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Aug
26
comment Is it possible to find the criminal with graph-theoretic methods?
There is one way to write a boolean expression to check that, isn't it? I'm just not sure if it would be: $$\neg d\land m\land \neg m\land \neg d\land d\land \neg g\land m\land M\land j\land \neg g$$ Or $$(\neg d\land m)\lor (\neg m\land \neg d)\lor (d\land \neg g)\lor (m\land M)\lor (j\land \neg g)$$
Aug
26
comment Is it possible to find the criminal with graph-theoretic methods?
I changed it, but it was what I meant with: Knowing everyone told a lie, who was the criminal?.
Aug
26
revised Is it possible to find the criminal with graph-theoretic methods?
added 13 characters in body
Aug
26
comment Is it possible to find the criminal with graph-theoretic methods?
No, each one told only one lie.
Aug
25
revised Is it possible to find the criminal with graph-theoretic methods?
added 626 characters in body
Aug
25
asked Is it possible to find the criminal with graph-theoretic methods?
Aug
24
asked $\forall a[P(a)\implies Q(a)]\wedge \forall a[Q(a)\implies P(a)]\stackrel{?}{\equiv} \forall a[[P(a)\implies Q(a)]\wedge [Q(a)\implies P(a)]]$
Aug
22
comment The method of exhaustion in tom apostol calculus vol $1$
You can skip this intro and keep reading it nonetheless. As you progress through the book, you may have developed more mathematical maturity. I tell you this because the first time I took this book, I knew very little mathematics and also found the introduction a little hard.
Aug
21
asked $1+2+3=\int_{0}^{\infty}t^3e^{-t} dt$?
Aug
18
comment On the inner workings of induction?
Good, that's the spirit. When I read about induction, everything seemed to be connected and it made no sense. After some thought, I realized that induction is one thing that can be used to prove ideas in another ideologically feasible frameworks, such as graphs, semirings (in this case), etc.
Aug
18
comment On the inner workings of induction?
No, I'm not trying to prove it's right. I'm trying to understand what are the laws that forbid me of writing $f(k+1)= f(k)+(k+1)$. And I guess it has something to do with the semiring axioms.
Aug
18
comment On the inner workings of induction?
I mean, to write $f(k+1)$ as $f(k)+(k+1)$, I need to have the idea of equivalent rewriting, and I guess this idea comes from the semiring axioms.
Aug
18
comment On the inner workings of induction?
Why doesn't induction works on semirings? After all, we're dealing with $\Bbb{N}$, addition and multiplication. And it follows these axioms.
Aug
18
comment On the inner workings of induction?
@KittyL Yes. The weird thing is that using the first $5$ numbers of that sequence, it yields preciselly the formula of the sum of the first $n$ positive integers. I'm curious about why does it gives precisely $\cfrac{x(x+1)}{2}$ and not some other random formula. But I'll ask this on another ocasion.
Aug
18
comment On the inner workings of induction?
@whacka Yes. I've also looked for some easy sequence (sharing some of the first terms) on OEIS. But as I read Farrugia's article I mentioned in the text, I had to use it because seemed fun.
Aug
18
revised On the inner workings of induction?
added 64 characters in body
Aug
18
comment On the inner workings of induction?
Sorry, @whacka. We edited at the same time.
Aug
18
revised On the inner workings of induction?
added 305 characters in body
Aug
18
asked On the inner workings of induction?
Aug
17
comment How do I prove this $\frac{dx^n}{dx}=nx^{n-1}$ is true for every $n\geq 1$ to convince my students?
@AndréNicolas That is the proof that is also given by Hairer/Wanner: Analysis by It's History. I guess that it's the less bureaucratic way to do it, since it rely only in the proof of the product rule, which is quite simple.