What are the methods of solving linear congruences? I have been trying very hard to solve this type of congruence equation:
$$ax = b \pmod n$$
and finally managed to actually solve a few by using properties like
$$\text{if}\quad
\begin{cases}
ax=b \pmod n \\
cx=d \pmod n
\end{cases}
\quad\text{then}\quad acx = bd \pmod n$$
but still there are some simple  congruences which I am not able to solve  like $$25x=15\pmod{29}.$$
I tried to make use of both the above and transitive property of congruences, but that is not working here.
Now, I wanted to ask if there is any other method to solve congruences. I am asking this because I have very little knowledge of congruences. I have Burton's book of number theory and that helped me much better than Zuckerman's text did. But, still there are some topics that I am not able to do yet.

$$25x\equiv 15\pmod{29}\tag{1}$$

$29$ is a prime number, an one can show (I think Gauss showed) that that implies every number in $\{1,2,3,\ldots,28\}$ must have a mod-$29$ multiplicative inverse within that set.  So let's find the inverse $y$ of $25$, so that $25y\equiv1\pmod{29}$, and then multiply both sides of $(1)$ by $y$, getting
$$
x\equiv15y\pmod{29}.
$$
We want:
$$
\begin{align}
25y & \equiv 1 \pmod {29} \\
25y & = 1 - 29z \\
25y+29z & = 1.
\end{align}
$$

If we divide $29$ by $25$, we get $1$, with remainder $4$:
$$
29-\{1\}25 = 4\tag{2}
$$
Now divide $25$ by $4$, getting $6$, with remainder $1$:
$$
25-\{6\}4 = 1\tag{3}
$$
Because $(2)$ is true, we can put $29-\{1\}25$ in place of $4$ in $(2)$:
$$
25-\{6\}(29-\{1\}25).
$$
Now collect the $29$s and $25$s:
$$
-\{6\}29 + \{7\}25 = 1.\tag{4}
$$
So we have $y=7$ and $z=-6$.

Thus $(4)$ tells us that
$$
7\cdot25\equiv 1\pmod{29}
$$

Thus multiply both sides of $(1)$ by $7$:
$$
x\equiv7\cdot15\equiv 18 \pmod{29}.
$$
See Calculating the Modular Multiplicative Inverse without all those strange looking symbols for the way to find the inverse of $322$ mod $701$.  It turns out to be $455$.

Solution using Euler's function:
enter image description here


Left inverse of a matrix and a full column rank Dr Strang in his book linear algebra and it's applications, pg 108 says ,when talking about the left inverse of a matrix( $m$ by $n$)

UNIQUENESS: For a full column rank $r=n . A x=b$ has at most one solution $x$ for every $b$ if and only if the columns are linearly independent. Then $A$ has an $n$ by $m$ left-inverse $B$ such that $B A=I_{n}$. This is possible only if $m \geq n$.

I understand why there can be at most one solution for a full column rank but  how does that  lead to $A$ having a left inverse?
I'd be grateful if someone could help or hint at the answer.

If $A$ is $m \times n$, then the following are equivalent:

$A$ has full column rank $n$
The columns of $A$ are linearly independent
The null space of $A$ is trivial
The map induced by $A$ is injective
$A$ has a left inverse

Proof that 1 $\iff$ 2:
Immediate from the definition of column rank.
Proof that 2 $\iff$ 3:
Observe that the vector $Ax$ is equal to the linear combination $\sum_{i=1}^{n}a_i x_i$, where $a_i$ is the $i$'th column of $A$, and $x_i$ is the $i$'th component of $x$.
In particular, $Ax = 0$ if and only if $\sum_{i=1}^{n}a_i x_i = 0$.
The null space of $A$ is trivial if and only if $x=0$ is the only solution to $Ax = 0$ which, by what we said above, is true if and only if $\sum_{i=1}^{n}a_i x_i = 0$ implies $x_i = 0$ for all $i$, which is true if and only if $a_1, a_2, \ldots, a_n$ are linearly independent.
Proof that 3 $\iff$ 4:
Suppose that $Ax = Ay$. Since $A$ is linear, this is equivalent to $Ax - Ay = A(x-y) = 0$. Therefore $x-y$ is in the null space of $A$. But the null space of $A$ is trivial, hence $x-y = 0$, so $x=y$. This shows that (the map induced by) $A$ is injective (one-to-one).
Conversely, suppose that $A$ is injective. Then $x=0$ is the unique vector such that $Ax = 0$. Therefore the null space of $A$ is trivial.
Proof that 2 (and equivalently 4) $\implies$ 5:
Let $e_1, e_2, \ldots, e_n$ be the canonical basis for $\mathbb R^n$, meaning that $e_i$ has a $1$ in the $i$'th component, and zeros everywhere else. Note that for each $i$ we have $a_i = Ae_i$, where again $a_i$ is the $i$th column of $A$. Moreover, since $A$ is injective, $e_i$ is the unique vector that is mapped by $A$ to $a_i$.
Now, since $a_1, a_2, \ldots, a_n$ are linearly independent, they are a basis for the column space of $A$, which can be extended to a basis $a_1,a_2,\ldots, a_n, b_1,b_2,\ldots,b_{m-n}$ for $\mathbb R^m$. Hence an arbitrary $y \in \mathbb R^m$ has a unique representation of the form $y = \sum_{i=1}^{n} c_i a_i + \sum_{j=1}^{m-n} d_j b_j$ where $c_i$ and $d_j$ are scalars.
Therefore we can define a linear map $g : \mathbb R^m \to \mathbb R^n$ by first setting $g(a_i) = e_i$ for each $i=1,2,\ldots,n$ and $g(b_j) = 0$ for each $j=1,2,\ldots,m-n$, and then extending $g$ linearly to all of $\mathbb R^m$:
$$g(y) = g\left(\sum_{i=1}^{n} c_i a_i + \sum_{j=1}^{m-n} d_j b_j \right) = \sum_{i=1}^{n} c_i g(a_i) + \sum_{j=1}^{m-n} d_j g(b_j) = \sum_{i=1}^{n} c_i g(a_i) = \sum_{i=1}^{n} c_i e_i$$
Then $g$ is a left inverse of $A$:
$$g(Ax) = g\left(\sum_{i=1}^{n}a_i x_i\right) = \sum_{i=1}^{n} x_i g(a_i) = \sum_{i=1}^{n} x_i e_i = x$$
Proof that 5 $\implies$ 3:
Suppose that $Ax = 0$. Let $g$ be a left inverse of $A$. Then $x = g(Ax) = 0$. This shows that the null space of $A$ is trivial.

As a side note, it turns out that 4 and 5 are equivalent for general functions, not just linear maps. If $f$ is any injective function, then it has a left inverse, and conversely if $f$ is any function that has a left inverse, then it is injective. There is a proof here, for example. Since you indicated in the comments that this is an unfamiliar fact, I did not use it in the proof above but instead constructed a left inverse explicitly.

Note that my proof shows why a left inverse of $A$ must exist if $A$ has full column rank, but it doesn't explicitly show how to compute the left inverse.
As Strang notes, one formula for a left inverse is $B = (A^T A)^{-1} A^T$. That this is a left inverse is clear by computing:
$$BA = ((A^T A)^{-1} A^T) A = (A^T A)^{-1} (A^T A) = I_n$$
But as you will have noted, Strang punts to a later chapter the proof that $A^T A$ is invertible when $A$ has full column rank. So that's not very satisfactory!
Also, computing Strang's left inverse is very inefficient because it involves inverting $A^T A$. This requires a lot of calculation, proportional to $n^3$ operations for an $n \times n$ matrix.
In practice, probably the best way to compute a left inverse is to perform row reduction on $A$ to bring it to the form
$$\begin{bmatrix} I_n \\ 0_{m-n \times n} \end{bmatrix}$$
where $I_n$ is the $n \times n$ identity matrix, and $0_{m-n \times n}$ is the $m - n \times n$ matrix consisting of all zeros. Row reduction to this form is possible if and only if the columns of $A$ are linearly independent.
Assuming you're familiar with row reduction, you probably know that each row operation can be expressed as an $m \times m$ elementary matrix of one of three forms, corresponding to the three row reduction operations (multiplying a row by a scalar, interchanging two rows, and adding a scalar multiple of one row to another). The row reduction procedure can then be expressed by left-multiplying $A$ by the corresponding elementary matrices. Assuming there are $k$ of these, we have:
$$E_k E_{k-1} \cdots E_2 E_1 A = \begin{bmatrix} I_n \\ 0_{m-n \times n} \end{bmatrix}$$
The product $E_k E_{k-1} \cdots E_2 E_1$ is easy to understand conceptually: it corresponds to the $k$ row operations used to bring $A$ into the reduced form. Fortunately, it's not necessary to compute $E_k E_{k-1} \cdots E_2 E_1$ as a product of $k$ matrices! Instead you compute it by starting with $I_{m}$ and performing the same row operations on it as you perform on $A$.
In any case, denoting $E_k E_{k-1} \cdots E_2 E_1$ by $B$, the above becomes
$$BA = \begin{bmatrix} I_n \\ 0_{m-n \times n} \end{bmatrix}$$
Note that $B$ is an $m \times m$ matrix. It is almost the left inverse we seek, except we want just $I_n$ on the right hand side and a left inverse should be $n \times m$, not $m \times m$. If $m > n$ then the right hand side has $m-n$ spare rows of zeros at the bottom. To get rid of these, we can simply remove the bottom $m-n$ rows of $B$ to get a $n \times m$ matrix $B'$ which satisfies $B'A = I_n$ and is therefore a left inverse of $A$, as desired!

Question: "I understand why there can be at most one solution for a full column rank but how does that lead to A having a left inverse? I'd be grateful if someone could help or hint at the answer."
Answer: Let $k$ be a real numbers and $V:=k^n, W:=k^m$. Since the equation $Ax=b$ has a unique solution for all $b\in W$ it follows the equation $Ax=0$ has a unique solution, namely $x=0$. Hence the map $A: V \rightarrow W$ is an injection and it follows $n \leq m$. We get an exact sequence of $k$-vector spaces
$$0 \rightarrow V \rightarrow W \rightarrow^p W/V \rightarrow 0$$
and we may choose a section $s$ of $p$. This is a $k$-linear map $s: W/V \rightarrow W$ with $ p \circ s = Id$. This gives an idempotent endomorphism of $W$: $u:=s \circ p$  with $u^2=u$. From this it follows we may write
$$W \cong V \oplus Im(u)$$
and the projection map $p_V: W \cong V \oplus Im(u) \rightarrow V$ is a left inverse to the inclusion map defined by the matrix $A$. If you choose a basis for $W$ you get a matrix $B$ with $BA=Id_n$


An uncountable family of measurable sets with positive measure Is there a family of measurable sets $\{E_\alpha\}_{\alpha \in A}$ , disjoint, uncountable and each one with positive measure?

It depends on your measure space.  For instance, if you take as your measure space $\mathbb{R}$ with counting measure, then the family of singletons works.

However, for a $\sigma$-finite measure space (in particular, for Lebesgue measure), the answer is no.  Suppose your measure space is $(X,\mu)$ and you can write $X = \bigcup_{n = 1}^\infty X_n$ where $\mu(X_n) < \infty$.  Let $\mathcal{E}$ be any family of pairwise disjoint sets, all of which have positive measure.  Let
$$\mathcal{E}_{n,k} = \{ E \in \mathcal{E} : \mu(E \cap X_n) \ge 1/k\}.$$
Then every $E \in \mathcal{E}$ is in at least one $\mathcal{E}_{n,k}$, so $\bigcup_{n,k=1}^\infty \mathcal{E}_{n,k} = \mathcal{E}$.  Now suppose $E_1, \dots, E_m$ are distinct sets from $\mathcal{E}_{n,k}$.  Since these sets are disjoint, we have
$$\frac{m}{k} \le \sum_{i=1}^m \mu(E_i \cap X_n) \le \mu(X_n)$$
so $m \le k \mu(X_n)$.  Thus the cardinality of $\mathcal{E}_{n,k}$ is at most $k \mu(X_n)$; in particular it is finite.  So we have shown that $\mathcal{E}$ is a countable union of finite sets, hence countable.

This is impossible for Lebesgue measure. For simplicity suppose we're on the interval $[0,1]$. One can find countably many of the sets whose union has measure 1, in which case the complement has measure zero.


Classification of maximally non-hamiltonian graphs? A graph is called maximally non-hamiltonian, if it does
 not contain a hamilton-cycle, but no edge can be added without creating a
 hamilton-cycle. In other words, every pair of non-adjacent vertices $(u/v)$
 can be connected with a hamiltonian path.

How can the maximally non-hamiltonian graphs be classified with known graph properties ?

Using the edge count. If there is a non-Hamiltonian graph $G$ on $n$ vertices, the maximal number of edges that $G$ can contain is ${n-1\choose 2} +1 = {n\choose 2} - (n-2).$


Determinate analytically the range of a given function Is there a general way to obtain analytically the range of a function?

For instance, for the function $f(x)=4+\sqrt{x-3}$, I proceed as follows:

If $y \in ran(f)$ then for some $x$, $4+\sqrt{x-3} = y$ then $y-4 = \sqrt{x-3}$ now since we only consider non negative values of the square root then $y-4 = \sqrt{x-3} \geq 0$ so $y \geq 4$ and $ran(f) = [4, \infty)$.

But for the function $f(x)= 2x^{3}-5\sqrt{x}$  I can't seem to find a similar procedure to obtain the range analytically, of course I can plot the function and I get that the range is $[-3.497, \infty)$

Since $f:\mathbb R\mapsto\mathbb R$, $x\geq0$ for $\sqrt x$ to be real. Also,
$$\lim_{x\to\infty}f(x)=\infty$$
and
$$f(0)=0,$$
therefore $f$ is bounded from below since it is continuous on ${({{0};{\infty}})}$.
$$f'(x)=\frac{\mathrm d}{\mathrm dx}f(x)=6x^2-\frac5{2\sqrt x}=\frac1{\sqrt x}\left(6x^{5/2}-\frac52\right);$$
hence the minima are given by
$$f'(x)=0\iff6x^{5/2}-\frac52=0\iff x^{5/2}=\frac5{12}\iff x=\left(5/12\right)^{\,2/5};$$
so the lower bound is either at $x=0$ or $x=\left(5/12\right)^{\,2/5}$;
$$f\left(\left(5/12\right)^{\,2/5}\right)=2\left(5/12\right)^{\,6/5}-5\left(5/12\right)^{\,1/5}=\frac{5}{6}\sqrt[5]{\frac5{12}}-5\sqrt[5]{\frac5{12}}=-\frac{25}6\sqrt[5]{\frac5{12}};$$
since $$-\frac{25}6\sqrt[5]{\frac5{12}}=f\left(\left(5/12\right)^{\,2/5}\right)<0=f(0),$$
the minimum at $\left(\left(5/12\right)^{\,2/5}\right)$ provides the lower bound, that is, the image of $f:\,{{[{{{{0}};{{\infty}}}})}}\mapsto\mathbb R$ such that $f(x)=2x^3-5\sqrt x$ is
$$\therefore\boxed{{{\left[{{{{-\frac{25}6\sqrt[5]{\frac5{12}}}};{{\infty}}}}\right)}}}\,.$$

Domain: $x\ge 0$
Slopes: $f'(x)=6x^2-\frac52x^{-\frac12}$.

Now what does it indicate when the slope is positive/negative/zero?

Finally, plot the function (loosely, but it's gonna be sufficiently informative about your problem) by hand!


If $x_n \ge y_n$ and $y_n$ is not bounded above then $x_n$ is not bounded above I'm not quite sure how to prove this property of limits with sequences, can someone help?

Show that if $x_n ≥ y_n$ and $y_n$ is not bounded above then $x_n$ is not bounded above.

I'm supposing that $(x_n), (y_n)$ are sequences in $\mathbf{R}$. Suppose that for all $n\in \mathbf{N}$, $x_n\ge y_n$, and $y_n$ is not bounded above. That is, there exists no $M\in \mathbf{R}$ such that $M\ge y_n$ for all $n\in \mathbf{N}$. If there existed an $N\in \mathbf{R}$ with the property that $N\ge x_n$ (an upper bound) for all $n\in \mathbf{N}$, then this would imply that $N\ge y_n$ for all $n\in \mathbf{N}$ - contrary to assumption.

You can say that a sequence is not bounded from above, if $\exists n,\not\exists M \in \mathbb{R} : y \le M_n$, just the negation of the definition of upper bound, but you can also say that "no matter what number you give me, there's always (at some place; with some value of $n$, an element of the sequence $y_n$ which is larger than the number you give me - so the sequence it's not bounded from above, because you'd be going over all the numbers (in $\mathbb{R}$):

$y_n$ is not bounded from above $\equiv \forall n,\exists M \in \mathbb{R}
: y_n >M$.

$ \Rightarrow \forall n,\exists M\in\mathbb{R}: x_n \ge y_n > M \Rightarrow x_n > M \Rightarrow x_n$ is not bounded from above.

$\square$


Radius of Convergence for Polynomial I have to find the radius of convergence for this one, but I haven’t found a solution for this type of a term.

$\mathbb f (x)$ = $1 + 7x^3 + 5x^4 + x^{13} + x^{2015}$

Could you help me out?

Being a finite sum, this one converges for every $x \in \Bbb C$, so its radius of convergence is $\infty$.

Pardon the bad English, but I struggle with this to and the comment above doesn't helped me to totally demonstrate the situation, so here is a more complete answer to anyone who has to demonstrate the radius of convergence for a polynomial.
What happens to the nth derivative of your function when n is greater than one plus your biggest exponent? all are 0.
So, you can write the polynomial as your polynomial plus the sum from 0 to infinity of your nth derivative of your function over n factorial times $(x-a)^n$ where a is your center.
Since the nth derivative at this point are all equal, you're multiplying a constant, so you can bring to the outside of the sum that constant (property of sums) and now you can make any test to show that the radius of convergence is infinite.


3-D equation of a circle I came across a sum but could not solve it as i dont know the 3d equations of a circle :
The sum is   If $A(3,-2,2)$ and $B(2,9,5)$ are the end points of a diameter of a circle,then the third pt that lies on the circle is

$5,6,1$
$5,6,-1$
$5,-6,1$
$-5,-6,-1$

Ans:

   2

Would appreciate if someone can explain the equation of a circle (whether parametric or a normal equation) before helping solve the sum.

P.S- The time limit for solving the sum was one minute

To solve this problem, we can use quite basic tools.

$1$. Locate the centre $C$ of the circle. This is the midpoint of the line segment you were given.

$2$. Compute the radius $r$ of the circle.

$3$. Which one of the candidate points is at distance $r$ from $C$?

Once you know the coordinates $(c_1,c_2,c_3)$ of the centre $C$, and the radius $r$, then the equation of the sphere with centre $C$, radius $r$ is given by
$$(x-c_1)^2+(y-c_2)^2+(z-c_3)^2=r^2.$$
A candidate point $P$ lies on a circle with the given line segment as a diameter if and only if it lies on the sphere with the above equation.

Or else we can use the Pythagorean Theorem. Find the square of the length of the diameter. For which of your candidate points $P$ is the sum of the squares of the distances from $P$ to the diameter ends equal to the square of the diameter?

Or else we can use perpendicularity directly. Let our given points be $A$ and $B$, and let $P$ be a candidate point. For $P$ to lie on a circle that has $AB$ as a diameter, we need the "dot product" of $A-P$ and $B-P$ to be $0$. That might be fast enough for the $1$ minute restriction.

hi this is hunde mine goes like this
let say
first it is in a x and y dimension
x^2+y^2=r^2
then rotate it in to z dimension along x axis
cos@=a/r)^2+y^2-z^2=r^2

this is hunde from Ethiopia


Correlation between three variables question I was asked this question regarding correlation recently, and although it seems intuitive, I still haven't worked out the answer satisfactorily. I hope you can help me out with this seemingly simple question.

Suppose I have three random variables $A$, $B$, $C$. Is it possible to have these three relationships satisfied?
$$
\mathrm{corr}[A,B] = 0.9
$$
$$
\mathrm{corr}[B,C] = 0.8
$$
$$
\mathrm{corr}[A,C] = 0.1
$$
My intuition is that it is not possible, although I can't see right now how I can prove this conclusively.

Here's an answer to the general question, which I wrote up a while ago. It's a common interview question.

The question goes like this: "Say you have X,Y,Z three random variables such that the correlation of X and Y is something and the correlation of Y and Z is something else, what are the possible correlations for X and Z in terms of the other two correlations?"

We'll give a complete answer to this question, using the Cauchy-Schwarz inequality and the fact that $\mathcal{L}^2$ is a Hilbert space.

The Cauchy-Schwarz inequality says that if x,y are two vectors in an inner product space, then

$$\lvert\langle x,y\rangle\rvert \leq \sqrt{\langle x,x\rangle\langle y,y\rangle}$$

This is used to justify the notion of an ''angle'' in abstract vector spaces, since it gives the constraint

$$-1 \leq \frac{\langle x,y\rangle}{\sqrt{\langle x,x\rangle\langle y,y\rangle}} \leq 1$$
which means we can interpret it as the cosine of the angle between the vectors x and y.

A Hilbert space is an infinite dimensional vector space with an inner product. The important thing for this post is that in a Hilbert space the inner product allows us to do geometry with the vectors, which in this case are random variables. We'll take for granted that the space of mean 0 random variables with variance 1 is a Hilbert space, with inner product $\mathbb{E}[XY]$. Note that, in particular

$$\frac{\langle X,Y\rangle}{\sqrt{\langle X,X\rangle\langle Y,Y\rangle}} = \text{Cor}(X,Y)$$

This often leads people to say that ''correlations are cosines'', which is intuitively true, but not formally correct, as they certainly aren't the cosines we naturally think of (this space is infinite dimensional), but all of the laws hold (like Pythagorean theorem, law of cosines) if we define them to be the negative of the cosines of the angle between two random variables, whose lengths we can think of as their standard deviations in this vector space.

Because this space is a Hilbert space, we can do all of the geometry that we did in high school, such as projecting vectors onto one another, doing orthogonal decomposition, etc. To solve this question, we use orthogonal decomposition, which is often called the ''uncorrelation trick'' in statistics and consists of writing a random variable as a function of another random variable plus a random variable that is uncorrelated with the second random variable. This is especially useful in the case of multivariate normal random variables, when two components being uncorrelated implies independence.

Okay, let's suppose that we know that the correlation of X and Y is $p_{xy}$, the correlation of Y and Z is $p_{yz}$, and we want to know the correlation of X and Z, which we'll call $p_{xz}$. Note that we don't lose generality by assuming mean 0 and variance 1 as scaling and translating vectors doesn't affect their correlations. We can then write that:

$$X = \langle X,Y\rangle Y + O^X_Y$$

$$Z = \langle Z,Y\rangle Y + O^Z_Y$$

where $\langle \cdot,\cdot\rangle$ stands for the inner product on the space and the $O$ are uncorrelated with Y. Then, we take the inner product of $X,Z$ which is the correlation we're looking for, since everything has variance 1. We have that

$$\langle X,Z\rangle = p_{xz} = \langle p_{xy}Y+O^X_Y,p_{zy}Y+O^Z_Y\rangle = p_{xy}p_{yz}+\langle O^X_Y,O^Z_Y\rangle$$

since the variance of Y is 1 and the other terms of this bilinear expansion are orthogonal and hence have 0 covariance. We can now apply the Cauchy-Schwarz inequality to the last term above to get that

$$p_{x,z} \leq p_{xy}p_{yz} + \sqrt{(1-p_{x,y}^2)(1-p_{y,z}^2)}$$

$$p_{x,z} \geq p_{xy}p_{yz} - \sqrt{(1-p_{x,y}^2)(1-p_{y,z}^2)}$$

where the fact that

$$\langle O^X_Y,O^X_Y\rangle = 1-p_{xy}^2$$

comes from the equation setting the variance of X equal to 1 or

$$1 = \langle X,X\rangle = \langle p_{xy}Y + O^X_Y,p_{xy}Y+O^X_Y\rangle = p_{xy}^2 + \langle O^X_Y,O^X_Y\rangle$$

and the exact same thing can be done for $O^Z_Y$.

So we have our answer. Sorry this was so long.

(Edit, @Did, from a 5 years old comment, 2018-09-23) This answer is wrong.

Perfectly possible, first set up your correlation matrix:
$$C = \left[\matrix{1 & 0.9 & 0.1 \\ 0.9 & 1 & 0.8 \\ 0.1 & 0.8 & 1}\right]$$
then, use one of the methods such as described here.

EDIT: As noticed below, because one of the eigenvalues is negative you will end up with a complex valued solution for the correlated random vector (using the eigenvector decomposition method -- Cholesky decomposition is not possible).


Is there any "real" number that may not actually be a real because we haven't found its Dedekind cut? I just watched a video that shows how real numbers are constructed using Dedekind cuts, and what I understood was that a real number is a subset of Q which, among a few other conditions, contains no greatest element. The video also shows an example of how this property can be proven for $\sqrt 2$. I also found these proofs for $pi$ and $e$.

I was wondering if there was another way to construct the reals using Turing machines, since the Dedekind cut feels a bit like a "by negation" kind of definition, and a Turing machine would be more of a procedural, actual way to construct them. But it turns out Turing machines are countable and reals are not, so there must be real numbers that can't be computed by Turing machines.

Which then led me to this other question, in which they give a few samples of non-computable real numbers. But now I wonder, given any of these numbers (say Chaitin's constant), can we prove that the Dedekind cut that corresponds to this number (and to every other number that we intuitively think of as a real number) has no greatest element? Or could it be the case that Chaitin's constant turns out to be a non-real number (which I have no idea what it'd even mean, since it definitely seems to be a point on the real line)?

I know there are other ways to construct real numbers that I haven't learned about yet, so perhaps one of them can be used to prove it?

The Dedekind cut $A$  representing a certain real number $\alpha$ consists of all rational numbers that are $<\alpha$. Among the numbers in $A$ there are arbitrarily precise rational approximations to $\alpha$, but also totally uninteresting numbers in reference to  the value of $\alpha$. But the latter play no rôle in the subsequent construction. The reason why we are carrying them along is that we want a single and simple mathematical object $A$ representing the number $\alpha$, without having to bother about extra data like error bounds.

The condition that such a cut should have no maximal element is not a "definition by negation", but of a purely technical nature. A cut defining an irrational number, like $\pi$, automatically has no maximal element. But we want to have the rational numbers  as ordinary real numbers as well. Without taking care a rational number $\rho$ could be represented by all rational numbers $<\rho$ as well as by all rational numbers $\leq\rho$. In order to avoid ambiguities one requires once and for all that the second kind of "cuts" is excluded.

Your reasoning about counting Turing machines versus counting real numbers has a certain weakness. You might imagine to make a list of all Turing machines; this can be done automatically, since the description of a TM is syntactically clear, and they would be numbered from 1,2,3,... and onwards. Having done that, you might imagine to examine each Turing machine, starting from number 1, then number 2, and so on, to isolate those that happen to compute real numbers, which obviously only some of them would do. If you determine that TM numbered n computes a real number, you would give that real number a label i(n) to show that it is the i(n)'th real number computable by a TM.

It leaves the small obstacle to have to determine whether a given TM computes a real number or not. You probably see where this is going. The solution to the halting problem says that you cannot even determine whether the given TM ever halts, much less whether it halts and outputs a real number description that you desire.

It follows that you cannot argue in this way that there is any "set of computable real numbers", nor that such a set would be countable.

Something like Chaitin's constant is obviously a well-defined real number, since it is given by a description of the individual decimals in its decimal expansion. It is easy in principle to construct a corresponding Dedekind cut or a Cauchy sequence. Not so much in practice.


How to write "let" in symbolic logic How do I write let in symbolic logic? For example, if I am in the middle of a proof and there is a variable which I can assign to an arbitrary value, what would I write? My best guess is:

$$ x := a $$

Would that be clear? Is there a better way to write "let $x$ equal $a$"?

There is no symbol in first-order logic that corresponds to "let", and formal proofs in first-order logic do not have a means to assign a value to a variable.

The phenomenon of setting a variable equal to a value is limited to natural-language proofs, in which it hard to improve on "Let $x$ equal $a$".

Of course these natural-language proofs can be formalized, and each formal deductive system has some way to deal with this type of reasoning, but none of the usual ones does so by literally allowing values to be assigned to variables.

Well, this an easy one! You, simply, do NOT write "let" in (symbolic) logic, as you do it in the vernacular (in your mother-tongue, for instance)!

Now, the so called "mathematical vernacular" [WOT, in Dutch] has been explained (and codified) long ago, by one of my Dutch math teachers, N. G. [ = Dick ] de Bruijn, of the Eindhoven Polytechnics [TUE], The Netherlands, in lectures (on "[De] Taal en Structuur van de Wiskunde" [= The language and structure of mathematics]), sometime during the late seventies, to be precise. (Cf., e.g., Euclides 55, 1979-1980.) --

There is even a nice book (in Dutch) on this, authored by one of his former PhD students and collaborators, Rob Nederpelt (Eindhoven) :

“De taal van de wiskunde, Een verkenning van wiskundig taalgebruik en logische redeneerpatronen” [The language of mathematics, An exploration of the mathematical use of language and the logical proof-patterns] (With an introduction by Dirk van Dalen), Versluis, Almere 1987.

Unfortunately, Rob's book has not been translated so far. But see his recent Cambridge-UP-monograph, co-authored with Herman Geuvers [Radboud Univ., Nijmegen], on "Automath" and the like: "Type Theory and Formal Proof. An Introduction", Cambridge University Press, Cambridge, 2014 [ISBN 978-1-107-03650-5, xxviii + 436 pages] etc.

Essentially, an Automath "language" / system (of proof checking mathematical texts) amounts to a very formal way of writing down a WOT- / "mathematical vernacular"-text, such as to be readable by a machine, a computer, say.

In Automath (and the like), the informal "let" goes either into

(1) a [meta] variable-declaration

or else into

(2) an explicit definition,

pointing out to two distinct -- rather tricky -- ways of manipulating textual substitutions in actual mathematics. (There is third way of doing "substitutions" in logic / mathematics, by applying the so-called CUT-rule of Gerhard Gentzen [1935]; this is just a way of codifying the usual practice of "proving by lemmas" in Euclid & so on.)

For the rest -- if (still) Dutchless and unable to read Rob's WOT-book of 1987 --, I'd suggest buying the recent monograph mentioned above: it is aimed at undergraduates and, as a bonus, it covers a lot of technical stuff you won't easily find elsewhere in print...


Calc 3 practice test question possibly involving chain rule and cross product? I have my final in a few hours, please please help.

Show that if F"(t) exists for all t then:
d/dt [F(t) X F'(t)] = F(t) X F"(t)

It is

$$\frac{d}{dt}(u(t)\times v(t))=u'(t)\times v(t)+u(t)\times v'(t).$$ Thus

$$\frac{d}{dt}(F(t)\times F'(t))=F'(t)\times F'(t)+F(t)\times F''(t).$$ Now, we have that $$F'(t)\times F'(t)=0.$$ Thus

$$\frac{d}{dt}(F(t)\times F'(t))=F(t)\times F''(t).$$


Derivatives of holomorphic function Let $f:U\rightarrow \mathbb{C}$ be a holomorphic function on a open and bounded subset $U$ of $\mathbb{C}$. Then suppose $|\frac{df}{dz}|<C$ for a certain constant $C>0$. Does this condition imply that all the norms of the other derivatives of $f$, $(|\frac{d^2f}{dz^2}|,|\frac{d^3f}{dz^3}|,\cdots$) are bounded?

I think the answer is yes: around $z_0\in U$ the derivative of the function is $\frac{df}{dz}(z)=a_1+2a_2(z-z_0)+3a_2(z-z_0)^2+\dots$ and if this sum is bounded in every point then I think it's safe to say that also the sum $\frac{d^2f}{dz^2}(z)=2a_2+6a_2(z-z_0)+\dots$ is bounded.. But this reasoning is very rough, so I was hoping someone could help me with a better argument

The claim is not true. A counter example: look at $f(z) = z^{3/2} + z$ over an open set $U = B_{2/3}(0) - (-\infty, 0] \subset \mathbb{C}$, where $B_{2/3} = \{z \in \mathbb{C}||z| < 2/3\}$. Then $f$ is holomorphic on $U$ with $df/dz = (3/2)z^{1/2} + 1/2$ therefore $1/2 < |df/dz|\leq 3/2|z| + 1/2 < 3/2(2/3) + 1 = 2$. But it is clear that $d^2f/dz^2 = 3/4z^{-1/2}$ would be unbounded as $z \rightarrow 0$ in $U$.

No, take for $U=\{|z|< 1\}$ the unit disc and let $f'(z)=\sum_{n\geq 1} \frac{1}{n^2} z^n$ so that $|f'(z)|<2$ uniformly on $U$. Then  $f''(z)=\sum_{n\geq 1}\frac{1}{n}z^{n-1}\rightarrow \infty$ as $z\rightarrow 1^-$. However, if you restrict yourself to $|x|\leq r<1$ then you may write down bounds for every higher order derivative on this smaller disc e.g. by using Cauchy's formula for the derivatives.


Why does $\frac{1}{x} < 4$ have two answers? Solving $\frac{1}{x} < 4$ gives me $x > \frac{1}{4}$. The book however states the answer is: $x < 0$ or $x > \frac{1}{4}$.

My questions are:

Why does this inequality has two answers (preferably the intuition behind it)?

When using Wolfram Alpha it gives me two answers, but when using $1 < 4x$ it only gives me one answer. Aren't the two forms equivalent?

You have to be careful when multiplying by $x$ since $x$ might be negative and hence flip the inequality. Suppose $x>0$. Then
$$\frac{1}{x}<4\iff4x>1\iff x>1/4.$$
If $x>0$ and $x>1/4$, then $x>1/4$.

Now suppose $x<0$. Then
$$\frac{1}{x}<4\iff4x<1\iff x<1/4.$$
If $x<0$ and $x<1/4$, then $x<0$.
So the solution set is $(-\infty,0)\cup(1/4, \infty).$

$$\frac{1}{x} < 4,$$
$$\frac{1}{x} - 4 < 0,$$
$$\frac{1 - 4x}{x} < 0,$$
$1 - 4x < 0$ or $x > 0$. Because if $1 - 4x$ is negative then $x$ must be positive. So we must write the solution in that way.


Proof of a property of fuzzy extension principle I'm looking for a proof of the following property of the fuzzy extension principle:

$$
B \supseteq f(f^{-1}(B))
$$

($f: X\rightarrow Y$ is an arbitrary crisp function, and $B$ is a fuzzy set). I think if f is a function, the two sides would be equal. But I see this property on page 47 of the book by Klir and Yuan.

Thank you.

$f\left(f^{-1}\left(B\right)\right)\subseteq B$ is always true, which
is not the case for $B\subseteq f\left(f^{-1}\left(B\right)\right)$
(see the answer of 5xgum). It is true for each $B$ if and only if
$f$ is surjective.

As you said, $f$ is a crisp function, so it will only work with those elements that ARE or ARE NOT in $X$, and the same for $Y$.

Note that $B:Y\to [0,1]$ and $f^{-1}$ will take those $y$ in $Y$ such that $B(y)=1$ (you can call that subset of $Y$, $B^{-1}(\{1\})$) and relate them with some $x$ in $X$ such that $f(x)=y$, if it does exist.

Now, the subset of $X$ made of those $x$'s will be returned to $Y$ with $f$, into a subset of $B^{-1}(\{1\})$, because if some $y$ in $B^{-1}(\{1\})$ actually doesn't have a preimage.

So, $f(f^{-1}(B)) \subset B^{-1}(\{1\}) \subset B$, because in fact it may have existed a $y_0$ in $Y$ such that $B(y_0)=\frac{1}{2}$, for say something, and it wouldn't even be in $B^{-1}(\{1\})$.

(Sorry for the mecca font; hope this helps you 1 year after you published it)


Is there an algebraic method to get at this limit? I know the limit of this function as it approaches $\pi$ is oscillating between $-1$, and $1$ meaning it does not exist.

  $$\lim_{x\to\pi}(\sin(\frac{1}{x-\pi}))$$

I was just wondering if graphing is considered an algebraic technique?  Because I don't know how to approach this algebraically.

I do not really get what you call algebraic methods (and non-algebraic methods), but to show that the function is oscillating between $-1$ and $1$, consider the following sequence (with $y$ being any real number)  :

$u_n = \frac{1}{y+2n\pi} + \pi$

You can easily check that $u_n \to_{n \to \infty} \pi$ and that $\sin(\frac{1}{u_n - \pi}) \to_{n \to \infty} \sin(y)$

A way to approach the limit is by using limit properties, squeeze theorem, and algebra techniques.

  $$\lim_{x\to\pi}(\sin(\frac{1}{x-\pi}))$$
  $$\lim_{x\to\pi}(\sin(\frac{1}{x-\pi})\times\frac{1}{\frac{x-\pi}{x-\pi}})$$
  $$\lim_{x\to\pi}(\frac{\sin(\frac{1}{x-\pi})}{\frac{1}{x-\pi}}\times\frac{1}{x-\pi})$$
  $$\lim_{x\to\pi}((x-\pi)(\sin{\frac{1}{x-\pi}})(\frac{1}{x-\pi})$$

Now applying limit properties:

  $$\lim_{x\to \pi}((x-\pi)(\sin(\frac{1}{x-\pi}))\times\lim_{x\to\pi}(\frac{1}{x-\pi})$$

Simplifying the limit on the left it can be done using squeeze theorem:

  $$f(x)\le g(x)\le h(x)$$
  $$-1\le \sin(\frac{1}{x-\pi})\le 1$$

Then,

  $$-(x-\pi)\le (x-\pi)\sin(\frac{1}{x-\pi})\le (x-\pi)$$

Afterwards a limit can be taken as x approaches $\pi$

  $$\lim_{x\to\pi}(-(x-\pi))\le \lim_{x\to\pi}((x-\pi)\sin(\frac{1}{x-\pi}))\le \lim_{x\to\pi}(x+\pi)$$

Which is equal to:

  $$0\le\lim_{x\to\pi}((x-\pi)\sin(\frac{1}{x-\pi}))\le 0$$

From this we can conclude that:

  $$(0)\times\lim_{x\to\pi}(\frac{1}{x-\pi})$$

Because the other limit has a similar form as:

  $$\lim_{x\to\pi}(\frac{1}{x-\pi})=\lim_{u\to0}(\frac{1}{u})$$

Therefore the limit is known to not exist therefore:

  $\lim_{x\to\pi}(\sin(\frac{1}{x-\pi}))$ does not exist.


Integrand of an integral equation. In the integral equation assuming $y\gg 0$ $$\large\int_{y^r}^{y^{2r}}f(x)dx=\frac {y^{2r}}2$$ what is a good candidate for $f(x)$ if $r\in\big(\frac12,1\big)$ holds?

Assume $f(x)>0$ and is monotone with $f(y^r)\rightarrow c= O(1)$ and $f(y^{2r})\rightarrow g(y^{2r})= O(y^{2r})$.

$f(x)=\frac x{y^{2r}}$ comes pretty close.

I'll put the problem into a more general context (the present problem
being a special case):
Let $g:{\Bbb R}_+\rightarrow  {\Bbb R}$ be  Lipschitz continuous
and consider the problem of finding $f$ so that:
$$ \int_{y^\alpha}^y f(x)\; dx = g(y) $$

Suppose first that $g(1)=0$ and let
$F$ be a primitive of $f$. Then we want to solve
$$ F(y) - F(y^\alpha)= g(y) $$
which has the formal solution:
$$ F(y) = g(y)+F(y^\alpha) = g(y) + g(y^\alpha) + g(y^{\alpha^2}) =
... =
  \sum_{k\geq 0} g(y^{\alpha^k}) .$$
As $k\rightarrow \infty$ we have $y^{\alpha^k}=1+\alpha^k \log(y)+ ...$ so a sufficient condition for the
series is that e.g. $g$ is Lipschitz continuous at $1$ and $g(1)=0$.
The solution in this case exists for all $y>0$.

Consider now for $y\neq 1$, $y>0$ the problem of finding $f_0$ so that:
$$ \int_{y^\alpha}^y f_0(x)\; dx = 1 $$
The solution is given by:
$$ F_0(y) = \frac{\log |\log y|}{-\log \alpha} $$
as is seen by inspection since:
$${\log |\log y|} - \log|\log (y^\alpha)| =
{\log |\log y|} - \log \alpha (|\log y|) = -\log \alpha  $$

The general solution is thus (and we need $y\neq 1$ when $g(1)\neq 0$):
$$ F(y) =
  \left( \sum_{k\geq 0} \left(g(y^{\alpha^k})-g(1) \right) \right)
   + g(1)   \left( \frac{\log |\log y|}{\log 1/\alpha} \right) .
$$

To find $f$ it suffices that $g$ is continuously differentiable in which case:
$$ f(y) = F'(y)=
  \frac{1}{y} \left( \sum_{k\geq 0} g'(y^{\alpha^k})
    \alpha^k  y^{\alpha^k} \right)
   + g(1)   \left( \frac{{\rm sign}(y)}{y |\log y|}
    \frac{1}{\log \frac{1}{\alpha}} \right) .
$$
The first sum converges when $g'$ is assumed continuous.

In the present context we are looking at
$\alpha=1/2$ and $g(y)=y/2$ (details omitted).


Make the dominated convergence theorem rigourous I have the following when I apply the dominated convergence theorem :

  Let's say I want to find : $\lim_{n \to +\infty}\int_0^1 t^n \mathrm{d}t$ using the DCT (I know we can just integrate but it's an example to show where I have a problem).

So we defined the set of function : $f_n : [0,1] \to \mathbb{R} : t \mapsto t^n$. We have : $f_n \leq 1$ and $f_n$ converges to the piewise continuous function : $g = 0$ on $[0,1[$ and $g(1) = 1$.

Now the problem is that when I interchange limit and integration :

$$\lim_{n \to \infty} \int_0^1 f_n = \int_0^1 \lim_{n \to \infty} f_n = \int_0^1 g$$

Then clearly $\int_0^1 g(x) \mathrm{d}x = 0$ but how can I show this rigourously ? Since $g(1) = 1$ I don't know how to prove it's equal to $0$.
Maybe I cant take an $\epsilon$ and say : $\int_0^1 g \leq  \int_0^{1-\epsilon} 0 + \int_{1-\epsilon}^1 1 $

But it seems a lot of effort to prove an obvious fact.

Thank you !

N.B : All these integrals are Riemann not Lesbegue

First, you have to understand that the Lebesgue integral is a generalization of the Riemann integral. What we mean by that is that if a function is Riemann integrable, it is Lebesgue integrable and the two notions of integration are he same, ie : $\int_a^b f(x)dx=\int_{[a,b]}f dm$ where $m$ is the Lebesgue measure. A powerful theorem for this equivalence is the following :
$f$ is Riemann integrable iff its set of discontinuities has Lebesgue measure 0. Now in your example, it is easy to see that you only have 1 point where the function is discontinuous and so the two notions agree.
Suppose now you dont know that this theorem exists and that you only care about Riemann integrals. The important thing you have to remember is that integration never depends on the value of a function at a point ! Namely, you can change countably many points of your function (for example, let the function be equal to $\pi$ at all rational points !) and the resulting value of the integral will be the same. Your idea is good ! Simply take an $\epsilon$ neighborhood around 1, integrate as you wanted to do and let $\epsilon\to 0$. The intuition behind that is that the area under a single point is always as small as you want it to be !
I hope I clarified your uncertainties, if you have any questions leave them in the comment.

PS : The power of the Lebesgue Integration theory lives in its powerful convergence theorems (monotone convergence and dominated convergence). They not only yield results that were previously unobtainable with the Riemann theory of integration but also reduce the labor in proving classical results (such as a function being Riemann integrable).

To answer your question: It's trivial to verify from the definition of the integral that $\int_0^1 f(t)\,dt=0$.

But it's very curious that you talk about DCT in the context of Riemann integrals! Because the statement usually known as "DCT" is false for the Riemann integral - hence one wonders exactly what theorem  you're referring to.

A weak version of DCT for integrals on $[0,1]$:

    DCT. Suppose $f_n$ and $g$ are integrable on $[0,1]$, $|f_n|\le g$, and $f(t)=\lim_nf_n(t)$ exists for every $t\in[0,1]$. Then $f$ is integrable on $[0,1]$ and $\int_0^1 f=\lim_n\int_0^1f_n$.

Counterexample for the Riemann integral: In general if $A$ is a set define the function $\chi_A$ by $$\chi_A(t)=\begin{cases}1,&(t\in A),
\\0,&(t\notin A).\end{cases}$$

Say $E=\{r_1,r_2,\dots\}$ is a countable dense subset of $[0,1]$. Let $E_n=\{r_1,\dots,r_n\}$. Let $f_n=\chi_{E_n}$, $g=1$.  Then $$\lim_nf_n(t)=f(t)=\chi_E(t)$$for every $t$, but $f$ is not Riemann integrable (because for example it is discontinuous everywhere).


How do you find the area of a parallelogram with the vertices? How do you find the area of a parallelogram with the following vertices; $A(4,2)$, $B(8,4)$, $C(9,6)$ and $D(13,8)$.

The absolute value of the cross product of two vectors $\vec{a}, \vec{b} \in \mathbb{R}^3$ spanning the parallelogram is its area:

$$A_\text{parallelogram}= \left|\vec{a}\times\vec{b}\right|$$

So in your case we have to write the points in $\mathbb{R}^2$ as vectors in $\mathbb{R}^3$ and apply the formula:

$\vec{AB} = \begin{pmatrix}8\\4\\0\end{pmatrix}  -\begin{pmatrix}4\\2\\0\end{pmatrix} =\begin{pmatrix}4\\2\\0\end{pmatrix}$

$\vec{AD} = \begin{pmatrix}13\\8\\0\end{pmatrix}  -\begin{pmatrix}4\\2\\0\end{pmatrix} =\begin{pmatrix}9\\6\\0\end{pmatrix}$

$A_\text{parallelogram}= \left|\vec{AB}\times\vec{AD}\right| = \left| \begin{pmatrix}4\\2\\0\end{pmatrix} \times \begin{pmatrix}9\\6\\0\end{pmatrix} \right| = \left|\begin{pmatrix}0\\0\\6\end{pmatrix}  \right| = 6$

You might have noticed that this simplifies to

$$A_\text{parallelogram}= (b_1 - a_1)(d_2-a_2)-(b_2-a_2)(d_1-a_1)$$
$$= (8 - 4)(8-2)-(4-2)(13-4)=-24-(-18)=6$$

I am just providing you with the simplest shortcut to doing this

Pick the first three points A(4,2), B(8, 4) and C(9, 6)

Negate point A to get (-4, -2) and add to the other two points B and C. Add x's and y's so you have a new point

(4, 2)
(5, 4)   Now use determinant to find the area.

16-10 = 6sq unit

Neglect any negative sign that arises


What is the advantage of measuring an angle in radian(s)? What is the advantage and use of measuring an angle is radian(s) compared to degree(s)? My book suddenly switched to radian(s) for measuring an angle in this grade and I do not know why.

Look at the following picture. It is a circle of radius $r=1$ and there is an angle $\alpha$ that cuts out an arc $c$ from the circumference of the circle.

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$

How long is this arc? Well, this is where radians are immensely beneficial. If you express the angle in radians, then the length of the arc is exactly $\alpha$. If $\alpha=0.123$, then the arcs length is $0.123$ too. Easy.

You know that the full angle in radians is $2\pi$. If you choose $\alpha=2\pi$ in the above image, you would describe an arc $c$ that is actually just the full circle. And you immediately see, that its circumference (here also the length of the arc) is $2\pi$ too. This is exactly what the formula for the circumference $U(r)=2\pi r$ would give you anyway.

Of course this is also useful for circles of radii other than $1$. If your circle has a radius $r$ and you look at an arc cut out by an angle $\alpha$ in radians, then the length of the arc is $\alpha r$. It just scales linearly.

This is the visual aspect. But there are more (mathematical) reasons. Ask yourself, what would be a natural unit to measure an angle in? Degree is not very natural. There is nothing special about the number $360^°$ but its high divisibility. You could measure an angle in the interval $[0,1]$. Or you can use its natural connection to the circle that I explained above.

However, there are many mathematical functions that take angles as inputs, e.g., $\sin(x),\cos(x)$, etc. But have you ever asked how to actually compute $\cos(x)$ without a calculator? There are formulas that give very good approximations, e.g.,

$$\cos(x)\approx 1-\frac12 x^2+\frac1{24}x^4.$$

But they only work for radians. You can write them down for other units, but they will never look so natural, not even for angles in the units $[0,1]$.

There is no "advantage" of radians over angles, for example. Think of it as an unit of measure. For example, kg vs pounds. The only difference is on the choice one makes, and not in what actually represents.

EDIT:

Do not be misled by the down votes. This is an answer from the perspective of a theoretical physicist.

I strongly encourage you to think in, again, units in physics. There is no advantage in using one units or the other, except perhaps your final numerical results and the application of what you want to give them. The theory does not change at all if you use degrees or radians, the angle is still the same angle viewed with different glasses! I have never used degrees since I started the career (and not from that I am going to say that one is advantageous one over the other, because I can google up examples where there is a wide application of degrees over radians), because I don't need them. Perhaps and engineer would need them more and that would make them advantageous for him.

I want to stress that the theory is not change by either using degrees or radians, and from this point of view the advantage type of feeling you have is not really relevant.

If you continue your studies in science, you will find specially in Physics, that most of the equations are given in free-unit form. Maxwell's equations describe the same phenomena no matter which units you choose (the only difference are a couple of prefactors in the equations, which will give you your chosen units). Newton's equations are still the same, whether you choose to measure forces or masses in  cgi or international unit system (or any other system unit).


Sum of functions bounded between 0 and 1? Let $y \in \mathbb{R}$ and $x \in \mathbb{R}$. I'm looking for two functions $f,g$ such that
$$
f(x)+g(y) \in [0,1]
$$
Do they exist? In positive case, do you have suggestions for what $f$ and $g$ could be? I'm not interested in $f,g$ taking a constant value or period functions.

The function $\tanh$ is neither constant nor periodic and takes values in the open interval $\ ]{-1},1[\ $. By choosing the constants $c_1$, $c_2$ properly you can then fabricate many pairs of functions
$$f_i(x):=c_i\bigl(1+\tanh(\alpha_i x+\beta_i)\bigr)\qquad(i=1,2)$$
that suit your needs.

What about $f(x)=g(x)=$$(-1)^x/2$?
$(tanh^2x/2)$ won't work because the range of $f+g$ is $[0,1)$, not $[0,1]$


Understanding Converse of Lagrange's theorem! Lagrange's theorem states that for any finite group $G$, the order (number of elements) of every subgroup $H$ of $G$ divides the order of $G$ .......(1)

The converse of Lagrange's theorem is if $x$ divides order of $G$ ,then there exists a sub group of order $x$.                 ......         (2)

If my statement is $p\to q$ then converse is $q\to p$.

i couldn't understand how converse of Lagrange's theorem comes...please explain with this definition of converse : $p\to q$ then converse is $q\to p$

Actually, you need to write the Lagrange theorem in $p \rightarrow q$ form. So, the Lagrange theorem is actually,

If $H$ is a subgroup of $G$, then order of $H$ divides order of $G$.

Here $p:$ $H$ is a subgroup of $G$

$q;$ Order of $H$ divides order of $G$.

Here, actually the existence of $H$ is in the hypothesis ($p$) and the number $m$ is actually the order of the subgroup which divides order of $G$.

So, the converse will be

If a number $m$ divides order of $G$, then there is a subgroup of order $m$.

I hope this gives you insights.

Take the group $A_4$, then group of all even permutations of $4$ elements. $o(A_4)$ is $12$, but it doesn't have any subgroup of order $6$. Please check it !!


Integral test for convergence? Why does the series' terms have to be non-negative to use the integral test?

Consider the series:

$$\sum_{n = 1}^{\infty}\frac{n\cos n - \sin n}{n^2}$$

Even though it has negative terms, why can't the integral test be used with the integral being:

$$\int_1^\infty\frac{x\cos x - \sin x}{x^2}$$

Since the series can be interpreted as the Riemann sum of the function above?

It's precisely the fact that it has infinitely-many negative terms; the Integral Test only works when the sequence $a_n$ we're taking the series of is the restriction of a positive non-increasing integrable function $f(x)$.  To understand why, it is useful to understand the general idea of the proof of the integral test.  At this site is a good illustration of the Riemann sums being taken and how they actually relate to the infinite series (this is one of the nice examples when the proof is essentially the picture).

Assume the following series:

$$ \sum_{n=1}^\infty \cos(\pi n) = \sum_{n=1}^\infty \pm 1$$
As you might know this series does not converge!

But if you look at the Riemann integral you have:

$$ \lim_{n\to\infty} \int_{1}^n \cos(x\pi)\text{d}x = 0 $$


Prove $\lim\limits_{x\to\infty}\frac{1}{x}\int_0^xf(t)dt=a$ Let $f:[0,\infty)\to\mathbb{R}$ continuous and $\lim\limits_{x\to\infty}f(x)=a$. Claim: $\lim\limits_{x\to\infty}\frac{1}{x}\int_0^xf(t)dt=a$.

My try: It is $\int_0^xf(t)dt=F(x)-F(0)$ (because of the fundamentaltheorem of calculus) and $\lim\limits_{x\to\infty}F(x)=ax+b$, because $\lim\limits_{x\to\infty}f(x)=a$.
What can I do next, or do you have an idea how to prove it? Regards

Use L'Hôpital's rule and the fundamental theorem of calculus. There are two cases.

Case 1: $\int_0^\infty f(t)dt = \pm \infty$. In this case, by L'Hôpital's rule and the FTC, $$\lim_{x \to \infty} \frac{1}{x} \int_0^x f(t)dt = \lim_{x \to \infty}\frac{\frac{d}{dx}\int_0^x f(t)dt}{\frac{d}{dx}x} = \lim_{x \to \infty}\frac{f(x)}{1} = a \text{.}$$

Case 2: $\int_0^\infty f(t)dt$ is finite. Then the limit in question is clearly 0, and we must show that $a = 0$ as well. However, if $a \neq 0$, then $f$ is eventually bounded away from 0, and so the improper integral would be infinite (a contradiction).

You're close: as $\;f\;$ is continuous it has a primitive $\;F\;$ , so

$$\frac1x\int_0^xf(t)dt=\frac{F(x)-F(0)}{x-0}\stackrel{MVT}=F'(c_x)=f(c_x)\;,\;\;c_x\in(0,x)$$

and now just take the limit $\;x\to\infty\;$ ... and be careful (for example, for different $\;x$'s one could possibly get the same $\;c_x\;$ . How do you go around this?)


Why does $\frac{z-i}{z+i}$ map the unit disk onto an open half-plane? I've been self studying complex analysis, and I read that there is a canonical conformal bijection
$$\varphi(z) = \frac{z-i}{z+i}$$
from the open unit disc to the open half-plane.
I see why $\varphi$ is analytic, and I was able to show that it is one to one by taking differences and counting zeros. I cannot see why $\varphi$ is onto, however, why is this true?

You can use the fact that the inverse function is $\varphi^{-1}(z)=-i\frac{z+1}{z-1}$, and that for $z$ in the left half plane, $|z-1|^2$ is always greater than $|z|^2+1$ (expand $(z-1)\overline{(z-1)}$, and see how $\text{Re}(z)$ appears). Meanwhile, $|z+1|^2 < |z|^2 + 1$ for similar reasons. So, the norm-square of $\varphi^{-1}(z)$ on the left half plane always has a larger denominator than numerator; thus its norm-square is always less than 1, and therefore so is its norm. So, the image of $\varphi^{-1}$ on the left half plane is contained in the unit disc, and we’re done.
Here’s a more geometric way of looking at it: $\varphi(z) = \frac{i-z}{-i-z}$, i.e. the ratio of the two “vectors” from $z$ to $\pm i$. This ratio is the complex number $s$ that makes a similar triangle with $1$ in the complex plane, upon identifying $i$ with $s$, $-i$ with $1$, and $z$ with the origin. We know that for any $s$ in the left open half plane, the angle $s$ makes with $1$ through the origin is between $\pi /2$ and $3\pi /2$. As such, since the triangles are similar, the angle through $0$ is the same as the angle through $z$ and so the point $z$ must lie within the unit circle. So, we’ve constructed a point in the unit disc that yields an arbitrary point in the left half plane.
If any step is unclear here let me know! :)

It is onto because you can write its inverse by inverting the matrix
$$\begin{pmatrix} 1 & -i \\ 1 & i\end{pmatrix}.$$


Sherlock Holmes Base Number Problem Sherlock Holmes and Dr. Watson recover a suitcase with a three-digit combination lock from a mathematician turned criminal. Embedded in the suitcase above the lock is the cryptic message "AT SEA BASE. SEAS EBB SEA: BASS. "
Dr. Watson comments, "This probably isn't about ocean fish. Perhaps it is an encrypted message. The colon suggests the addition problem $SEAS + EBB + SEA = BASS$, where each letter represents a distinct digit, and the word 'BASE' implies that the problem is in a different base."
Holmes calmly turns the combination lock and opens the suitcase. While Dr. Watson gapes in surprise, Holmes replies, "You were indeed right, and the answer was just the value of the word $SEA$, interpreted as decimal digits." What was the lock combination?
I'm finding this problem really hard. I think this is a Mathcounts problem. I have tried setting the base to different numbers and then bashing it but I got nowhere. Help!

Suppose that each letter represents a distinct digit, and that the problem is in a different base, say $x$. Then we can represent the problem as follows:
\begin{eqnarray*}
Sx^3+&Ex^2+&Ax+&S&\\
     &Ex^2+&Bx+&B&\\
     &Sx^2+&Ex+&A&\quad+\\
\hline
Bx^3+&Ax^2+&Sx+&S&
\end{eqnarray*}
In each column we have at most $2$ carries, so from this it follows that
\begin{eqnarray*}
S+B+A &=& S+k_1x,\\
A+B+E+k_1&=& S+k_2x,\\
E+E+S+k_2&=& A+k_3x,\\
S+k_3 &=& B,
\end{eqnarray*}
where $0\leq k_1,k_2,k_3\leq2$, and $0\leq A,B,E,S<x$ are distinct.
The first equation shows that $B+A=k_1x$, so $k_1=1$ because $0\leq A,B<x$ are distinct, so
$$B=x-A\tag{1}.$$
Then in the second equation we see that $k_2\geq1$ because $A+B=x$, and we find that
$$E=S+(k_2-1)x-1.$$
Because $0\leq E<x$ we have either $k_2=1$ and $E=S-1$, or $k_2=2$ and $S=0$ and $E=x-1$. In the latter case, the third equation shows that
$$A+k_3x=E+E+S+k_2=(x-1)+(x-1)+0+2=2x,$$
and hence $k_3=2$ and $A=0=S$, contradicting the assumption that the digits are distinct. Therefore $k_2=1$ and
$$E=S-1.\tag{2}$$
The fourth equation shows that $k_3\neq0$ because $S\neq B$, and the third equation now yields
$$A=3S-2-k_3x.$$
Plugging in $A=x-B$ and $S=B-k_3$ and rearranging then yields
$$4B+1=(k_3+1)(x+3),$$
which shows that $k_3$ is even, so $k_3=2$. Then
\begin{eqnarray*}
B&=&\frac{3x+8}{4}&=&\frac34x+2,\\
A&=&x-B&=&\frac14x-2,\\
S&=&B-2&=&\frac34x,\\
E&=&B-3&=&\frac34x-1,\\
\end{eqnarray*}
which shows that $x$ is divisible by $4$ and $x\geq12$ because $B<x$. Because $S,E,A<10$ it follows that $x=12$, and $(S,E,A)=(8,7,1)$.

The lock combination was $871$. This can be proved as follows. First, since $S$, $E$ and $A$ have to be interpreted as digits, there are $\leqslant 9$. Let $N$ be the unknown base. One has
\begin{matrix}
&S&E&A&S\\
+&&E&B&B\\
+&&S&E&A\\
\hline
&B&A&S&S
\end{matrix}
which gives successively

$S + B + A = S + N$, whence $A + B = N$ and $A > 0$,
$1 + A + B + E = S + N$, whence $S = E + 1$
(a) $1 + E + E + S = A + N$ and $B = S + 1$ or (b) $1 + E + E + S = A + 2N$ and $B = S + 2$.

Case (a) would lead to $2 + 3E = 2A + E + 2$, whence $A = E$ and this would not be a code. Case (b) leads to $2 + 3E = 3A + 2E + 6$, whence $E = 3A + 4$. Since $E \leqslant 9$ and $A > 0$, one has necessarily $A = 1$, $E = 7$, $B = 10$, $S = 8$ and $N = 11$ (the base). Thus the code $SEA$ is $871$.


Equivalence of focus-focus and focus-directix definitions of ellipse without leaving the plane Take a look at the following two definitions of ellipse:

For some fixed points $F_1,F_2$ and real number $2a>|F_1F_2|$ an ellipse is the locus of points $P$ such that $|F_1P|+|F_2P|=2a$.
For some fixed point $F$, line $d$ and number $e<1$ an ellipse is the locus of points $P$ such that $|FP|$ is $e$ times the distance from $F$ to $d$.

These two definitions can be easily shown equivalent using Dandelin spheres (which, in fact, also estabilishes that ellipse can be defined as a kind of conic section). However, for some time, I have been wondering if there is a way to show these definitions equivalent while "staying on the plane", i.e. without Dandelin spheres, cones etc.
My question here is: Is there any direct proof of equivalence of the above two definitions of ellipse? Also with "direct" I mean one without using the equation of an ellipse.
Thanks in advance.

$\frac{r1}{d1} =\frac{r2}{d2} = \frac{r1+r2}{d1+d2} = e$; Obviously $d_1+d_2$ is constant, so r1+r2 is also constant


Does $\liminf_{r\to 0+} f'(r) \geq 0$ implies the existence of $f(0 +)$? Let $f$ be a real-valued differentiable function on $(0,1)$, and suppose that
\begin{equation*}
\sup_{0<r<1} |f(r)| < \infty,
\end{equation*}
and
\begin{equation*}
\liminf_{r \to 0+} \,f'(r) \geq 0.
\end{equation*}
Can we conclude that $f(0+)$ exists?

Hint: Let $g(r)= r+f(r).$ Then $g'(r)> 0$ for $r$ in some $(0,a).$

Your first condition says $f$ is bounded on $(0,1)$.

Case I:    $\underset{r \rightarrow 0+}{\text{lim inf}} f'(r)>0$

In this case the lim inf condition says that eventually, for some $\epsilon>0$ sufficiently small, $f'$ is positive on $(0,\epsilon)$, which means $f$ is increasing on $(0,\epsilon)$.

$f$ bounded and increasing on $(0,\epsilon)$ implies $f(0+):= \underset{r\rightarrow 0+}{\text{lim }} f(r) = \underset{r\in(0,\epsilon)}{\inf}f(r) \in \mathbb{R}$.

Case II:    $\underset{r \rightarrow 0+}{\text{lim inf}} f'(r)=0$

...

(It is not clear to me whether $f(0+)$ exists in this case.)


Find $ P(Z>X+Y)$ where $X,Y,Z \sim U(0,1)$ independently I'm trying to follow a line in a derivation for $P(Z>X+Y)$ where $X,Y,Z$ are independent continuous random variables distributed uniformly on $(0,1)$.

I've already derived the pdf of $X+Y$ using the convolution theorem, but there's a line in the answer that says:

$P(Z>X+Y) = \mathbb{E}[\ P(Z>X+Y\ |\  X+Y )\  ]$ where $\mathbb{E}$ is the expectation.

I'm not familiar with this result. Could anyone give a pointer to a similar result if one exists?

Thanks.

This is not an answer to your question about the justification for the
equation that is puzzling you, but I think the geometrical method
described below for solving the problem that may give you a different
insight into the calculation of the desired probability $P\{Z > X+Y\}$.

The random point $(X,Y,Z)$ is uniformly distributed in the interior of the unit cube with diagonally opposite vertices $(0,0,0)$ and $(1,1,1)$.  The cube has unit volume and so the probability that $(X,Y,Z)$ is in some region is just the volume of that region.  Thus, $P\{Z > X+Y\}$ is the volume of the tetrahedron with vertices $(0,0,0)$, $(1,0,1)$, $(0,1,1)$ and $(0,0,1)$.  If we think of this
as an inverted pyramid whose base is the right triangle with vertices
$(1,0,1)$, $(0,1,1)$ and $(0,0,1)$ and apex $(0,0,0)$ is at
altitude $1$ "above" the base,
then since the base has area $\frac{1}{2}$, we get
the volume as
$$P\{Z > X+Y\} = \frac{1}{3}\times (\text{area of base})\times(\text{altitude})
= \frac{1}{3}\times \frac{{3}}{2}\times1 = \frac{1}{6}.$$
Of course, if you have already computed the density of $X+Y$, then it is
straightforward to use the result given by Artiom Fiodorov to get
$$P\{Z > X+Y\}= \int_0^2{P}(Z>v)f_{X+Y}(v)\;dv
= \int_0^1(1-v)\cdot v\;dv =
\left.\frac{v^2}{2}-\frac{v^3}{3}\right|_0^1 = \frac{1}{6}.$$

I don't know if this helps since Dilip has given the answer, but the distribution of X+Y is triangular on [0,2] (isosceles with peak at X+Y = 1). So P(Z>X+Y) is the probability that a uniform on [0,1] is larger than  the triangular random variable on [0,2]. If X+Y>1 then Z cannot be >X+Y and the probability that X+Y is greater than 1 is 1/2. Now this is where taking the expectation fo the conditional probability helps in my proof.
P{Z>X+Y) =E[P(Z>X+Y|X+Y)]= ∫u P(Z>u|X+Y=u)du =∫u P(Z>u)du where u is integrated from 0 to 1.  The condition X+Y=u gets dropped because Z is independent of X+Y. P(Z>u)=1-u for 0<=u<=1.
hence P(Z>X+Y) =∫u(1-u)du  = 1/6. Just as Dilip showed.


Sum of two gamma/Erlang random variables $\Gamma(m,\lambda)$ and $\Gamma(n, \mu)$ with integer numbers $m \neq n, \lambda \neq \mu$ The gamma distribution with parameters $m > 0$ and $\lambda > 0$ (denoted $\Gamma(m, \lambda)$) has density function
$$f(x) = \frac{\lambda e^{-\lambda x} (\lambda x)^{m - 1}}{\Gamma(m)}, x > 0$$
Given two independent gamma random variables $X = \Gamma(m,\lambda)$ and $Y = \Gamma(n, \mu)$ with integer numbers $m \neq n, \lambda \neq \mu$, what is the density function of their sum $X + Y = \Gamma(m,\lambda) + \Gamma(n,\mu)$?
Notice that both $X$ and $Y$ are also Erlang distribution since $m,n$ are positive integers.
My attempt
First, I searched for well-known results about gamma distribution and I got:
(1) If $\lambda = \mu$, the sum random variable is a Gamma distribution $\sim \Gamma(m+n, \lambda)$ (See Math.SE).
(2) $\Gamma(m, \lambda)$ (or $\Gamma(n, \mu)$) is the sum of $m$ (or $n$) independent exponential random variables each having rate $\lambda$ (or $\mu$). The hypoexponential distribution is related to the sum of independent exponential random variables. However, it require all the rates distinct.
(3) This site is devoted to the problem of sums of gamma random variables. In section 3.1, it claims that if $m$ and $n$ are integer numbers (which is my case), the density  function can be expressed in terms of elementary functions (proved in section 3.4). The answer is likely buried under a haystack of formulas (however, I failed to find it; you are recommended to have a try).
Then, I try to calculate it:
$$f_{X+Y}(a) = \int_{0}^{a} f_{X}(a-y) f_{Y}(y) dy \\ = \int_{0}^{a} \frac{\lambda e^{-\lambda (a-y)} (\lambda (a-y))^{m-1}}{\Gamma(m)} \frac{\mu e^{-\mu y} (\mu y)^{n-1}}{\Gamma(n)} dy \\ = e^{-\lambda a} \frac{\lambda^m \mu^n}{\Gamma(m) \Gamma(n)} \int_{0}^{a} e^{(\lambda - \mu) y} (a-y)^{m-1} y^{n-1} dy$$
Here, I am stuck with the integral and gain nothing ... Therefore,

How to compute the density function of $\Gamma(m,\lambda) + \Gamma(n,\mu)$ with integer numbers $m \neq n, \lambda \neq \mu$?

Added: The answers assuming $m = n$ ($\lambda \neq \mu$) are also appreciated.

A closed form expression is provided in the following paper.

SV Amari, RB Misra, Closed-form expressions for distribution of sum of exponential random variables, IEEE Transactions on Reliability, 46 (4), 519-522.

see eqn 3.383(1) of book tables of integrals Gradshteyn and Ryzhik, it reads

$$\int\limits_0^u x^{\nu-1}(u-x)^{\mu-1}e^{\beta x}\,\mathrm{d}x=
B(\mu,\nu)\overline{u}^{\mu+\nu-1} {}_1F_1(\nu;\mu+\nu;\beta u)\\
[\operatorname{Re}\mu>0,\,\operatorname{Re}\nu>0].$$


lower bounding exponential functions I am struggling to understand constant lower bounds for some special form of exponential functions. I am aware of the following result:
$\lim_{n\to \infty}(1-\frac{x}{n})^n = e^{-x}$.

But how do I get constant lower bounds like

$$\left(1-\frac{2^x}{10n}\right)^{\Large\frac{n}{2^x}} \ge 2^{-\large\frac{1}{10}}$$

or

$$\left(1-\frac{2^x}{10n}\right)^{\Large\frac{n}{2^{(x-1)}}} \ge 4^{-\large\frac{1}{5}}$$

I have found the following estimation:

$4^{-x} \le 1-x \le 2^{-x}$

where the first inequality hold for $0 \lt x \le \frac{1}{2}$ and the second for $0 \lt x \lt 1$.

With some basic arithmetic rules we can get the bounds asked for in the question.

(For others it might also be helpful if someone could provide a proof)

You can find the minimum value of the function that is on the left in your expression

$$\left(1-\frac{2^x}{10n}\right)^{\Large\frac{n}{2^x}} \ge 2^{-\large\frac{1}{10}}$$

Remember that to find the minimum value only you have that to find the derivate and to equal it to 0


For prime $p$, normal subgroups of $SL(2, \mathbb Z/p\mathbb Z)$ remains normal in $GL(2, \mathbb Z/p\mathbb Z)$? Let $p$ be a prime. If $G$ is a normal subgroup of $SL(2, \mathbb Z/p\mathbb Z)$ , then is $G$ also normal in $GL(2, \mathbb Z/p\mathbb Z)$

For $p\ge5$, $SL(2,p)/\{\pm I\}$ is simple, so the only nontrivial
normal subgroup of $SL(2,p)$ is $\{\pm I\}$ which is normal in $GL(2,p)$.
You'll then have to treat the cases $p=2$ and $p=3$ separately.

Note the matrix (with $a\neq 0$)
$$m_a=\left[ {\begin{array}{cc}
a &0\\
0 &1
\end{array}}\right]$$
has determinant $a$ and inverse
$$\left[ {\begin{array}{cc}
a^{-1} &0\\
0 &1
\end{array}}\right]$$
in $GL(2, \mathbb Z/p\mathbb Z)$. Therefore every matrix $g\in GL(2, \mathbb Z/p\mathbb Z)$ can be written as $m_as$ (where $a=\det g$) for some $s\in SL(2, \mathbb Z/p\mathbb Z)$.

Now for all $s\in SL$, $sGs^{-1}=G$. Then for any $g\in GL$, let $g=m_as$, so $gGg^{-1}=m_asGs^{-1}m_a^{-1}=m_aGm_a^{-1}=G$ (somehow)

Can someone help me complete this proof :(


How to find coordinates of point in 3D given few points How to find coordinates of point in 3D given coordinates of few(more than 3 non linear/noncoplaner) points and distance from desired point using MATLAB?

suppose given five points are:
    [2.6643    1.4664    0.3019]
    [7.1360    1.4661    1.7333]
    [5.3205   -3.1543    0.5792]
    [3.1621   -1.1489   -2.4490]
    [8.0418   -0.6020   -0.1050]

and their respective distances from point(x,y,z) are:
    0.0376,
    0.0418,
    0.1882,
    0.0753,
    0.0538

How can i find x, y and z using MATLAB?

Write the sum of the square distances as a function, then use one of its minimization routines.


How much set theory is necessary for serious logic? I'm currently studying logic at my university and I have been trying to squeeze in as much set theory on the side as possible. Considering that I am spending quite a lot of time studying set theory independently from my logic coursework I have begun to wonder if it would be a better to devote my extra time to just the logic. Should I expect to encounter necessary set theory for logic in the logic texts that I will use as I advance or is it necessary to study set theory as well as logic to develop a firm grasp of logic?

There's logic and logic! And when does it start counting as "serious"?

For a first math logic course on first-order logic (and I'm including here things like the completeness proof for "ordinary", i.e. countable, languages) you need no serious set theory.

To be sure, it is entirely  standard to use set theoretic notation and say e.g. that a model has a domain which is a set equipped with functions and relations where those are also identified with sets of tuples from the domain. But actually, this set-theoretic talk is doing little real work at the outset (and even can be positively misleading). For example, instead of saying that -- in a certain case -- the domain of quantification is the set (singular) of numbers, we can instead say simply that the quantifiers run over the numbers (plural), and so on.

But once you get into e.g. the elements of model theory (compactness, upward and downward L-S theorems and their applications, and a little more), then you'll need to know something about ordinals and cardinals and something about the axiom of choice: so this is where you'll begin to need to know a little set theory.

But how much? Not a lot at the outset. Take a look at the Teach Yourself Logic Guide you can download from http:// logicmatters.net/tyl. Just one of the elementary books mentioned at the beginning of the section on Beginning Set Theory will give you more than enough to get into the other parts of a first course in mathematical logic  [Of course, as Asaf says in his comments, as you get further into serious model theory you get more and more embroiled with serious set theory!]

You could say that set theory is logic - u have ontological axioms (like existence, pairing etc.) guaranteeing the existence of certain objects and comprehension essentially identifying formulas and sets. So what is the difference between
$$\{x|S(x)\}\text{ and }S(x)$$

or $\{x\in A|S(x)\}$ and $A(x) \land S(x)$?

The thing with set theory is the following: that what it axiomatizes - the idea of something "belongs to" or "is part of" something - and how it axiomatizes it appears to be very familiar and somehow natural to our intuition. This seems to suggest that set theory is the appropriate base for all mathematics.

But despite that natural feeling with set theory, that claim seems to lack any further justification. Indeed more and more people think that taking set theory in any case for granted as a base for certain theories (like quantum physics or quantum gravity) might be just not appropriate and unjustified and involving it anyway might be the reason for some severe difficulties we face with these theories.

This is one motivation for being interested also in more general concepts like topos theory or typed $\lambda$-calculus - interestingly also theoretical physicists (f.i. C. Isham).

..for more views beyond set theory and logic(= introductory lectures on formal languages and systems, grammars, (in-)completeness + correctness, compactnes, Herbrand, Löwenheim-Skolem, computability, model theory, etc. ?) on logic u could f.i. see

Barnes + Mack: An algebraic introduction to mathematical logic

C. Isham: `What is a Thing?': Topos Theory in the Foundations of Physics


Which of the following about $f(x)$ is correct? Let $f:\mathbb R \to \mathbb R$ be defined by
$$f(x) =     \left\{
\begin{array}{1l}
      \dfrac{\sin x}{x} & x \neq 0 \\
      1 & x = 0 \\
\end{array}
\right.$$
Then:

$f$ is not continuous.
$f$ is continuous but not differentiable.
$f$ is differentiable

  Solution:
  Clearly $f(x)$ is continuous at $x=0$ and $\lim_{x\to 0^{+}}\frac{\sin x}{x}=1=\lim_{x\to 0^{-}}\frac{\sin x}{x}$.

Now, about differentiability at $x=0$,
$\lim_{x\to 0^{+}}\frac{\frac{\sin h}{h}-1}{h}=\frac{\sin h}{h^2}-\frac{1}{h}$ and $\lim_{x\to 0^{-}}\frac{1-\frac{\sin (-h)}{-h}}{h}=-\frac{\sin h}{h^2}+\frac{1}{h}$ .
Now both these limits should go to $0$ as $f'(0)=0$. But they are not going to zero. Hence, the function is continuous but not differentiable.

:Option (3) is correct

$\lim_{h\to 0^{+}}\frac{\frac{\sin h}{h}-1}{h}$ $(\frac{0}{0}) $ form.
Now use L'Hospital's rule we get
$\lim_{h \to 0^{+}}\frac{\frac{h\cos h-\sin h}{h^2}}{1}
 =\lim_{h \to 0^{+}}\frac{h\cos h-\sin h}{h^2}(\frac{0}{0})form$

again using L'Hospital's rule,
$\lim_{h \to 0^+}(-\frac{sinh}{2})=0$

Similarly using L'Hospital's rule $ \lim_{h\to 0^{-}}\frac{\frac{\sin h}{h}-1}{h}(\frac{0}{0} form)=0$
Thus $$\lim_{h\to 0}\frac{\frac{\sin h}{h}-1}{h}=0$$
Thus when $x\ne 0,f'(x)=\frac{x\cos {x}-\sin {x}}{x^2}$
And when $x=0 ,f'(x)=0$

Actually $f$ is infinitely differentiable

Your limit is right $ \dfrac{\sin(h)}{h^2}=\frac{h-h^3/6}{h^2}+o(h) $ developing $\sin$ asyptotically at second order so $\dfrac{\sin(h)}{h^2}-\dfrac{1}{h}=-h/6+o(h)$ Actually you're dealing with $sinc$ (say cardinal sinus function). $f$ would by infinitely differentiable. In order to show it you can develop it into  a serie.

$$ \forall x\in \mathbb{R}\ , \  f(x)=\sum_{n=0}^\infty \dfrac{(-1)^nx^{2n}}{(2n+1)!}$$

so  $f$ is infinitely differentiable on its circle of convergence  (here $\mathbb{R}$  because the radius of convergence equal to $\infty$)


If $2^{2013}-2^{2012}-2^{2011}+2^{2010} = k \cdot 2^{2010}$, find $k$ Hey this is just a question i was having fun with but couldnt solve for some reason. Would love if you can help me solve it thankyou!:

If $2^{2013}-2^{2012}-2^{2011}+2^{2010} = k\times2^{2010}$.  find k

Well there is of course the option of simplifying via a calculator but if you want to do it by hand, first I would pull out the factor of $2^{2010}$ out of the left hand side to get
$$2^{2010}(2^3-2^2-2+1)=k\cdot 2^{2010}$$
Then simplify and divide both sides by $2^{2010}$ to get
$$\frac{2^{2010}\cdot 3}{2^{2010}}=k$$
Which yields
$$3=k$$


Injective or one-to-one? What is the difference? What is the difference between the terms 'injective' and 'one-to-one', 'surjective' and 'onto', and 'bijective' and 'isomorphic'?

Let $f:A\to B$

Injective/one-to-one: $f(x)=f(y)\implies x=y$

Surjective/onto: for all $b\in B$, there is an $x\in A$ with $f(x)=b$ (basically, all elements of the codomain are represented by $f$)

Bijective: $f$ is both injective and surjective

Isomorphism: a bijective map between two algebraic structures

Injective and Surjective has been given already, the differens between isomorphism and bijection is that for a bijection $f$ versus an isomorphism $\phi$
we have that $\phi(a\cdot b)=\phi(a)\cdot\phi(b)$ while for for $f$ do we not have it, for a bijection we can have $f(a\cdot b)\neq f(a)\cdot f(b)$

Examples of them are $f:R_+\to R_+$ where $f(x)=x^2$, this is a bijection but
$$f(a+b)=(a+b)^2 =a^2+2ab+b^2 \neq a^2+b^2 = f(a) + f(b)$$

Let $\phi:Z\to 2Z$ be an isomorphism with $\phi(a)=2a$ then
$$\phi(a+b)=2(a+b)=2a+2b=\phi(a)+\phi(b)$$


Finding $\lim\limits_{x\to\infty}x^{3/2}\cdot\sin\frac1x$ How would I go about solving this problem? Not really sure where to start.

Put $y = \dfrac{1}{x} \implies x = \dfrac{1}{y}\implies L = \displaystyle \lim_{y \to 0} \dfrac{\sin y}{y\sqrt{y}}= \infty$

Let $h=\frac{1}{x}$

As $x\to \infty ,h\to 0$.

So $x^{\frac{3}{2}}{\sin{\frac{1}{x}}}=\dfrac{\sin h}{h^{\frac{3}{2}}}$

Put $h^{\frac{3}{2}}=t\implies h=t^{\frac{2}{3}}$

So $\dfrac{\sin h}{h^{\frac{3}{2}}}=\dfrac{\sin t^{\frac{2}{3}}}{t}=(\dfrac{0}{0})\text{as} t\to 0$

Applying L'Hospital's Rule ,we get $\dfrac{\cos t^{\frac{2}{3}}\times\frac{2}{3}}{t^{\frac{1}{3}}}$

which tends to $-\infty $ as $t\to 0^{-}$ and tends to $\infty $ as $t\to 0^{+}$

So the limit does not exist.


Prove that $\varlimsup_{n\to\infty}(\max\{x_n,y_n\})=\max\{\varlimsup_{n\to\infty}x_n,\,\varlimsup_{n\to\infty}y_n\}$ Prove, that $\varlimsup_{n \to \infty}(\max\{x_n, y_n\}) = \max\{\varlimsup_{n \to \infty} x_n,\, \varlimsup_{n \to \infty} y_n\}$.
Let $(x_n)$ and $(y_n)$ be bounded number sequences. Does also the following equality always stand $\varlimsup_{n \to \infty} (\min\{x_n, y_n\}) = \min\{\varlimsup_{n \to \infty} x_n, \,\varlimsup_{n \to \infty}y_n\}$?
In the proof, should I use upper limit monotonicity and/or upper limit description with partial sequences? Any tips or advice? I'm a little stuck and don't know where to begin.

You can write the max-function as $\frac{x_{n}+y_{n}+|x_{n}-y_{n}|}{2}$ . Show continuity and then u can pull limes into the max-function.


Is this a paradox in the Cantor set? In a previous question here Can we define the Cantor Set in this way?

we defined a family of sets $ \left\{ C_0,C_1,C_2,C_3,\dots \right\}$

We can call this set $S_1$ , where the values of these elements is

$C_0 = \left\{ 0.0   \right\}$

$C_1 = \left\{ 0.0 , 0.2  \right\}$

$C_2 = \left\{ 0.0 , 0.2 ,0.02 , 0.22 \right\}$

$C_3 = \left\{ 0.0 , 0.2 ,0.02 , 0.22 ,0.002 , 0.202 ,0.022 , 0.222 \right\}$

and so on...

We can then take the union of $\bigcup S_1$ and get some countable set $X$

We will define a set  $S_2$ to be all the members of  $S_1$ that are not redundant.

$C_0$ is redundant because it can be removed from $S_1$ and the union will still be the same.

In fact all the elements of $S_1$ are redundant so our set

$\bigcup S_2 = \left\{ \not C_0,\not C_1,\not C_2,\not C_3,\dots   \right\}  = \varnothing \neq X \to$ Contradiction

We removed all the sets from the union that did not contribute any information, and all the information disappeared.

What went wrong here?

First, note that this isn't really a question about the Cantor set, just about the combinatorics of infinite sets in general.

E.g. we can recast it in terms of sets of natural numbers: let $A_n=\{1, 2, 3, ..., n\}$, and think about $\bigcup A_n$. Or, even easier, we could just take $B_n=\{0\}$ (that's not a typo) and think about $\bigcup B_n$.

Now as to the answer to your question, nothing went wrong except your intuition for how infinite sets behave. Even though removing any finite number of the $A_n$s leaves the union of the rest unchanged, that's no reason to believe that removing infinitely many of the $A_n$s should do the same. There are lots of times we see a "phase transition" between finite behavior and infinite behavior - e.g. changing finitely many terms of a divergent series to $0$ leaves the result divergent, but if you change infinitely many terms to $0$ it might converge.

Casting this in terms of the set of natural numbers, there seems to be no contradiction if $ \mathbb{N} \in S_1 $.

Let:$$  S_1 =  \left\{ C_m =  \left\{ n \in \mathbb{N} : n \le m  \right\} : m \in \mathbb{N} \right\} $$
Then:$$C_0=  \left\{ 0 \right\}$$

$$C_1=  \left\{ 0,1 \right\}$$

$$C_2=  \left\{ 0,1,2 \right\}$$

$$\dots$$

$$ S_1 = \left\{ C_0,C_1,C_2,C_3,\dots , \mathbb{N} \right\} $$

$$ \mathbb{N} \in S_1 $$

$$\bigcup S_1 = \mathbb{N}  $$

Let:$$ S_2 = \left\{ C_n\not\subset (X \in  S_1) : n \in \mathbb{N}  \right\} $$

Then:$$ S_2 = \left\{ \not C_0,\not C_1,\not C_2,\not C_3,\dots , \mathbb{N} \right\} $$

$$S_2 = \left\{\mathbb{N} \right\} $$

$$\bigcup S_2 = \mathbb{N}  $$

It only gets ugly if $ \mathbb{N} \notin S_1 $


Determining if $973$ is prime Without a calculator, determine if $973$ is prime or not

I was given this question to solve. I know $973$ is not prime. I was told a strategy to solve whether a number is prime or not is to test all the numbers less than the square root of $973$

So I would have to test till $32$

and i find $1,7,139$ and $973$ are factors of this number.  Basically, what I want to find out is are there any other strategies to solve this question? then i wouldn't have to check till 1-32 to see if any of the numbers are factors.

If you're good at arithmetic, you can try this without a calculator $:)$

Wilson's theorem:

$p$ prime $\iff$ $(p-1)! \equiv -1 \pmod p$.

Else, pocklington's test could be fast.


What does apostrophe as a suffix denote? I was just curious as to what "$'$" denotes; i.e. $x' = y$, as in $x'(t) = x(t)$ which has the solution $x(t) = c_1\;e^t$.

I've found out that it has something to do with differential equations, but I can't seem to find any information specifically on "$x'$".

If someone could provide a source to such information, it would be much appreciated.

Typically, $x'$ refers to some derivative of $x$ with respect to a given variable. It is often used in contexts where the derivative being taken is clear, for ease of notation.

In the equation you list, for example, $x'(t) = x(t)$ is the same thing as writing $\frac{d}{dt}x(t) = x(t)$.

It is a symbol of derivative, it means $\displaystyle \frac{dx}{dt} = y$. You can find the solution by separation of variables technique.


Duality principle in boolean algebra All the definitions I came across so far stated, that if a statement is true, then also its dual statement is true and this dual statement is obtained by changing + for ., 0 for 1 and vice versa.

However when I say 1+1, whose dual statement according to the above is 0.0, I get opposite results, that is:

1 + 1 = 1
0 . 0 = 0

How should I understand this duality principle ?

"$1 + 1 = 1$" is a statement (a boolean statement, in fact), and indeed, $1 + 1 = 1$ happens to be a true statement.

Likewise, the entire statement "$0 \cdot 0 = 0$" is a true statement, since $0 \cdot 0$ correctly evaluates to false: and this is exactly what "$0 \cdot 0 = 0$" asserts, so it is a correct (true) statement about the falsity of $0 \cdot 0$.

The duality principle ensures that "if we exchange every symbol by its dual in a formula, we get the dual result".

Everywhere we see 1, change to 0.
Everywhere we see 0, change to 1.
Similarly, + to $\cdot$, and $\cdot$ to +.

More examples:

(a) 0 . 1 = 0: is a true statement asserting that "false and true evaluates to false"

(b) 1 + 0 = 1: is the dual of (a): it is a true statement asserting that "true or false evaluates true."

(c) 1 . 1 = 1: it is a true statement asserting that "true and true evaluates to true".

(d) 0 + 0 = 0: (d) is the dual of (c): it is a true statement asserting, correctly, that "false or false evaluates to false".

In my humble opinion, it is more helpful to describe the duality relationship in terms of variables. Example:
Expression: x + x = x
Dual of expression: x * x = x
x could be 1 or 0. In all iterations, the expressions are correct.
Good Luck!


Fail time with joint density $f_{X,Y}(x,y)=6e^{-x}e^{-2y}, \quad 0<x<y< \infty$ The following is the problem I am solving for.

Let $X$ and $Y$ be the fail time for a machine with two components.  The component $Y$ will start working if and only if component $X$ fails.  The machine fails when component $Y$ fails as well.  Given the joint density function
$$f_{X,Y}(x,y)=6e^{-x}e^{-2y}, \quad 0<x<y< \infty$$
find the expected time of failure of the machine.

My method was to find
$$\int_0^{\infty} \int_x^{\infty} (x+y)6e^{-x}e^{-2y} dydx$$
which ended up being equal to $ 7 \over 6$.
However, the answer was supposedly approximately 0.83.
I am not quite sure how to solve this problem.
Can someone explain me what is going on?

Hint:

The answer is not approximately 0.83, but $\frac{5}{6}$.  Use the following formulation  $$\int_{0}^{\infty} \int_{0}^{y} (y)(6e^{-x}e^{-2y})dxdy$$.  A tedious integration by parts and you will land up with the answer $\frac{5}{6}$


How to solve the differential equation $y'=y(1-y)$. Up until now, we simply rearranged and integrated both sides, so
$$y'=y(1-y)$$
$$\frac{dy}{dx}=y(1-y)$$
$$\frac{dy}{y(1-y)}=dx$$
$$\int\frac{dy}{y(1-y)}=\int dx$$
With partial fraction decomposition one gets
$$\int\frac{dy}{y} +\int\frac{dy}{(1-y)}=\int dx$$
$$\log|y| + C_1 + \log|1-y|+C_2=x+C_3$$
$$\log|y| + \log|1-y|=x+C_4$$
$$|y||1-y|=C_5e^x$$
How do I continue from here?

Edit: It should be
$$\int\frac{dy}{(1-y)}=-\log|1-y|+C$$

You have $|y||1-y|=C_5e^x$, and $C_5$ must be positive because it's $e$ raised to a power.  This implies $y(1-y)=C_6 e^x$, where $C_6$ need not be positive.

However, when you divided by $y(1-y)$, you tacitly assumed $y\ne0$ and $1-y\ne 0$.  That means you need to check separately to see if there are solutions consistent with those.

However: You neglected the chain rule and should have instead
$$
\frac y{1-y} = C_6 e^x.
$$
That implies:
$$
y = (1-y)C_6e^x
$$
$$
y = C_6e^x - yC_6 e^x
$$
$$
y+yC_6 e^x = C_6 e^x
$$
$$
y(1+C_6 e^x) = C_6e^x
$$
$$
y = \text{etc.}\cdots
$$

A hint: notice that $y(x) = 1$ and $y(x) = 0$ are constant solutions to the DE. Existence and uniqueness then implies that no solution can ever cross these values.

Using this, you could simplify all those absolute values, considering each case... Do you think you can take it from here?


Is it true that any set which contains elements with contradictory properties is the empty set? Here's what I've "discovered" but can't verify.
Let's take a set $A$ defined as "the set of all $x$ which don't belong to $A$".
Obviously this definition wouldn't hold in particular axiomatic systems, but it would in others.
Let's analyze this in NST. $X\in A$ iff $X\not\in A$.
This is a logical contradiction, but it still doesn't prove that $A$ doesn't exist. In fact, I would argue that $A$ is the empty set.
If put all of the elements of $A$ into the empty set, we get $(X∈A \text { iff } X∉A) \text { iff } X\in \emptyset$.
This is a tautology for whatever $X$ and $A$. Now, if we switch two of the atomic sentences, we get $X\not\in A \text { iff } (X\in A \text { iff } X\in\emptyset$). We know this to also always be true, which means that either $A$ doesn't exist or $A$ is the empty set.
Therefore, the empty set does contain every element belonging to any set it doesn't belong to.
Is there some error in my proof?

The statement $$(*)\quad(X∈A \text { iff } X∉A) \text { iff } X\in \emptyset$$ does not mean what you seem to think it means (though I'm not sure exactly what you think it means).  If $A=\{X:X\not\in A\}$, then $(X∈A \text { iff } X∉A)$ must be true for all $X$.  When you assert $(*)$ you are asserting that $(X∈A \text { iff } X∉A)$ is true if $X\in\emptyset$, but also that it is false if $X\not\in\emptyset$.  So if $A$ actually is equal to $\{X:X\not\in A\}$, then this statement (with an implicit universal quantifier on $X$) is false, since $(X∈A \text { iff } X∉A)$ would be true for all $X$, even $X$ such that $X\not\in\emptyset$.
Again, I'm not entirely sure what your train of thought is, but it sounds like you are confusing $$A=\{X:X\not\in A\}$$ with $$A=\{X:X∈A \text { iff } X∉A\}.$$  If $A$ was supposed to satisfy the latter equation, then $A=\emptyset$ would be equivalent to $(*)$ (and indeed, your argument then correctly shows that $A=\emptyset$ satisfies the latter equation).  But the first equation is quite different.

NO. If we assume $\exists x: [x\in A \land [P(x) \land \neg P(x)]]$ we can easily obtain a contradiction. Therefore, such a set $A$ cannot exist.
EDIT: In your example, if we assume $\exists x: [x\in A \land [x\in A \iff x \notin A]]$ we can similary obtain a contradiction. Therefore, in this case, too, such a set $A$ cannot exist.
BTW, when you say that a set contains any elements, you rule out that it is empty.
EDIT: Also, if you assume $x\in A \iff x\notin A$, this is already a contradiction. We can only conclude that there cannot exist such a set $A$ and element/object $x$.
We can, however, define an empty set $E$ given set A as follows:
$\forall x: [x \in E \iff x \in A \land x \notin A]$


$\cos(\theta_n) \to \cos(\theta)$ and $\sin(\theta_n) \to \sin(\theta)$. How to show that $\theta_n \to \theta$? $\cos(\theta_n) \to \cos(\theta)$ and $\sin(\theta_n) \to \sin(\theta)$ where $\theta_n$ , $\theta \in (-\pi, \pi)$.

How to show that $\theta_n \to \theta$ ?

Can I use the continuity of $\cos^{-1}$ or $\sin^{-1}$ (only one of them) to get the result?

If not, then how to proceed?

Edit:

As we know, range of $\arcsin $ is $[-\pi/2, \pi/2]$, we can use continuity of $\arcsin $ to get $\theta_n \to \theta$ when they belongs to  $[-\pi/2, \pi/2]$, in other cases use continuity of $\arccos$.

Is this argument correct?

Consider the map $f \colon (-\pi,\pi) \rightarrow \mathbb{R}^2$ given by $f(\theta) = (\cos(\theta), \sin(\theta))$. It maps the interval $(-\pi,\pi)$ bijectively onto the the unit circle minus the point $(-1,0)$ and you can explicitly write the inverse as

$$ f^{-1}(x,y)= 2 \arctan \left(  \frac{y}{1 + x} \right) $$

(see the entry on atan2 in wikipedia).

From this expression, it is clear that $f^{-1}$ is continuous and so if $f(\theta_n) \to (\cos \theta, \sin \theta)$ then
$$f^{-1}(f(\theta_n)) = \theta_n \to f^{-1}(f(\theta)) = \theta. $$

Observe that any convergent subsequence $\{\theta_{n_{k}}\}_{k=1}^{\infty}$ satisfies $\theta_{n_{k}}\rightarrow \theta$ as $k\rightarrow \infty$.

Thus, as the limit superior and the limit inferior are the same, $\theta_{n}\rightarrow \theta$ as $n\rightarrow \infty$.


Given real numbers: define integers? I have only a basic understanding of mathematics, and I was wondering and could not find a satisfying answer to the following:

  Integer numbers are just special cases (a subset) of real numbers. Imagine a world where you know only real numbers. How are integers defined using mathematical operations?

Knowing only the set of complex numbers $a + bi$, I could define real numbers as complex numbers where $b = 0$. Knowing only the set of real numbers, I would have no idea how to define the set of integer numbers.

While searching for an answer, most definitions of integer numbers talk about real numbers that don't have a fractional part in their notation. Although correct, this talks about notation, assumes that we know about integers already (the part left of the decimal separator), and it does not use any mathematical operations for the definition. Do we even know what integer numbers are, mathematically speaking?

There are several ways to interpret this question.

The naive way would be simply to give some sort of definition to the set of integers. This is not very difficult, we can recognize $1$ in the real numbers, because it is the unique number $x$ such that $r\cdot x=r$ for all $r\in\mathbb R$.

Now consider the set obtained by repeatedly adding $1$ to itself, or subtracting $1$ from itself. Namely $\{0,1,-1,1+1,-1-1,1+1+1,-1-1-1,\ldots\}$. One can show that this set is indeed the integers. Every integer is a finite summation of $1$ or the additive inverse of such set.

One could also define, like Nameless did, what is being an inductive set, then define $\mathbb N$ as the least inductive set, and $\mathbb Z$ as the least set containing $\mathbb N$ and closed under addition and subtraction.

However one could also interpret this question by asking "Is the set $\mathbb Z$ first-order definable in $\mathbb R$ in the language of ordered fields?", namely if you live inside the real numbers, is there some formula in the language $0,1,+,\cdot,<$ such that only integers satisfy it?

The answer to this question is negative. The integers are not first-order definable in $\mathbb R$. This is a nontrivial result in model theory. But it is important to note it, because it is a perfectly valid interpretation of the question, and it results in a completely different answer than the above one.

In particular it helps understanding first-order definability vs. second-order definability, and internal- vs. external-definability.

I am adding some more to the answer because of a recent comment by the OP to the original question. I feel that some clarification is needed to my answer here.

First we need to understand what does "define" mean from a logic-oriented point of view. It means that there is a language such that the real numbers interpret the language in a coherent way, and there is a formula in one free variable which is true if and only if we plug an integer into it.

For example we cannot define $\mathbb R$ within $\mathbb C$ if we only have $0,1,+,\times$, but we can do that if we also have the conjugation map in our vocabulary - as shown in the question.

Naively speaking, when we approach to mathematics we may think that everything is available to us, which is true to some extent. But when we want to talk about logic, and in particular first-order logic, then we need to first understand that only things within a particular language are available to us and we cannot expect people to guess what this language is if we don't specify that.

This question did not specify the language, which makes it not unreasonable to think that we are talking about the real numbers in the language of ordered fields. So in our language we only have $0,1,+,\times,<$ (and equality, we always have equality). In this language we cannot define the integers within the real numbers with a first-order formula.

Ah, but what is a first-order formula? Well, generally in a formula we can have variables which are objects of our structure (in this case, real numbers) and we can have sets of real numbers and we can have sets of sets of real numbers and so on. First-order formulas allow us only to quantify over objects, that is real numbers. So variables in first-order logic are elements of the universe, which in our case means simply real numbers. Second-order logic would allow us to quantify over sets (of real numbers) as well, but not sets of sets of real numbers, and so on.

So for example, we can write a definition for $2$ using a first-order formula, e.g. $x=1+1$. There is a unique element which satisfies this property and this is $2$. And we can write the definition of an inductive set using a second-order formula, $0\in A\land\forall x(x\in A\rightarrow x+1\in A)$.

But as it turns out we cannot express the property of being an inductive set (and we certainly cannot express the property of being the minimal inductive set) in first-order logic when we restrict ourselves only to the real numbers as an ordered field. The proof, as I remarked, is not trivial.

The comment I referred to says:

  @WillieWong I don't know the real number field: to me real numbers form a line. – Virtlink

Which gives yet another way to interpret this approach. We can consider the real numbers simply as an ordered set. We ignore the fact it is a field, and we simply look at the order.

But this language has even less expressive powers than that of an ordered field, for example we cannot even define the addition and multiplication. In fact we can't even define $0$ and $1$ in this language. We only have the order to work with, and that's really not much to work with.

It is much easier to show that simply as an ordered set all the results about undefinability hold, I won't get into this but I'll just point out that definability is "immune to automorphisms", and $\mathbb R$ has plenty of automorphisms which preserve the order and move every element.

Beyond that one runs into philosophy of mathematics pretty quick. What are the real numbers? Are they sets? Are they any structure which interpret a particular language in a particular way? Do we construct the integers from the real numbers or do we construct the real numbers from the integers? (First we have to construct the rational numbers, of course.)

Those are interesting questions, but essentially moot if one simply wishes to talk about first-order definability in a particular language or another. But if one is approaching this in a "naive" way which allows higher order quantification, and the usage of any function we know about then the answer becomes quite simple (although it is possible to run into circularity if one is not careful enough).

I hope this explains, amongst other things, why I began my answer with "There are several ways to interpret this question". We simply can see the real numbers as different structure in different languages, and we may or may not allow formulas to use sets of real numbers. In each of these interpretation of the question we may have a different answer, and different reasons for this answer to be true!

(I should stop writing this monograph now, if you're read this far - you're a braver [wo]man than I am!)

But wait, there's more!

True Definition of the Real Numbers
FO-definability of the integers in (Q, +, <)
What is definability in First-Order Logic?

How about the values (Image) of
$$\lfloor x\rfloor:=x-\frac{1}{2}+\frac{1}{\pi}\sum_{k=1}^\infty\frac{\sin(2\pi k x)}{k}$$

But this is nonsense; we sum over the positive integers, and such, we can just define the integers as
$$x_0:=0\\x_{k+1}=x_k+1\\
x_{-k}=-x_k$$


Let $f:[a,b]\to\mathbb R$ be Riemann integrable and $f>0$. Prove that $\int_a^bf>0$. (Without Measure theory) I've been struggling with this for a while, and I have a couple of leads that kind of got me nowhere:

At first I thought that if $f$ is continuous somewhere then the integral will be $>0$. So, if the integral was $0$ then that would mean it would need to be nowhere continuous. That seemed unlikely to me, but I couldn't prove the existence of a point at which it is continuous.

For the integral to be $0$ it would necessitate that for any sub interval of $[a,b]$ the function's infimum would have to be $0$. Also seems weird for $f>0$. Again, got me nowhere.

I should mention that I'm aware that it's possible to prove this by defining "Measure" and all that, but I don't want to go there. I'm wondering if there are more elementary tools to show the above.

Thank you!

Sketch: For a bounded function $f$, we define the oscillation of $f$ at $a$ by $$\mathrm{osc}f(a) = \limsup_{x\to a} f(x) - \liminf_{x\to a} f(x). $$ Then it is easy to show that $f$ is continuous at $a$ if and only if $\mathrm{osc}f(a) = 0$.

Now let us given $\epsilon > 0$ and $\delta > 0$. Then we can find a partition $P$ such that $U(f,P)-L(f,P) < \epsilon$. Thus the set $S = S(\epsilon, \delta)$ denotes the collections of subintervals $I$, formed by the points of $P$, where $\sup_{I} f - \inf_{I} f > \delta$ satisfies

$$ \epsilon > U(f,P)-L(f,P) > \sum_{I\in S} \delta |I| $$

Now, choose $(\epsilon_n, \delta_n)$ such that $\delta_n \downarrow 0$ and

$$ l := \sum_{n=1}^{\infty} \frac{\epsilon_n}{\delta_n} < b-a. $$

Choose $r >1$ such that $rl < b-a$. For each closed interval $I \in S(\epsilon_n, \delta_n)$, we choose an open interval $J$ containing $I$ such that $|J| = r|I|$. Finally, let $\mathcal{U}$ be the family formed by collecting all such open intervals $J$. If $f$ has no point of continuity, then each $x \in [a, b]$ is a point where $f$ has positive oscillation and hence lies in some $J\in \mathcal{U}$. Thus $\mathcal{U}$ is an open cover of $[a,b]$, and hence it has a finite subcover. But the sum of the length of the open intervals in that subcover cannot exceed $rl < b-a$, a contradiction! Therefore $f$ must have a point of continuity.

Of course, this proof also hides the idea of measure, though indirectly and elusively.

What does it mean for it to be Riemann integrable? Why $should$ it be positive?  The Riemann sum is defined as
$$\sum_{i=0}^{n-1} f(t_i)(x_{i+1}-x_i) : t_i \in (x_{i+1},x_i)$$
over a partition $a=x_0 <x_1 < x_2 < \cdots < x_n = b$

and what is the Riemann integral?  In essence, we let $n\to \infty$ (our partition becomes very fine).  So essentially your question is logically equivalent to asking why the above sum is always positive, well that's because $f(t_i)>0$, so naturally the sum is too... and the integral version just gives us the best solution (but it will always be positive for the same reason).