Let $d \geq 2$ and $K$ some positive integer. Consider distinct points $\theta_1, \ldots, \theta_K\in \mathbb{T}^d$ and (not necessarily distinct) $z_1, \ldots, z_K \in \{-1,1\}$, does there exist an eigenfunction of the Laplacian $f: \mathbb{T}^d \to \mathbb{R}$ such that $sgn(f(\theta_i)) = z_i...
oh my bad iam sorry , i was saying for tall man that are weak
i see in real life we see tall man that are weak , but this statement say if a person is tall that it can not be weak ( as implication is used to designate this relation)
i find it hard to deduce when i have to analyse a statement like i did above to tell that whether is it true or not and when a statement is taken as argument
like say P , Q as premises and R as conclusion
here conculsion is " if a man is strong then he is tall"
P; a man is strong
Q: a man is tall
so using this to check whether ihe argument is right we check whether the ( p^q ) => ( p=>q) is tautology
and it is tautology thus this tells that argument is correct which we have seen that argument is wrong
so basically i know how to analyse statement and arguments correctly
but i do not know when to tell whether a statement is argument or not
Hi, when considering the cayley graph of $\mathbb{Z}_p \times \mathbb{Z}_p$, with respect to a generating set $S={s, t}$, where $|s|=|t|=p$, $p$ is a prime, is there a way to denote a Hamiltonian cycle in a notation form in terms of $s$ and $t$?
Like, a Hamiltonian cycle in this Cayley graph is s t s^2 t... or something like that?
'Suppose $I^k$ is $k$-cell in $\Bbb R^k$, consisting of all $\vec {x}=(x_1,\ldots,x_k)$ such that $$a_i\le x_i\le b_i\,\,\,(i=1,\ldots,k)\tag 1$$. $I^j$ is $j$-cell in $\Bbb R^j$ defined by the first $j$ inequalities (1). $f$ real continuous function on $I^k$. Put $f=f_k$ and define $f_{k-1}$ on $I^{k-1}$ by $$f_{k-1}(x_1,\ldots,x_{k-1})=\int_{a_k}^{b_k}f_k(x_1,\ldots,x_{k-1},x_k)\,dx_k.$$ The uniform continuity of $f_k$ on $I_k$ shows that $f_{k-1}$ is continuous on $I^{k-1}$.' Why is $f_{k-1}$ continuous?
Let $(X,d)$ be a metric space and $A$ be a subset of $X$. Then the
following statements are equivalent.
$A$ is nowhere dense.
$\overline{A}$ doesn't contain any non-empty open set.
Each non-empty open set has a non-empty open subset which is disjoint from
$\overline{A}$
Each ...
@299792458 If it is a question that is of interest to someone doing mathematical research, post on MO. If it is a question you can't answer, say an exercise in an undergradute textbook or graduate textbook, post on MSE.
eg , the statement " if a man is tall , then he is strong " is false because , a man can be tall but weak in this case the statement is false but the case could happen
as you could see the possible combination of above two atomic statement can be
This statement is false because a man can be tall and weak. However, there are still tall and strong man. The use of "valid argument" is not even a question, because those are the wrong words to use in English.
Arguments are classed as "valid or invalid" and "sound or unsound"---validity refers to "whether the reasoning really is logical, irrespective of the premises", and soundness refers to "whether the premises are true".
So if I, say, take the premise that 2 + 2 = 5, and deduce (by noticing that 3 = 2 + 1, 6 = 5 + 1, and applying axioms of arithmetic) the conclusion that 2 + 3 = 6, then I have constructed a valid, but unsound, argument.
You would either have to know that every tall man is strong (either by empirical discovery or by some sort of logical deduction, which, good luck), or that, more easily, there exists a tall man who isn't strong.
Or that no men are tall, I guess---which is secretly an example of the first thing I said in the previous message
But it's hard to extrapolate general rules from single examples. I mean, you can say, "to determine that P -> Q is true, it is enough to show either that P is false, or that, when P is true, Q must also be true", but that's kind of just restating the definition, to an extent.
That's difficult to know. Sometimes people mean "both valid and sound" when they say "correct". But some people just mean "I believe the conclusions of the argument are true", which is...sloppy.
If someone says a mathematical argument is "correct" they generally mean the former, I believe---with whatever axioms are being taken yadda yadda blah blah blah.
If it doesn't rain, but the ground still gets wet (say someone has a sprinkler), that doesn't mean "if it rains, the ground gets wet" is now suddenly false.
i.e. if "rains = false", but "ground is wet = true", that doesn't mean "rain => ground is wet" is false. It's still true in this scenario, because "rain => ground is wet" isn't saying anything about what happens if it doesn't rain.
The exact same thing is happening. If, hypothetically, ignoring the fact that this doesn't happen, we assume that somehow your heart doesn't stop but you still die, that doesn't mean "heart stops => die" is now suddenly false.
What would make this statement false is if someone's heart did stop but they didn't die. Which does happen. That's what makes the statement "heart stops => die" false.
That's the only thing, in fact, that could make that statement false. Just like the only way "rains => ground is wet" can be false is if it rains, but the ground doesn't get wet.
Another way to think about it is to think about the example of theorems in math, like Rolle's theorem.
"If you have a function f that is continuous on a closed interval [a,b] and differentiable on the open interval (a,b), and if f(a) = f(b), then there is a point c in (a,b) such that f'(c) = 0."
What if you have a function that's not continuous on a closed interval? Or not differentiable on the associated open interval? Or for which f(a) doesn't equal f(b)?
That f might, or might not, have a c for which f'(c) = 0.
The theorem is still true, even if its hypothesis is false in that case.
What makes Rolle's theorem true is the fact that it's never the case that "a function fits all those criteria, but doesn't have a point with zero derivative".
Implications can be tricky business because of this. I know that this particular point of logic confused me for a long time.
the thing associated with this "if your heart stops, then you die" that we used implication to tell the relation between the stopping of heart and death and i was told that the implication is false when a situation occur where " p is true and q is false " as it contradicts implication
Impossible "in real life" or not (and I would argue that "false" is the wrong word to use for an event that can't happen, because events aren't true or false), you do have to consider it.
And if, by some miracle, someone died but their heart beat forever, it would not make "heart stops => die" false.
Because when you say "heart stops => die", who cares if their heart doesn't stop? You aren't talking about that case at all, so nothing that happens if a heart doesn't stop can disprove the implication.
"heart stops => die" is false because there are people who have had their hearts stop momentarily, but they lived.
But "death does not occur" doesn't matter to what you're saying, so if death doesn't happen and your heart still beats, then the implication isn't made false.
What would make it false is if someone died, but their heart beat forever (see how this is the same thing I said didn't matter in the other case? well now we've switched the implication around)---and this is impossible, as you noted, so the implication "death => heart stops" is true.
Another thing that can be helpful: P -> Q is logically equivalent to "not-P, or Q (or both)".
The negation of this statement is "P and not-Q"---so the thing that makes an implication false is P being true and not-Q being true (i.e. P true and Q false).
A caution I'd give is that outside of mathematics and logic, things usually aren't cut and dry.
For example, the "rains => ground is wet" example? Well, there are times where water falls from the sky, but evaporates before it hits the ground. If you call that "rain", then the implication is false, but if you say "rain is only when the water makes it all the way to the ground", then the implication is true.
@Noob The reason we define implication in this way is precisely because we don't want theorems to break in dumb ways.
We don't want a non-continuous function to make Rolle's theorem false, because we aren't talking about non-continuous functions in that theorem.
But this reasoning does hold more generally. Any "if-then" statement isn't a lie if the "if" doesn't happen.
So we define implications as we do, both because it makes theorems simpler to work with, and because it corresponds to our intuition about truth and lying in social situations (although, again, sometimes this logic can be very tricky).
but when you take same anaolgy over " the heart stops but person does not declare dead " is false while the implication " heart => death occur " also suggest false thus no contradiction hence "heart =>death " should be true but it not
People's hearts can stop momentarily. That's partly what defibrillators are good for. In that case they don't die.
Unless you want to say "but they might be declared legally dead", or "I meant stopping forever"---but that comes down to, again, implications aren't cut and dry in the real world.
The same sentence might be true or false to someone who takes different definitions for the words used in the sentence.
"All Christians believe in the Holy Trinity" is obviously true if you think Unitarians aren't Christians, and obviously false if you think they are.
why cant you use " hearts does not stop person died" which is false where in implication, p = false , q = true means true which contradicts thus implication is wrong
"A heart doesn't stop but someone dies" is an impossible situation in real life. We can debate whether that's fair to call "false" or not (I think calling it "false" is a category error, but, oh well).
Let me call that statement R, just for simplicity. So R is "heart doesn't stop but they die".
the thing that to argue that " heart stops death does not occur " is false does not make any sense and this is the only way we can say that " heart stops => death occur " is correct
@Jasper - I figured it wasn't research level, so I posted it here on Math.SE instead of MO. Anyhow, I am dropping a link here also, in case anyone feels it maybe worth their time. :)
There exists a popular model in the Physics of heavy quark bound systems, called the Cornell potential model, in which the inter-quark potential is modeled to vary with radial distance $r$ as
$$V(r) = - \frac{\kappa}{r} + \frac{r}{a^2}$$
The mathematical problem is reduced to solving the radi...
@Noob Say for the sake of argument that zombies are possible. And that zombies have beating hearts (seems like they'd need to to move). But they're still dead (if we define stuff right or whatever).
Those zombies existing wouldn't have anything to do with "heart stops => die".
@fragle i think i got your point and about implications and have no doubt about it but to come across such example is more than coming accrosss implicaitons
Don't get hung up on examples that are confusing. It's a bad idea to just attack a confusing example over and over until the concept itself becomes confusing.
Better to use the clear examples to deeply solidify your notion of an idea or topic, and then use that foundation to come back to the confusing examples later.
i can tell you i have no problem over it but somehow the notion of " heart stops => death occur " was just a fragment that i created to solidify my concept and indeed it helped
Is there a neccessary and sufficient condition for the equivalence of pointwise convergence almost everywhere and $L^p$-convergence? I know for example that the dominated convergence theorem gives a sufficient condition, but is there a way to completely characterise the relationship between the two?
Thought: for the complex numbers, the distributive property just means "if you rotate a parallelogram, you get another parallelogram"
(Well, I suppose this is backwards, in a sense - you usually define complex multiplication in terms of the distributive property, and then show that it's related to rotations)
Find the number of tuples (x,y,z) not necessarily unique selected from {1,2,3,...n} such that z>=max(x,y). My attempt- Let r be the max of (x,y). So the other among the two can be selected from {1,2,...,r} and z must be from {r,r+1,...,n-1,n}. So the total possibilities=$\sum_{r=1}^{n}r(n+1-r)$. Correct?
If f:[a,b] -> R has f' = 0 at some point in [a,b] and f' > 0 for any other point. Is it true that f not only increasing, but also strictly increasing on [a,b]?
I've read a proof at the end of this document that any nonabelian group of order $6$ is isomorphic to $S_3$, but it feels clunky to me.
I want to try the following instead:
Let $G$ be a nonabelian group of order $6$. By Cauchy's theorem or the Sylow theorems, there is a element of order $2$, l...
I'm reading Arturo's answer, and I am wonder why letting $G$ act on $G/H$ gives us a homomorphism from $G$ to $S_3$? What general fact about group actions is he appealing to?
here's a result I came across in reading papers lately which I'm trying to prove
Suppose $M$ is a 4-by-4 symmetric matrix with ones on the diagonal. I can identify each such matrix by the 6 matrix elements $(M_{12},M_{13},M_{14},M_{23},M_{24},M_{34})$ above the diagonal
The challenge is this: For what set of $(M_{13},M_{14},M_{23},M_{24})$ do there exist $M_{12},M_{34}$ such that $M$ is positive semi-definite?
Hello! Excuse my stupidity but since i wasnt able to decipher from the description ill just ask(never hurted no one). Is anyone here good with Matlab programming?
Anonymous
8:05 PM
@Phil FWIW there's a dedicated Matlab room on Stack Overflow.
Here's a nice puzzle: Given $s$ ($s > 0$) points in the plane such that every three of them are contained in a disk of radius $1$. Prove that all $s$ points are contained in a disk of radius $1$.
Anyone have a nice proof that the Klein four-group is the only possible subgroup of order $4$ to $A_4$? I am thinking that a three cycle $\sigma$ in $A_4$ has order $3$ so you'd have to include a self inverse cycle, i.e. a composition $\tau$ of two transpositions and since one of the elements in $\tau$ is not in $\sigma$, the composition $\sigma \tau \notin \langle \sigma \rangle$?
Pick the triples $\{T_n\}_{n=1}^{\binom{s}{3}}$ in any order; associate each triple $T_n$ with a corresponding point $C_n$ such that $C_n$ is the center of the disk. Such point exists by hypothesis.
How does one prove that the smallest multiple of $m$ divisible by $n$ is $m \cdot \frac{n}{gcd(m,n)}$? I tried playing around with Bezout's lemma, but I couldn't see how to proceed.
@Blue is there anyway to speak to an admin of this chats? Because i created an account on the portugues version of the website and so i have 83 reputation. In this line of thought, i can access a lot of chats like this one, but when i tried to access the link you sent me, it redirected me to another account with the same name but lower reputation.
@user193319: note that $\frac{n}{\operatorname{gcd}(m,n)}$ is an integer so we have in fact a multiple. Now suppose that $k < \frac{m}{\operatorname{gcd}(m,n)}$ and $mk = nl$ for some integer $l$. But then $mk < \frac{m}{\operatorname{gcd}(m,n)} \implies k\operatorname{gcd}(m,n) < 1$. But $\operatorname{gcd}(m,n) \geq 1$ and then $k = 0$. But we don't want the trivial multiple 0, so we get that $k$ can't be less than our number. So it follows.
@Anush No. If $f$ is defined over $\Bbb R$ it's clear that, taking limits for $\pm \infty$, its range is $(20,18; 0)$, whose integral values are $\{1,\dots, 20\}$.
@LucasHenrique eu coloquei uma questao de Matlab e o usuario blue recomendou-me outro chat que é sobre esse tema. Porem quando tento aceder a esse site dizme que nao tenho reputação suficiente,apesar de eu estar neste chat e ter quase 100 ed reputaçao. Quando vou ver o usuario com que estou a aceder ao chat dizme que é um com o mesmo nome porem com muito menos reputação. Ja coloquei uma questao no meta porem se me pudesse ajudar agradecia
@Phil puts. Realmente não sei como te ajudar, desculpe
I'll suppose you're taking $m, n > 0$. Let $l = \operatorname{min}\{r\in \Bbb Z_{\geq 0}: m|r \land n|r\}$. Without loss of generality $m \geq n$. We know that since $\frac{m}{\operatorname{gcd}(m,n)}$ and $\frac{n}{\operatorname{gcd}(m,n)}$ are both integers, $\frac{mn}{\operatorname{gcd}(m,n)}$ is a candidate.
here is an assertion tell me if correct . A proposition that is made by use of logical connecticve "implication " is true if the truth table of the proposition for differnet truth values of its atomic statements matches standard truth table
Hm. Define $d = \operatorname{gcd}(m,n)$. Let $m' = m/d$, $n' = n/d$ be coprime numbers (proof below). Now observe that $m'n'd$ is such desired a number if you manipulate the fractions. Use that pair of coprimes divide $l \implies$ product of coprimes divide $l$. Also use that $d$ is coprime to both $m'$ and $n'$ from $m = m'd| l = m'n'k \implies d |k$ so you finally get $ l = m'n'dk'$. $k \geq 1$ and it follows.
@user193319 I think it can be proved using Bezout's. So $\gcd(m,n) = d$ implies $mx+ny = d$. Introduce $ a = mn/d$, which clearly is an integer and a common multiple of $m$ and $n$. We need to show that $a$ is the least common multiple. Let $b$ be another common multiple of $m$ and $n$. Then $\dfrac{b}{a} = \dfrac{bd}{mn} = \dfrac{b}{n}x + \dfrac{b}{m}y$ is also an integer, so $a = mn/\gcd(m,n)$ must be the least common multiple.
like discussed the implication " death occur => heart stops " is false ( almost all say it true ) because the situation when premise is false i,e " death does not occur " and conclusion is true " hear stops " should be true by the defination of implication but its false that " death does not occur when heart stops" hence it contradicts the situation
when premise is false and conclusion is true
but now some say we do not care as when premise is false the implication is true
so what iam saying is , that is it possible that the way we defined implication when " premise is false then implicaition is correct " could lead to false of the implicaiton
If the condition fails, the implication is vacuously true. For an implication $A\Rightarrow B$, you can have $A$ false and $B$ true, but you can also have both $A$ and $B$ false, so you can't transform this into $B\rightarrow\lnot A$ as it seems you're trying to.
It doesn't matter. The implication only asserts "if A, then B"; it doesn't care about whether that condition is actually met and neither does it care about what happens if the condition is not met.
Let $(X,d)$ be a metric space and $A$ be a subset of $X$. Then the
following statements are equivalent.
$A$ is nowhere dense.
$\overline{A}$ doesn't contain any non-empty open set.
Each non-empty open set has a non-empty open subset which is disjoint from
$\overline{A}$
Each ...
There exists a popular model in the Physics of heavy quark bound systems, called the Cornell potential model, in which the inter-quark potential is modeled to vary with radial distance $r$ as
$$V(r) = - \frac{\kappa}{r} + \frac{r}{a^2}$$
The mathematical problem is reduced to solving the radi...
I've read a proof at the end of this document that any nonabelian group of order $6$ is isomorphic to $S_3$, but it feels clunky to me.
I want to try the following instead:
Let $G$ be a nonabelian group of order $6$. By Cauchy's theorem or the Sylow theorems, there is a element of order $2$, l...
