On the Wikipedia page on Cardinal Numbers, Cardinal Arithmetic including multiplication is defined. For finite cardinals there is multiplication by zero, but for infinite cardinals only defines multiplication for nonzero cardinals. Is multiplication of an infinite cardinal by zero undefined? If s...
Well eh... the no cloning and no communication theorem is really because of the hermitian properties of linear operators and tensor products, nothing physics really going on there
if you look at the proof, the whole thing is mathematical
(con't) I think the strongest of the division by zero no-go theorems I have proved is the following:
Theorem: Given any division by zero magma $M_D$ where $0^2\neq 0$. Then it is maximally nonassociative (i.e. All possible associators terms will not be the identity map)
(B)I note that if a function is continuous and injective, then very often the function is monotonic, and thus strictly monotonic, this is nearly always true?
That is long as the domain and range are connected, and/or are real valued, closed and bound intervals, that this is correct, and that th...
Can you give an example of a function that is injective, but non continuous and of course not strictly monotonic. The point is to illustrate the importance of continuity
It would preferably be defined on an interval
I could find examples but only on unions of disjoint intervals
@Astyx maybe that's an equivalent condition? i learned that complete= there is no orthonormal set that (properly) contains the complete set of orthonrmal vectors.
@AlessandroCodenotti if i want to show that $S_d$ is contained in some homogeneous ideal $a$ , i think its enough to show that $x_i ^d$ is in $a$ , am i right?
cool, my course is called "commutative algebra and AG" so we just started AG a week ago
so if i assume $\sqrt{a} = S_+$ i want to show that $S_d \subset a$ for some $d\gt 0$. so i know that $x_0,..,x_n \in \sqrt{a}$ so $x_i^m \in a$ for some $m$
now i want to show that $S_{m(n+1)} \subset a$
ah. got it, it is enough to check that $\Pi x_i ^{k_i} \in a$ where $\sum k_i = m(n+1)$ @LeakyNun
Let $\sum a_n$ be a convergent series over reals. If we remove infinitely many terms $a_n$, does new shrunk series convergent? (I know this is true for absolutely convergent series.) @LeakyNun
In fact if the series is not absolutely convergent (i think that's called semiconvergent) then you can always remove an infinite number of terms and get a divergent series
Because that's what happening here, $q^-$ and $q^+$ have nothing in between if I understood correctly Leaky's order. It's like a subset of $2\cdot\Bbb R$ as order
A dense set has to have any point in the parent set to be at least a limit point or a point in the set itself. That means any point in $O$ has to have some convergent net to it from $D$
Order topology of the reals with the additional relation $q^- < q^+$?
hmm...
The basis of the order topology generates the same open sets as the open interval topology of the reals. So since there always exists some sequence in $\Bbb{Q}^+$ that will converge to its closure (which is going to be a copy called $\Bbb{R}^+$), and likewise for $\Bbb{Q}^-$, then both $\Bbb{Q}^-$ and $\Bbb{Q}^+$ should be dense in $\Bbb{R}$
hmm... where does the sequence $q_i^-$ where $q_i^-$ grows to indefinitely large in value, converge to. There is no smallest element $r$ such that $r < q^+$ for all $q^+$
not sure whether an analogous case holds for DLO with endpoints though (e.g. isomorphic to a disjoint union of half open intervals of rational points (i.e. sets of the form $[a,b) \cap \Bbb{Q}$))
A simple explanation would say that the slope of the tangents would be equal at that point but what about curves like x^3 and x^5 which have same slope at x=0 but they dont 'touch' at that point. So, what is happening?
I agree... but lets not make touch=intersect. Lets define 'touch' as the two curves lying locally on opposite sides of the tangent at the point of 'touch'
@tatan yes man. If you have like $f(x)$ and $g(x)$ then check for roots of $f(x)- g(x) = 0$. It should have only a single root so that they touch each other.
@tatan yes man. If you have like $f(x)$ and $g(x)$ then check for roots of $f(x)- g(x) = 0$. It should have only a single root so that they touch each other.
@Abcd You would expect a root of multiplicity $> 1$ though, if they not only intersect but are also tangent. This is because $f(x_1) = g(x_1)$ and $f'(x_1) = g'(x_1)$ in that case, so $(f - g)(x_1) = (f - g)'(x_1) = 0$, symptomatic of $f - g$ having a double root at $x_1$.
@LeakyNun Idk, I just remember when I told Ted I was working on a derivative topic, and he asked me if my class had done concavity yet. I said no, but I assumed there was a reason why he asked, which is why I mentioned it now.
Right, I figured. But the point and the line determine a plane, and you find the (perpendicular) distance in that plane, regardless. I don't understand what question they're asking you.
In all my years of teaching such things, I've never seen such silliness.
So all I can think of is this: Move the plane parallel so that it contains $P$. Find the intersection $Q$ of that new plane with $L$. Find the distance from $P$ to $Q$. I have no idea why anyone would care.
This is from Pinter's abstract algebra book. I am not getting this part clearly: " *From one point of view the set of the integers, with addition and multiplication, forms a ring (that is, it satisfies the axioms stated previously). From another point of view it is an ordered set, and satisfies special axioms of ordering. On a different level, the positive integers form the basis of “recursion theory,” which singles out the particular way positive integers may be constructed, beginning with 1 and adding 1 each time* ."
@UnknownMathMan Recursion theory is also known as computability theory and is a branch of mathematical logic.
@UnknownMathMan You do not need to know any recursion theory to study algebra. Just read that passage and move on, no need to worry about recursion theory.
@WillHunting Ok, I just wanted to know, how forming ring and satisfying ordering implies(as it is claimed to form the basis of 'recursion theory' and it is said that recursion theory implies) positive int. can be constructed with 1 and adding 1 each time.
(B)I note that if a function is continuous and injective, then very often the function is monotonic, and thus strictly monotonic, this is nearly always true?
That is long as the domain and range are connected, and/or are real valued, closed and bound intervals, that this is correct, and that th...
