prove theorems based on those axioms. The systems that result may describe the real world fairly accurately, or they may bear very little resemblance to the real world. Regardless, the mathematics is valid (as far as we can tellsilly Gödel).
The Pythagorean theorem doesn't say, "There exist straight lines, and putting three line segments together such that a right triangle is formed, the side lengths are related such that the square of the hypotenuse is equal to the sum of the squares of the other two sides." Rather, based on the axioms of Euclidean geometry, which assume the existence of straight lines and allow the construction of triangles, the Pythagorean theorem states that if a triangle contains a right angle, then its side lengths are related as stated before.
Propositional logic may be sufficient to prove the theorem, so Bayesian logic may be sufficient also. But you've yet to show how Bayes handles logics of higher orders than propositional logic, which are required for much of mathematics. Therefore, the point stands that "all mathematics is reducible to Bayes" doesn't seem to hold water. In any case, if your claim were true, Bayes' theorem still wouldn't deserve the highest praise, given that it's derived from the formula for conditional probability. Thus your ultimate claim seems to be that the Kolmogorov axioms (or something equivalent) provide a complete axiomatization of mathematics, which is problematic for reasons ajb mentioned before.
Give Bayes credit where it's due, but as ydoaPs indirectly mentioned in his OP, recognize its limitations as well.