> Given their non-constructive nature "real" numbers are unsurprisingly totally incompatible with computation.
It is funny you say that when Turing defined Turing machines to compute real numbers (like π for example). In its original definition, a number was computable if its Turing machine did not stop. Which makes sense since π does not have a finite decimal expansion.
Today, we usually define Turing machines to decide problems and a problem is decidable if for every input its Turing machine stops with a ``yes'' or ``no'' answer. I guess this is what makes people think what you said in the quote above. Maybe this definition is more intuitive but this conclusion from it could not be more wrong.
Think about it for a second, if the computable numbers were countable there would be no uncomputable problem (Turing actually used the classic cantor diagonal argument to prove that there were uncomputable numbers)
The set of computable numbers is actually countable (see ref. linked above). It has to be by definition because the set of finite computer-programs is itself countable.
This is the whole point of the un-reality of "real" numbers: "all" of it (= measure 1) is uncomputable except a "tiny" measure-0 set.
he was a physics and math major and did not know eigenvectors and eigenvalues? i would like to know how is this possible. can someone explain it to me?
He is a bit older. Linear algebra is also very old, but it didn't really become the field we know today until the 1950s. I would add that in 2025 it is cheap to buy a computer that can solve large linear systems, but that certainly wasn't true in 1975, so linear algebra was less applicable in the real world.
I am not too familiar with the pedagogical history of linear algebra, but I've been reading some advanced undergraduate geometry texts from the 30s-60s and linear algebra was generally not an assumed prerequisite. There was a particular separation between the studies of "two and three dimensional vector spaces over R" (largely geometric) versus "finite dimensional vector spaces over a field" (entirely algebraic), and determinants were presented directly as volume computations. These days undergraduates mostly treat R^2 and R^3 algebraically, maybe at the expense of geometric understanding. (E.g. Euler's rotation theorem is easily proved when restated as a theorem about matrices over R^3 with determinant +1, but Euler's original statement and proof using spherical trigonometry is deeper.)
I was a double major, one in physics, in the 80s. After the three semester engineering physics classes, intro QM was taught spring sophomore year. We used Liboff. In addition, it was required for all physics, chem and engineering majors to take math 20(5?) which was linear algebra.
And given that most of basic QM was formalized by 1930 and relies upon eigenvectors, hard to see any physics course taught since that time not having it.
Whoa, Liboff, that book... I only vaguely remember it now (took QM in 1988). I took "math for mathematicians" (Math 25) instead of "math for physicists" (Math 22?), but remember my classmates who took first year "math for physicists" got eigenvectors very quickly right off the bat in the pre-published book they used https://www.cambridge.org/core/books/course-in-mathematics-f...
I was introduced to eigenvectors in a math course on linear algebra. They seemed esoteric but I could prove theorems and stuff… cool but kind of forgettable.
Then I took quantum mechanics. That’s where I learned eigensystems. That’s where their utility and beauty were beaten into me, problem set by problem set. In quantum mechanics, eigensystems are ubiquitous: from using ladder operators to solve the harmonic oscillator in an elegant way, to what quantum numbers actually are, to the reason behind the Heisenberg uncertainty principle, and to the so many different ways to use perturbation theory to explain atomic and molecular spectra.
You can do the basics of quantum mechanics without explicit linear algebra, and many intro physical chemistry texts aren’t able to assume the math as a pre-requisite and have to do that. But it’s tedious and awkward, like trying to learn physics without calculus.
Same boat as the author here, except I switched from physics to software after year three.
I had never heard of them until I was _years_ into software engineering. I think this is more common than you may think. I had never dealt with linear algebra in a formal setting, despite leveraging a lot of the concepts, until then.
I asked myself the same thing. The article said “learn” Linear Algebra, not “review” Linear Algebra. Do some undergrad math programs not teach Linear Algebra?
In my experience (maths degrees at two universities with >15000 students, one of which I now teach at) group theory and abstract algebra more generally are much more likely to be part of a maths degree than topology. I've never heard anyone describe topology that way.
Speaking as a CS/Math dual major from the late 1980s, the explanations in the textbook were ... uselessly bad where I went (SFU). So while I know the math involved, I didn't know the terms until I retook it from a teacher who actually taught worth anything, years later.
(I still don't use the terms, they're not very ... well, they're awkward and while my embedded work sometimes calls for the math, the terms ... meh. There are clearer ways to put things!).
did not read the whole document. just the proof of theorem 1 and it has a minor error. it is not obvious that C(n) != x. that is because it is not true. to see that, consider a map C(n) such that
C(n) = 0.10000000...
that is: c_1(n) = 1 and c_i(n) = 0 for all other i.
If you let \overline{c}_1(n) = 0 and c_i(n) = 9 for all other i,
then we have that c_i(n) != \overline{c}(c)_i(n) for all i but C(n) = x.
This is a common mistake people make. To fix it, one needs to be more careful when defining \overline{c}_i.
> To fix it, one needs to be more careful when defining \overline{c}_i.
I think the usual way to fix it is actually to be more careful when defining C(n) - that is, to specify that the value of C(n) is a sequence of digits, with the sequences in 1:1 correspondence with [0,1], not itself a value that belongs to the interval [0,1]. Then you can just say that trailing zeroes, or trailing nines, aren't allowed, giving you a unique representation for every value in [0,1]. At that point c-bar will work as defined.
The genus is insufficient to determine if an object is equivalent to the other. Orientability distinguishes the mobius strip and the torus, a torus is orientable whereas a mobius strip is not. Therefore, topologically speaking they are not equivalent.
you can make a mobius strip with paper. then get a pencil and try to orient it in the mobius strip. that is, make it normal to the paper then move it around. you will see that if you go though the strip and go back to the starting point the pencil will be in the other direction. thus, the orientation is not continuous so the surface is not orientable
It is funny you say that when Turing defined Turing machines to compute real numbers (like π for example). In its original definition, a number was computable if its Turing machine did not stop. Which makes sense since π does not have a finite decimal expansion.
Today, we usually define Turing machines to decide problems and a problem is decidable if for every input its Turing machine stops with a ``yes'' or ``no'' answer. I guess this is what makes people think what you said in the quote above. Maybe this definition is more intuitive but this conclusion from it could not be more wrong.
Think about it for a second, if the computable numbers were countable there would be no uncomputable problem (Turing actually used the classic cantor diagonal argument to prove that there were uncomputable numbers)