Always a bad idea to demand that people who are stronger than you should be even stronger than they already are.
Lessons In Twisted Logic From A Millennial Regressivist
I find a lot of contradictions in life that I prefer sharing in the rudest, most G-rated way possible.
Wednesday, July 22, 2026
Tuesday, July 21, 2026
Does the Omniscient Viewpoint Matter?
Here again I'm arguing against moral relativism because it's something I like to do. Perhaps someone has used this argument before and I've just never seen it.
Suppose we have a court case, and the man is found guilty by the jury because the evidence is compelling and strong. But also suppose the man is innocent despite the indication of the evidence (kind of like a Sherlock Holmes story). The jury just voted him guilty because of their limited perception, but really did their best given their lack of omniscience. You can't bash their integrity for doing what they could with the tools they had.
Now moral relativists would classify the man on trial as immoral because his moral status was determined by consensus after some due diligence. Even more so if the general public pretty much agrees with the verdict. But what about the omniscient viewpoint - you know, the only one that can even potentially detect his innocence? Does that viewpoint mean anything?
Friday, July 17, 2026
We Can’t Say Something Positive is Impossible???
I’ve spent some time lately looking at Godel’s ontological proof and trying to comprehend it the best way I can. Axiom 5 seems the most troubling to me out of the whole argument, but today I want to focus on just Theorem 1 of the proof and the two axioms leading up to it. Here it is below, copied from Wikipedia for quick reference:
Theorem 1 says if a property is positive, then it’s possible there exists such a property. That seems reasonable. Positive things do exist in this world after all. But if we look a little more deeply at this statement, this claim is a bit stronger than it looks, because we’re saying we cannot definitively claim ANY positive property is completely impossible in all potential worlds. We can’t call any positive activity completely impossible.
And Godel seems to use a slick method to claim this, using the vacuous truth rules, which I’m always suspicious of because I think they were just made to keep the truth value system closed to two values instead of opening up a third undecided value.
First Godel assumes Theorem 1 isn’t true in an
attempt to derive a contradiction. He
says “Phi is a positive property AND it’s necessary that for all x, NOT phi of
x,” which is the opposite of what Theorem 1 says. This makes sense. To negate an if-then statement you assert the
premise and the opposite of the conclusion.
After that he notices that the portion of Axiom 1 stating
“it’s necessary that for all x, phi of x” opposes a portion of his assumption. Plus this portion of Axiom 1 is the beginning
of a conditional. So his assumption to
start the proof of the theorem creates a vacuous truth in Axiom 1 where the
axiom claims phi of x. Because of this,
by the rules of vacuous truths, the if phi of x then psi of x statement in
Axiom 1 is ALWAYS true, no matter what value psi of x takes.
So Godel just takes that psi of x and makes it equal
negative phi of x. So the output of
Axiom 1 after this substitution says if (phi is a positive property) and (it’s
necessary that for all x, if phi of x then NOT phi of x) then NOT phi is a
positive property. Kind of a nasty
double conditional. The conditional
inside the conditional was a vacuous truth.
But the outer conditional has its premise consisting of two true
statements, one being phi is a positive property, which we already asserted as
soon as we started the proof by contradiction, and the vacuous truth. So both parts of the conditional premise being
true will indeed yield the result of NOT phi being a positive property being a
true statement.
All that mess is not easy to write out. And we did it all just to get the statement
“NOT phi is a positive property.” Now we
can finally go to Axiom 2, and claim that since NOT phi is a positive property,
then that implies phi is NOT a positive property. Easy enough.
And of course now we have just concluded phi is NOT a
positive property, even though the very first thing we did at the beginning of
proof by contradiction was to assume “Phi is a positive property AND it’s
necessary that for all x, NOT phi of x.”
So we have phi being a positive property and phi NOT being a positive
property. This is a contradiction, and
completes our proof by contradiction. We
have shown that if phi is a positive property then it’s possible there exists
some x with that property.
But of course I am very suspicious of a vacuous truth being
used. So I asked my Gemini app if it
could produce an example of a similar scenario where this proof structure is
followed, and the axioms seem reasonable, but the conclusion seems like
nonsense. Mostly as an attempt to
discredit the use of vacuous truths in proofs in the future. It cleverly suggested to plug in the phi
variable as “impossible crime” and the P variable as “legal.” It also suggested phi as invisible socks and
P as fashionable, and phi as boiling ice cream and P as delicious. These seem CLOSE to fitting the axioms, but I
still don’t feel they fit the axioms comfortably enough to my satisfaction when
I sit down and try to plug them in.
Tuesday, June 30, 2026
A pretty picture involving the Gaussian Integral
The Gaussian integral is a neat and very easy representation of the bell curve with the area under the curve exactly equal to the square root of pi. Nice simple-looking expression. The usual bell curve formula looks a lot worse since it's a family of curves and it also sets the area under the curve equal to one instead of the square root of pi. Or something of that nature.
Inspired by that expression, I used Desmos to plot the picture below. The darker shape is just a semi-circle with radius equal to the fraction one over the 4th root of pi. But I designed the plot so that the dark semi-circle area in the picture was EXACTLY equal to the light orange area between the semi-circle and the bell curve. I think that precise relationship makes a geometric picture such as this look quite nice, even though it's difficult to intuitively tell the areas are indeed equivalent. The picture kind of looks like a pretty sunset if you fill it in with the right colors. Both areas should be equal to the square root of pi over two if I didn't make any errors.
Monday, June 29, 2026
Why is the number e in the bell curve equation?
When I think of bell curves, I tend to think more of Galton boards than a graphical representation of test score averages. Galton boards are those fun things with all the marbles in them that you flip over and most of them gather in the middle while a few go off to the sides. Big enough Galton boards can approximate a bell curve fairly accurately. Really interesting seeing how even though each ball follows an unpredictable path, altogether they form a shape that's quite predictable, if you have enough of them.
Anyhow, Galton boards have binomial coefficients in their mathematical calculations - like many things in probability do - and of course binomial coefficients have factorials. But what's interesting is factorials can be represented by Striling's approximation. This estimates n factorial to be pretty close to (root (2*pi*n))*(n/e)^n. Hopefully I typed all that out correctly, but I'm sure it's not hard to find online.
Well anyhow, that's how we see an e appear, if we trust formulas like that and feel comfortable looking up the derivations of those formulas later. Stirling's approximation is involved in the approximation of a factorial, and if you plug those approximations into the binomial coefficients in the formulas similar to Galton board type models, you get a formula called the De Moivre-Laplace Theorem.
The Galton board shows how a large accumulation of randomized left-right choices from a centralized point (which I guess we could think of as a mean) forms a bell curve, and the De Moivre-Laplace theorem gives the formula showing the e in that equation.
Friday, June 26, 2026
A Look at Why Anti-Derivatives May be Tougher than Derivatives
The definition of the derivative is the limit as x goes to zero of (F(x+dx)-F(x))/dx. If you're given a function, it's not too bad to plug it into this thing, do some algebra, take the limit, and see what happens.
But try going backwards. Try taking a function, then doing algebraic manipulations to get another function into this form. Seems to be much tougher, and I guess there are a lot more potential options you could do, but then getting them to simplify in the right way seems somewhat daunting if you're like me and not all that great at algebraic manipulation.
Let's take something simple like 2x for example, and try getting its anti-derivative by algebraically manipulating it to look like something that fits the definition of the derivative.
2x
First let's add dx to it. Since dx will just be going to zero when the limit is taken, I guess it's kind of like adding zero.
2x+dx
Then multiply by 1, but put 1 in the form of dx over dx.
(2x+dx)(dx/dx)
Multiply the top dx through.
(2xdx+dx^2)/dx
Add zero to the numerator, with zero in the form of x squared minus itself.
(2xdx+dx^2+x^2 – x^2)/dx
Shift one of the x squareds to the front.
(x^2+2xdx+dx^2 – x^2)/dx
Realize the first three terms can combine.
((x+dx)^2 – x^2)/dx
Then realize that we now have the function x^2 in the form of the limit as x goes to zero of (F(x+dx)-F(x))/dx.
This implies that the derivative of x^2 is 2x, which means the anti-derivative of 2x is x^2.
That was a LOT of algebraic manipulation I would have NEVER thought of doing if I didn't do this problem backwards to begin with. And this is one of the easiest anti-derivatives. Imagine if we tried something harder.

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