Fundie: Math Proves the Resurrection

[sketerpot]

a = x [true for some a's and x's]
a+a = a+x [add a to both sides]
2a = a+x [a+a = 2a]
2a-2x = a+x-2x [subtract 2x from both sides]
2(a-x) = a+x-2x [2a-2x = 2(a-x)]
2(a-x) = a-x [x-2x = -x]
2 = 1 [divide both sides by a-x]


Yet another example of what you can prove with mathematics. There is, naturally, a fallacy here, see if you can't find it.

[Spoler alert!]





Since a=x, a-x = 0, therefore dividing both sides by a-x is undefined.

[Kuroneko]

'Prove' is a bid of a wrong word in these kinds of situations, since it does involve improper operations. Still, some example are still amusing; one that does not involve the same somewhat overused trick as the the above follows:

Thanks to Mr. Sorresso for editing this image.

[brianeyci]

What is that, some kind of perverted integration by parts? All I know is that one plus the integral of cotangent theta with respect to theta is not equal to the integral of cotangent theta with respect to theta (first term of second line compared with first term of first line).

At first glance I don't even have to look at the "trick" to know it doesn't work, one only needs to look at the first equality and the result equality and see that one plus something is not equal to something (you don't even need calculus). I feel ashamed that I can't see it lol.

The "old" trick seems more "hidden" because you have to know not to divide by zero, while with yours you just need to know that 0 + 1 =/= 0 .

Brian

[Shadowhawk]

How does this guy survive his drive to work?

1. The stoplight has a one in two chance of being green. That is, it's either red or green. (We'll ignore the yellow because it's heathen)
2. Your car is either moving or it's not; it's a one in two chance it's moving.
3. The evidence that your car is moving is an argument for the light being green.
4. The chance that your car is not moving based on previous measurements is 1 in 10.
5. Therefore, there is a 99.9% chance the light is green.

I think I translated the formula properly.

[Lord of the Farce]

This stunning conclusion was made based on a series of complex calculations grounded in the following logic:

1. The probably of God's existence is one in two. That is, God either exists or doesn't.

2. The probability that God became incarnate, that is embodied in human form, is also one in two.

3. The evidence for God's existence is an argument for the resurrection.

4. The chance of Christ's resurrection not being reported by the gospels has a probability of one in 10.

5. Considering all these factors together, there is a one in 1,000 chance that the resurrection is not true.

Doesn't most of these points depend on "AND" conditions? So rather than the first two points being 50%, it becomes 25%? So on, so forth?

Swinburne said in a lecture he gave at the Australian Catholic University in Melbourne.

Yikes, to think that I might have even seen this idiot face-to-face.

[wautd]

Doesn't most of these points depend on "AND" conditions? So rather than the first two points being 50%, it becomes 25%? So on, so forth?

I'm pretty sure he made every mistake thats in the book regarding statistical research

[Hawkwings]

This stunning conclusion was made based on a series of complex calculations grounded in the following logic:

Wow, they make it sound so... technical-y and science-y, it's no wonder that it got on CNN.

[Kuroneko]

What is that, some kind of perverted integration by parts?

Perhaps. It does involve integration by parts, yes, but as to whether or not it is perverted... no comment.

All I know is that one plus the integral of cotangent theta with respect to theta is not equal to the integral of cotangent theta with respect to theta (first term of second line compared with first term of first line).

No.

At first glance I don't even have to look at the "trick" to know it doesn't work, one only needs to look at the first equality and the result equality and see that one plus something is not equal to something (you don't even need calculus). I feel ashamed that I can't see it lol. ... The "old" trick seems more "hidden" because you have to know not to divide by zero, while with yours you just need to know that 0 + 1 =/= 0 :twisted:.

Wrong reason, I'm afraid.

[brianeyci]

Perhaps. It does involve integration by parts, yes, but as to whether or not it is perverted... no comment.

Now you did it, I got to try it... u = 1/sinT, dv = d(sinT)/dT... damn it all works out, 0 = 1 .

What about this... sin(pi/2) = 0, therefore there is 0/0 and that term is undefined?

Brian

[Kuroneko]

Now you did it, I got to try it... u = 1/sinT, dv = d(sinT)/dT... damn it all works out, 0 = 1 :wtf:.

Indeed (although dv = d(sin θ), no 1/dθ factor). Identifying the source of error is a great question for calculus students. If you do not think it's the integration by parts, where else could it be?

What about this... sin(pi/2) = 0, therefore there is 0/0 and that term is undefined?

I think you meant sin(kπ) = 0. As with improper integrals, such cases are handled as lim_{x→kπ}[sin(x)/sin(x)] = cos(kπ)/cos(kπ) = 1 (cf. L'Hospital's rule).

[Grog]

'Prove' is a bid of a wrong word in these kinds of situations, since it does involve improper operations. Still, some example are still amusing; one that does not involve the same somewhat overused trick as the the above follows:
[image]
Thanks to Mr. Sorresso for editing this image.

Don't you have to do the integrals over some interval (otherwise it is 0 by definition?)? If you do that the 1 (the sin/sin) disappears.

[Rye]

By the way, ever hear the common fundie argument that "evolution has been disproven via math?" I have. Fundies must think that mathematics is their biggest theological ally.

I've heard an argument like that from a sophist I've talked to on the net (that basically boiled down to Xeno's paradox), but that's an argument he came up with. What are you talking about?

[Surlethe]

Don't you have to do the integrals over some interval (otherwise it is 0 by definition?)? If you do that the 1 (the sin/sin) disappears.

There are two types of integrals. The ones you're thinking about are termed "definite integrals", and involve breaking up the area under a curve into successively smaller rectangles to give a greater and greater approximation of the area, eventually limiting n -> infinity (there's a good introductory explanation here. The ones in Kuroneko's dilemma are "indefinite integrals", and are more aptly termed "antiderivatives". Indefinite integration is the inverse function to differentiation; thus, if du/dv = U(v), to find u you integrate U with respect to v: du = U(v) dv; u = Int (U dv). The basic difference between definite and indefinite integrals is this: definite integrals are numbers. Indefinite integrals are functions. Indefinite and definite integrals are linked through the Fundamental Theorem, which defines a definite integral over an interval in terms of indefinite integrals of the function at the endpoints.

[Surlethe]

'Prove' is a bid of a wrong word in these kinds of situations, since it does involve improper operations. Still, some example are still amusing; one that does not involve the same somewhat overused trick as the the above follows:
<snibby>
Thanks to Mr. Sorresso for editing this image.

Just a wild stab in the dark: could the problem specifically have to do with the lack of an integration constant in the third expression?

Reasoning from general principles: since there are an infinite number of indefinite integrals for a particular function, the integrals in the first expression may not necessarily be equal to each other; thus, to set them equal to zero is ... inept, to say the least.

[brianeyci]

Reasoning from general principles: since there are an infinite number of indefinite integrals for a particular function, the integrals in the first expression may not necessarily be equal to each other; thus, to set them equal to zero is ... inept, to say the least.

Damn how could I not see this, of course. Constant c1 and constant c2 are not necessarily equal, therefore the integrals are not equal. I feel shamed .

Just a wild stab in the dark: could the problem specifically have to do with the lack of an integration constant in the third expression?

The third expression is after he does the integration by parts correct? At that point there is no need to introduce the integration constant until you evaluate the integral. Unless you are talking about the third line rather than the third expression, which then you are right and there's a missing c1 and c2 constants.

Brian

[Grog]

Don't you have to do the integrals over some interval (otherwise it is 0 by definition?)? If you do that the 1 (the sin/sin) disappears.

There are two types of integrals. The ones you're thinking about are termed "definite integrals", and involve breaking up the area under a curve into successively smaller rectangles to give a greater and greater approximation of the area, eventually limiting n -> infinity (there's a good introductory explanation here. The ones in Kuroneko's dilemma are "indefinite integrals", and are more aptly termed "antiderivatives". Indefinite integration is the inverse function to differentiation; thus, if du/dv = U(v), to find u you integrate U with respect to v: du = U(v) dv; u = Int (U dv). The basic difference between definite and indefinite integrals is this: definite integrals are numbers. Indefinite integrals are functions. Indefinite and definite integrals are linked through the Fundamental Theorem, which defines a definite integral over an interval in terms of indefinite integrals of the function at the endpoints.

Oh I always called those primitive functions or something like that. Didn't he then simply forgot to add a constant to it (if F(x) is an antiderivative of f(x) then so is F(x)+C, C is a constant)?

[Grog]

Ghetto edit: damn you already said that.

[Surlethe]

Oh I always called those primitive functions or something like that. Didn't he then simply forgot to add a constant to it (if F(x) is an antiderivative of f(x) then so is F(x)+C, C is a constant)?

Exactly. Both F(x) and F(x) + C are antiderivatives of f(x), but they're not equal.

For an image, think of the antiderivative of 2x: x^2 + C. These parabolas are all the antiderivative of 2x, but are never equal:

The third expression is after he does the integration by parts correct? At that point there is no need to introduce the integration constant until you evaluate the integral. Unless you are talking about the third line rather than the third expression, which then you are right and there's a missing c1 and c2 constants.

I was talking about the third expression; you're right.

[drachefly]

I suppose 50% might be a reasonable Bayesian prior if attempting to iteratively determine the relative probabilities of things working out the way they did given various models. It is after all the minimum information statement.

Let's see. He could use Bayes' theorem and say, "If our god exists, then what actually happened is just about the only way it could have happened. Probability of History given God = 100%. If our god doesn't exist, then what actually happened would have been wildly improbable. Probability of History given no God = 0.01%. Normalize with a prior probability of 50%, and we get..."

In that case the 50% isn't the objectionable part, it's the 100% and the 0.01%.

[Max]

This thread has officially cause me to have a brain aneurism.

[Kuroneko]

Indefinite integrals are functions. Indefinite and definite integrals are linked through the Fundamental Theorem, which defines a definite integral over an interval in terms of indefinite integrals of the function at the endpoints.

No. Indefinite integrals are equivalence classes of functions. If they were simply functions, then the above attempt at showing 0 = 1 would have been correct, since f(x) - f(x) = 0 for any f:R→R.

Reasoning from general principles: since there are an infinite number of indefinite integrals for a particular function, the integrals in the first expression may not necessarily be equal to each other; thus, to set them equal to zero is ... inept, to say the least. :P

Precisely correct.

I suppose 50% might be a reasonable Bayesian prior if attempting to iteratively determine the relative probabilities of things working out the way they did given various models. It is after all the minimum information statement. Let's see. He could use Bayes' theorem and say, "If our god exists, then what actually happened is just about the only way it could have happened. Probability of History given God = 100%. If our god doesn't exist, then what actually happened would have been wildly improbable. Probability of History given no God = 0.01%. Normalize with a prior probability of 50%, and we get..."

But such an approach is even worse than what was in the OP. Since P(G|H) = P(H|G)P(G)/P(H) = P(G)/P(H) [P(H|G) = 1 assumed above], it follows that a prior probability of P(G) = 1/2 with P(H) < 1/2 give a probability measure greater than 1 for P(G|H). Hence, this approach is logically inconsistent rather than simply having unjustified probability measures.

[Kuroneko]

I suppose 50% might be a reasonable Bayesian prior if attempting to iteratively determine the relative probabilities of things working out the way they did given various models. It is after all the minimum information statement. Let's see. He could use Bayes' theorem and say, "If our god exists, then what actually happened is just about the only way it could have happened. Probability of History given God = 100%. If our god doesn't exist, then what actually happened would have been wildly improbable. Probability of History given no God = 0.01%. Normalize with a prior probability of 50%, and we get..."

But such an approach is even worse than what was in the OP. Since P(G|H) = P(H|G)P(G)/P(H) = P(G)/P(H) [P(H|G) = 1 assumed above], it follows that a prior probability of P(G) = 1/2 with P(H) < 1/2 give a probability measure greater than 1 for P(G|H). Hence, this approach is logically inconsistent rather than simply having unjustified probability measures.

That's not representative of your statements; mea culpa. However, applying Bayes' theorem on both sides of P(H|G)>P(H|¬G) and rearranging gives P(G|H)/P(¬G|H) > P(G)/P(¬G). For this to turn into P(G|H) > P(¬G|H), which is what is wanted, it must be the case that P(G) ≥ P(¬G), i.e., probability prior to evidence, but there is no reason to believe that.

[Surlethe]

Indefinite integrals are functions. Indefinite and definite integrals are linked through the Fundamental Theorem, which defines a definite integral over an interval in terms of indefinite integrals of the function at the endpoints.

No. Indefinite integrals are equivalence classes of functions. If they were simply functions, then the above attempt at showing 0 = 1 would have been correct, since f(x) - f(x) = 0 for any f:R?R.

I stand corrected.

Reasoning from general principles: since there are an infinite number of indefinite integrals for a particular function, the integrals in the first expression may not necessarily be equal to each other; thus, to set them equal to zero is ... inept, to say the least.

Precisely correct.

Do I get a cookie?

[drachefly]

... applying Bayes' theorem on both sides of P(H|G)>P(H|¬G) and rearranging gives P(G|H)/P(¬G|H) > P(G)/P(¬G). For this to turn into P(G|H) > P(¬G|H), which is what is wanted, it must be the case that P(G) ? P(¬G), i.e., probability prior to evidence, but there is no reason to believe that.

You have taken the stronger form of what I said and greatly weakened it here. Recall, I was positing that they would put forth some concrete figure of how much more likely history was with God in place, not merely say that it was genericaly 'more' likely.

The claims I was supposin they might make, in order:
P(H|G) = X P(H|¬G), with X >> 1 (your version: P(H|G)>P(H|¬G) )
P(G|H)/P(¬G|H) = X P(G)/P(¬G) ( your version: P(G|H)/P(¬G|H) > P(G)/P(¬G) )
they then only need to suppose P(G)/P(¬G) > 1/X
instead of P(G)/P(¬G) > 1

So yeah, the 50% figure is just as bad. Ok. However, merely getting P(G)/P(¬G) < 1/2 or something like that just wouldn't be an adequate counterargument, if X > 2.

[Kuroneko]

You have taken the stronger form of what I said and greatly weakened it here. Recall, I was positing that they would put forth some concrete figure of how much more likely history was with God in place, not merely say that it was genericaly 'more' likely.

The very first premise of the OP is equivalent to X = 2. That is simply not their argument, which I assumed you attempted to mirror. But even the premise that P(H|¬G) is small is very suspect; in a truly infinite universe, P(H|U) = 1, provided all of its areas are physically compatible with a given history. Unless, of course, one moves the question to the universe itself rather than a particular history within the (given) universe, but then it becomes the standard teleological argument that does not deserve to be hailed as anything substantially new (implied in the OP).