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Articles by Noah Schweber
Avoiding Medvedev reductions inside a linear order
Abstract While every endpointed interval I in a linear order J is, considered as a linear order in its own right, trivially Muchnik-reducible to J itself, this fails for Medvedev-reductions. We construct an extreme example of this: a linear order in which no endpointed interval is Medvedev-reducible to any other, even allowing parameters, except when the two intervals have finite difference.
Examples of the difference between Topological Spaces and Condensed Sets
Here's a one-paragraph answer: Topological spaces formalize the idea of spaces with a notion of "nearness" of points. However, they fail to handle the idea of "points that are infinitely near, but distinct" in a useful way. Condensed sets handle this idea in a useful way. Let me give a few examples. Say you have a nice space like the line $X=\mathbb R$, and acting on it you have a group $G$.
Opposite of Fermat's Last Theorem?
Clearly we must have $x, y<z$. Suppose $x\not=y$; WLOG, suppose $x<y$. Then this can be rewritten as $$a^x(1+a^{y-x})=a^xa^{z-x},$$ with $z-x, y-x>0$. This gives $$1+ a^{y-x}=a^{z-x};$$ but the right hand side is divisible by $a$, while the left hand side is not. So we must have $x=y$. But then this yields $$2a^x=a^z,$$ which in turn yields $$2=a^{z-x}$$ with $z-x>0$. So there are no nontrivial solutions.
Plantinga's logical argument for mind-body dualism
I think this is barely relevant to this site, but ultimately that it is relevant, and the reason is that it is a good motivataion for modal logic.
calculus - "multiplicative inverse in the modulo of the larger number" what does that mean? - Mathematics Stack Exchange
while I am reading this artical I have read the following paragraph: The interesting thing is that if two numbers have a $\gcd$ of $1$, then the smaller of the two numbers has a multiplicative inverse in the modulo of the larger number. It is expressed in the following equation: and he gives the following example: Lets work in the set $\mathbb{Z_9}$, then $4\in\mathbb{Z_9}$ and $\gcd(4,9)=1$. Therefore $4$ has a multiplicative inverse (written $4^{−1}$) in $\bmod9$, which is $7$.
logic - How can the axiom of choice be called "axiom" if it is false in Cohen's model? - Mathematics Stack Exchange
The existence of a model of a statement does not mean that statement is "true" (whatever that means; see below). For example, the Poincare disk is a model of Euclid's first four postulates plus the negation of the parallel postulate; this does not mean that the parallel postulate is "false." What having a model of a set of statements does mean, is: that set of statements is consistent. In particular, if there is a model of $T\cup\{ eg p\}$, then $p$ cannot be a consequence of $T$.
why do remainders show cyclic pattern?
First, let me address a simpler question: why is there a pattern to the remainders at all? Suppose I have any positive integer $b$ (in your case $b=7$). Then remainders respect multiplication: if $x, x'$ are positive integers which yield the same remainder when divded by $b$, and $y, y'$ are positive integers which yield the same remainder when divded by $b$, then $xy$ and $x'y'$ yield the same remainder when divided by $b$.
Must an algorithm that decides a problem in NP also produce a solution?
I think I have a basic misunderstanding in the definition of a decision problem. It's widely believed that a proof of P=NP would break all modern cryptography, for example:- I'm not sure I understand why. For example, 3-SAT is NP-Complete. So if we had an algorithm that could decide satisfiable or unsatisfiable in polynomial time, we could prove P=NP. Correct?
Can we speak of derivatives of sets?
You're absolutely right that we can think of a sequence of (not necessarily nested) sets as a function from $\mathbb{N}$ to $Sets$, and that - to the extent that we can think about $\mathbb{N}$ and $Sets$ as spaces - we might have a notion of derivative. However, it turns out that $\mathbb{N}$ is too bad for this to work nicely.
Is there a name for this property?
(2) is a consequence of the ring axioms (see https://en.wikipedia.org/wiki/Ring_(mathematics)); in general, an element $x$ with the property that $x*y=x$ for every $y$ is called an annihilator for the operation $*$. For (1), the set of subsets of $X$ (for $X$ fixed) forms a Boolean algebra (see https://en.wikipedia.org/wiki/Boolean_algebra; this is the structure provided by intersection, union, and complement).
Example of a ring with infinitely many zero divisors and finitely many invertible elements
I am preparing to my abstract algebra exam and I try to find an example of such ring (rather simple if possible). Does it even exist? Thank you in advance.
Are there fewer positive integers than all integers?
Before I address it directly, let me make a general point: When talking about infinity, it's important to make all your notions precise. Basically, a lot of the time, if we ask vague questions the answer winds up being "it depends." And this isn't to say that these vague questions are uninteresting; on the contrary, the point is that they're so interesting, that they are really many questions secretly rolled into one. So on to your question.
Gödels incompleteness vs incompleteness
This has been nagging me, and might be an unfit question, but still: I've been taught that completeness of a theory $T$ means that for any sentence $\varphi$ in the language of the theory, we have either $T \vdash \varphi$ or $T \vdash eg\varphi$. If this property doesn't hold (so if there is some $\varphi$ that the theory says nothing about), we have an incomplete theory.
Are there any Turing-undecidable problems whose undecidability is independent of the Halting problem?
We can define an equivalence relation on decision problems: $A\equiv_T B$ if $A$ computes $B$ and $B$ computes $A$. This defines the partial ordering of Turing degrees: decision problems ordered by relative reducibility. The properties of this partial order have been extensively studied. Some results include: Given any nonzero (= not the degree of the computable sets) Turing degree $d$, there is a $\hat{d}$ which is incomparable with $d$ (this answers your question).
Is this visual analogy to Gödel's incompleteness theorem accurate?
Well, first of all I wouldn't call them "theorem" $\infty, ?$, since by definition something's only a theorem if it's provable. :) But this is a very minor criticism.
Both ways seems right to me
Solution 1 (we solv it this way in the exam today) $\int {2e^x \over (e^x-1)(e^x+1)}dx = \int {(\frac{A}{e^x-1} + \frac{B}{e^x+1})dx} \qquad $from this, $A=1$ and $B=1$. Then, Let $\ e^x=u \ $ and $\ e^xdx=du \ $. So the integral changes to $\ \int{2\over u^2-1}du \ $. Then, $\int{({A \over u-1}+{B \over u+1})}du \qquad$ from this, $A=1$ and $B=-1$.
elementary set theory - Is the powerset of the reals any "more uncountable" (in some sense) than the reals are? - Mathematics Stack Exchange
I know that $\mathbb{N}$ is countable and has cardinality $\aleph_0$, and that $\mathbb{R}$ has cardinality $2^{\aleph_0} = \text{C}$ and is uncountable. Are sets with cardinalities greater than $\text{C}$ (like $2^{\mathbb{R}}$, for instance) "more uncountable" in some sense than the reals are? Edit: I am familiar with the proof of the fact that there is no bijection from a set to its powerset.
terminology - What is an adjective for "weaker than weak"? - Mathematics Stack Exchange
I defined a notion (say, some kind of equivalence) in three forms, the first implies the second, which in turn implies the third. I would like to use "strong", (nothing), and "weak" to describe them. But some one of great importance already used "weak" for something strongly related to the second form. To be compatible with previous studies, I have to refer to the first as "strong", to the second as "weak". The third is weaker than the weak. I need an adjective to describe it.
Are the two meanings of "undecidable" related?
$\begingroup$ I am usually confused by questions of the type "could such and such a problem be undecidable", because as far as I know there are two distinct possible meanings of "undecidable". I regard the following as the standard definition: Let $P(n)$ be a statement like "$n$ has at least three prime factors" or "the $n$th Turing machine halts" (we can also forget the statement and consider just its "truth value" function $P:\mathbb N\to\{0,1\}$).
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