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0

We can find the inverse Laplace transform by taking Mellin inversion formula. In the problem, we have a simple pole at $s=-2$ and a pole of order two at $s=0$. \begin{align} h(t) &= \mathcal{L}^{-1}\{F(s)\}\\ &=\frac{1}{2\pi i}\int_{\gamma - i\infty}^{\gamma + i\infty}\frac{e^{st}}{s^2(s+2)}ds\\ &= \lim_{s\to ...


2

Ok, so let us go for something famous: as you know $(\mathbb{N},+)$ and $(\mathbb{N},\cdot)$ are both (commutative) monoids. $(\mathbb{N},+)$ has one only idempotent, the identity: $0+0=0$, $(\mathbb{N},\cdot)$ instead has two: $0\cdot 0=0$ and $1\cdot 1=1$. $1$ is the identity here and, if we take it away from $\mathbb{N}$, we observe that ...


0

I will employ here again the same web resource by Eric Moorhouse I linked yesterday in this answer to a related question. This finite group and this other have both order $27$: since there's no $n$ such that $n!=27$, they are not symmetric; since there's no $m$ such that $2m=27$, they are not dihedral. Just a glance at the bottom right corner of the two ...


0

Take any $2\times2$ non-singular matrix $A$ with real entries. Regard the lattice point $(a\ b)\in\mathbf{Z}\times \mathbf{Z}$ as a vector $v$. Now take all the transformed vectors $vA$ (matrix multiplication of row vector $v$ by $A$) as $v$ varies over lattice of integers. This is also another example; choosing $A$ to be diagonal matrix with entries, ...


1

Yes, both of them are correct, provided you want them to be a subspace of a real vector space. If $a,b,c,d$ are integers then $(a,b)+(c,d)=(a+c,b+d)$ is also an integer. So we have closure under addition. Now for the other condition, if $(a,b)\in\mathbb{Z}^2$ then $(-a,-b)\in\mathbb{Z}^2$. However if you take a scalar from $\mathbb{R}$ say $\pi^2$, then for ...


1

Discrete just have to be separated from one another, finite means the total number must be $<\infty$. So $f(x)=x$ has no discontinuities at all, and $f(x)={1\over x}$ has one discontinuity at $0$ and again, this number is finite. On the other hand the floor function $f(x)=[x]$ (definition below if you haven't seen it) has infinitely many of them--one at ...


0

I think one of the most surprising results of this type is the following (and I could just be very naive): Kervaire and Milnor showed that diffeomorphism classes of oriented exotic spheres form the non-trivial elements of a finite abelian group under the connected sum for dimension not equal to $4$.


0

A set of permutations closed under composition. These were historically the first thing called groups, every group algebra is instantiated by some set of permutations closed under composition and most of the theorems you see in group theory books have been developed to understand permutations closed under composition better and their relationship with ...


0

Since associativity is not immediate to be verified for finite structures whose Cayley table is given, it is probably not obvious that magmas like this one or this other are groups. It is easier instead to catch at a first glance that both have identity or that none of these two is commutative.


3

Sandpile groups have the curious feature that the identity element is very complicated. See http://www.ams.org/notices/201008/rtx100800976p.pdf.


1

How about Cantor's proof that transcendental numbers exist? (Because the set of algebraic numbers is countable, and the set of real numbers is uncountable.) If I remember correctly, this proof appeared only soon after Liouville had provided an explicit construction (namely $\sum 10^{-n!}$). I know this example is hinted by other answers given, but I think ...


0

My favourite one: $[0, 1]$ is compact, i.e. every open cover of $[0, 1]$ has a finite subcover. Proof: Suppose for a contradiction that there is an open cover $\mathcal{U}$ which does not admit any finite subcover. Thus, either $\left[ 0, \frac{1}{2} \right]$ or $\left[ \frac{1}{2}, 1 \right]$ cannot be covered with a finite number of sets from ...


2

Consider the probability space $((0,1),\mathcal{B}(0,1))$ endowed with the Lebesgue measure $\lambda$ and the random variables $$X(\omega) := 1_{(0,1/2)}(\omega) \qquad \qquad Y(\omega) := 1_{(1/2,1)}(\omega), \qquad \omega \in (0,1).$$ Then $X \sim Y$. Set $X_n(\omega) := Y(\omega)$ for all $n \in \mathbb{N}, \omega \in (0,1)$. $X_n \to X$ in ...


4

You can recognize any minor-closed graph class in cubic time (Robertson–Seymour theorem), but it might be undecidable to actually find such an algorithm. Since the minor-ordering is a wqo, any minor-closed graph class $\mathcal{G}$ has a finite set of forbidden minors $\mathcal{H}$. Hence, if we want to check whether a graph $G$ is in $\mathcal{G}$, we ...


5

Baire's category theorem can be used to establish non-constructive existence proofs in a parsimonious and elegant way. Theorem (Baire)$\phantom{---}$No complete metric space is a union of countably many nowhere dense subsets. As Folland (1999, p. 162) puts it, “[Baire's] category theorem is often used to prove existence results: One shows that objects ...


4

In cryptography there are many times when I wish to prove that I have some private key, but need to make it impossible (or really difficult) for you to construct that key yourself. Take factoring, the basis of RSA. Claim: There exist integers P and Q such that P * Q = 651041 Proof: If 651041 is prime, then 2^651040 mod 651041 = 1 2^651040 ...


6

Many existence theorems related to algebraically closed fields are inherently non-constructive. 1) Every field has an algebraic closure. Try to make this explicit for $\mathbf C(x)$ or for $\mathbf Q_p$. 2) Any two algebraically closed fields with the same characteristic and uncountable cardinality are isomorphic as fields. To take a special case, any ...


1

One can use Stone duality to construct many counterexamples. It is not to hard to show that a Boolean algebra $B$ is complete if and only if the Stone space $S(B)$ is extremally disconnected. Furthermore, a Boolean algebra $B$ is $\sigma$-complete (recall that a Boolean algebra is $\sigma$-complete iff every countable subset has a least upper bound) if and ...


0

I think that there are different kinds of examples, some cited in the other answers. To see the difference I give three examples. 1) The number $a$ so defined: $a=1$ if Goldbach's Conjecture ($G$) is provable in ZFC, $a=0$ if $\neg G$ is provable in ZFC, a=2 if $G$ is undecidable in ZFC. The value of this number is not known today, but it can eventually ...


0

Uncountability of the real numbers. Since the set of all finitely-describable numbers (in any formal language) is countable, by definition uncountability implies the existence of numbers that are not finitely describable. To illustrate my point, here are some examples of finitely-describable numbers: Any natural number or integer Any rational number Any ...


3

The dihedral group of order $8$ - the symmetries of a square is one such. It has the identity element and two elements of order $4$ (clockwise and counterclockwise quarter turns). The other five elements are four reflections about axes of symmetry and a half turn - all of order 2.


10

Consider $D_4$ Dihedral group, non-abelian, $5$ elements of order $2$.


11

The intermediate value theorem, in the form that for a continuous function on $f:[a,b]\to\Bbb R$ with $f(a)<0<f(b)$ there exists $x\in(a,b)$ with $f(x)=0$, is not valid constructively. See for instance here or here. The fundamental difficulty with the obvious classical proof of the IVT is that one cannot prove constructively that for $c\in(a,b)$ ...


56

Some digit occurs infinitely often in the decimal expansion of $\pi$.


24

Let $f(x)= e^{-x^2}$. Then $$ f(x+\tfrac1x)=e^{-x^2-2-x^{-2}}=f(x)\cdot e^{-2}\cdot e^{-x^{-2}}$$so that the quotient tends to $e^{-2}$ as $x\to\infty$.


46

Claim: There exist irrational $x,y$ such that $x^y$ is rational. Proof: If $\sqrt2^{\sqrt2}$ is rational, take $x=y=\sqrt 2$. Otherwise take $x=\sqrt2^{\sqrt2}, y=\sqrt2$, so that $x^y=2$.


19

There is a standard kind of example here. Let $T$ be any open question that requires substantial work to resolve. Consider: There is a natural number $n$ such that $n = 0$ if and only if $T$. That statement is clearly true, via a "simple" non-constructive proof: if $n = 0$ is not an example, then $n = 1$ is an example. But if we could produce an ...


11

Definition: A real number $x$ is called normal if for every $b>1$ the digits $0,1,2,...,b-1$ are equally distributed in the $b$-adic expansion of $x$. Theorem: Normal numbers exist. The standard proof is non-constructive as it proceeds by showing that the set of non-normal numbers has Lebesgue-measure zero, hence must have non-empty complement. ...


0

The Cayley table $$\begin{array}{c|ccccccc} \ast & 0 & 1 & 2 & 3 & 4 & 5 & 6\\\hline 0 & 4 & 2 & 6 & 0 & 4 & 5 & 6\\ 1 & 3 & 5 & 1 & 6 & 4 & 5 & 6\\ 2 & 0 & 5 & 2 & 6 & 4 & 5 & 6\\ 3 & 4 & 1 & 6 & 3 & 4 & 5 & 6\\ 4 ...


1

Why not some properties of powers of (natural) numbers: $$\forall\,a,b,m,n\in\mathbb{N}\setminus\{0\},\quad a^m\cdot a^n=a^{m+n},\quad (a^m)^n=a^{mn},\quad a^n\cdot b^n=(ab)^n.$$ Another one, which is not totally elementary as you meant it but I think may be very formative in secondary school level is the following: If $p$ is a polynomial in variable ...


3

$$\varphi:Z\rightarrow Z_8; \quad \varphi(n)=\bar n $$ $J(Z)=0$. so $\varphi(J(Z))=0$ but $J(Z_8)= \langle 2 \rangle$


0

The set of truth-values $M=\{\mathbf T,\mathbf F\}$ with the operation $\to$; that is, $\mathbf T\to\mathbf F=\mathbf F$ and $\mathbf T\to\mathbf T=\mathbf F\to\mathbf T=\mathbf F\to\mathbf F=\mathbf T$. Then $$\mathbf F\to(\mathbf F\to\mathbf F)=\mathbf F\to\mathbf T=\mathbf T$$ while $$(\mathbf F\to\mathbf F)\to\mathbf F=\mathbf T\to\mathbf F=\mathbf F.$$


1

As @davcha pointed out already, it is easy to work out an example among finite magmas. Look at the Cayley table: $$\begin{array}{c|cccc} \cdot & 0 & 1 & 2 & 3\\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 1 & 1 \\ 2 & 2 & 2 & 0 & 2 \\ 3 & 3 & 3 & 3 & 0 \end{array}$$ over the set ...


2

Let $S$ be any set. On $S$ we define $+$ the following way $$x+y=x$$ Then $(S,+)$ is a semigroup which has your property. If $S$ is infinite, your equation has infinitelly many solutions. I think you can also add an identity to this $(S,+)$ to make it into a monoid with the required property.


1

As pointed out already in the first two comments to your question, yes, it exists. Consider for example the finite magma with Cayley table: $$\begin{array}{c|cccc} + & 0 & 1 & 2 & 3\\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 \\ 2 & 2 & 2 & 2 & 2 \\ 3 & 3 & 3 & 3 & 3 ...


0

Hint: Assuming Riemann integration, how do you justify the existence of integrals like $$ \int\limits_{0}^{x/n}f(t)\,\mathrm{d}t\,? $$


0

Correct. As a different counter-example, consider that the boundary of $S$ may coincide with the boundary of $R$, as in the case where $R=[0,2]$ and $S=[1,2]$. A ball around $2$ contains no points in the exterior of $S$.


0

Yes, this is correct. As for an example of a set with empty interior, consider $\Bbb Z$ the set of integers, as a subset of $\Bbb R$.


0

Notice that if $(X_n)$ and $(Y_n)$ are weakly convergent, then the sequence $(X_n+Y_n)_{n\geqslant 1}$ is tight, hence we can extract a weakly convergent subsequence. Therefore, the problem may come from the non-uniqueness of the potential limiting distributions. Consider $(\xi_i)_{i\geqslant 1}$ a sequence of i.i.d. centered random variables with unit ...


1

Let me give another, potentially more common example: the upper-half circle of radius $1/2$ with center at $1/2$, i.e. $$f(x)=\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}$$


0

Sure. Presuming you allow one-sided limits at interval endpoints in your limit of differentiability (otherwise any otherwise differentiable function is automatically an example because it fails to be differentiable at $0, 1$), you can take as an example the function $f$ defined by $0$ on $[0, \frac{1}{3}]$, $3x - 1$ on $[\frac{1}{3}, \frac{2}{3}]$, and $1$ ...


1

The idea is fine, but you may want to make the formalism a bit more rigorous: Let $1 > \epsilon > 0$. We must show that $\forall\delta > 0 \exists x\in[0,1], N\in\mathbb N$ such that $|x|<\delta$ but $|f_N(x)| > \epsilon$. For this pick $N = \lceil \frac1\delta \rceil +1$ and $x = \frac1N < \delta$. Then $|f_N(x)| = 1 > \epsilon$ as ...


0

Answer based on the hints given by Nate Eldredge.But this one I obtained from one more book . $f_{n}(x)=n^{2}x$ if $ 0 \leq x \leq \frac{1}{n} , = $-n^{2}x+2n$ if $ \frac{1}{n} <x< \frac{2}{n}$ =$0$ if $\frac{2}{n} \leq x \leq 1$ . Here we have to start with $n=2$. Graph of this function will look like triangle with changing base.


1

Let $\kappa$ be any uncountable cardinal. Let $D_\kappa$ be the discrete space of power $\kappa$, let $p$ and $q$ be distinct points not in $D_\kappa$, and let $X=D_\kappa\cup\{p,q\}$. Let $P_0$ be a countably infinite subset of $D_\kappa$, and let $Q_0=D_\kappa\setminus P_0$. Finally, let $\{P_1,Q_1\}$ be a partition of $D_\kappa$ into two sets of power ...


0

As other people have pointed out, L'Hospital's rule does not explicitly require that $h$ be locally nonzero at $x_0$. However, in practice, if $h$ has zeroes in every deleted neighborhood of $x_0$, then either The function $h$ is not differentiable in any deleted neighborhood of $x_0$, or The derivative $h'$ has zeroes in every deleted neighborhood of ...


3

If $A$ is a finite dimensional algebra, then every simple $A$-module is finite dimensional, since it is a quotient of the regular module (if $S$ is a simple left module, then for any $0\neq s\in S$, $a\mapsto as$ is a non-zero homomorphism $A\to S$, which is surjective since $S$ is simple).


2

It depends on the base field you are working with. For a field like the reals or complex numbers, where every positive element is a square, the answer is yes. There are two changes of basis that make both into the Gram matrix with 1's, 0's and -1's on the diagonal according to index, and by transitivity both matrices are cogredient. For an ordered field ...



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