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0

Lets call the point you found $X$ Via Thales' theorem (https://en.wikipedia.org/wiki/Thales%27_theorem ) you can proof that that $\angle AXB $ and $\angle AXC $ are right angles (with bases $AB$ and $AC$ ) that means that $AX$ is an altitude line. see https://en.wikipedia.org/wiki/Altitude_%28triangle%29 so you could call point X the foot of the ...


1

The only holomorphic functions that have this property are monomials $cz^n$. Indeed, the property implies (by continuity) that $|x|\le |y|\implies |f(x)|\le |f(y)|$, which in turn yields $|x|=|y|\implies |f(x)|=|f(y)|$. So, for every unimodular constant $\zeta $ the function $f(\zeta z)/f(z)$ has constant modulus, hence is constant. So, $f(\zeta z)\equiv ...


3

Formal Defintion: Given sets $A$ and $B$, the Cartesian product of $A$ and $B$, denoted $A × B$ and read “$A\: \mbox{cross} \: B$,” is the set of all ordered pairs $(a, b)$, where $a$ is in $A$ and $b$ is in $B$. Symbolically: $A × B = \left\{(a, b) \: | \:a \: \in \: A \: \mbox{and} \: b \in B\right\}$ Thus we have: $\mathbb{R} × \mathbb{R}$ is the set ...


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They do not seem incompatible to me since they talk about different types of 'totality'. The first definition takes 2 sets $X, Y$. While the second definition uses only one set. (It's a binary relation over one set, wikipedia speaks of endorelation) You could transform the first definition so that it uses one set: A relation $R \subset X\times X$ is ...


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Consider the relation $(n, 2n) \in R \lor (2n, n) \in R$. This would not be total if the $Y$ consisted only of the identical objects (members) in $X$.


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When you are talking about sets, and you say the sequence is "increasing" or "non-decreasing", it just means that you have the containment $A_{1} \subset A_{2} \subset A_{3} \subset \dots \subset A_{n} \subset A_{n + 1} \subset \dots$ (usually, these are all proper subsets if you say "increasing" while they would not all necessarily have to be proper if you ...


3

It means that they satisfy $A_n \subset A_{n+1}$. The limit is not a limit in the $\epsilon$-$\delta$ sense. It just means $\cup_{n=1}^\infty A_n$. For example, take $A_n = [0,n]$, then $\cup_{n=1}^\infty A_n = [0, \infty)$. From a probability perspective, there is a real limit associated with these nested sets. Suppose $p$ is the probability measure. ...


4

The word I've most commonly seen used for a point at which a function is continuous but not differentiable is "kink", as in: "The function $f(x) = |x|$ has a kink at the origin."


1

No, there is no such phrase. Simply reword your sentence: "Condition $Y$ ensure that $f$ is continuous but not smooth at $x_0$, which makes it fulfill $X$". No need for arcane terminology here.


3

Versine is $\displaystyle1 - \cos(\theta) $ or $ \displaystyle2\sin^2\left(\frac\theta2\right)$ 'versin' is the Versine here, 'ver' by itself is nothing. It is short for 'versed sine'. See here.


1

I'm fairly certain it's just pseudocode. If you were to make it pretty, it would look like $$\sum_{n,k}\frac{T(n,k)x^2y^k}{n!}=1+\ln\left(\sum_{n=0}^{\infty}{\frac{(4+y)^{\binom{n}{2}}x^n}{n!}}\right)$$ This is the exponential generating function for the sequence. The formula that is displayed on the site is actually ...


0

Of course not. This would be a useless notion (so there is no need for such terminology). Most functions are not surjective or injective (in the sense that if you take a "random" one, then you cannot expect it to be surjective or injective), mind you. We are mostly interested in the special class of mappings called injections and surjections, and for this ...


5

As far as I know, there isn't. The concept of a "non-surjective and non-injective function" just doesn't generally arise often enough to need a special term.


3

Yes, although you may find it more straightforward to prove the positive form of the statement: By definition, a set is countably infinite if it has a bijection with $\mathbb N$. If $X$ is countably infinite and has a bijection with $Y$, then composing gives us a bijection between $Y$ and $\mathbb N$, so $Y$ is also countably infinite.


3

No, $1$ is not a prime number, though human mathematicians were slow to recognize this fact. The most important reason that $1$ is not prime (or composite, for that matter) is that it is its own inverse; just the fact that it has an inverse in $\mathbb{Z}$ speaks volumes.


1

If you have $H = X\oplus Y$, then the projection onto $X$ is not well-defined without $Y$. The reason for this is that the splitting $H = X\oplus Y$ is not unique. That said, given $X$ there are many spaces $Y$ such that $H=X\oplus Y$. For $H=\mathbb R^2$, $X=span\pmatrix{1\\0}$, $Y_t=span\pmatrix{t \\1}$, we have $H=X\oplus Y_t$ for all $t\in\mathbb R$. ...


1

Let $m$ and $n$ be "perpendicular but skew" lines. I suppose that you need to introduce a line parallel to $m$, call it $m_1$, such that $m_1$ and $n$ are coplanar. If $m_1$ and $n$ are coplanar they can be perpendicular to each other. So, now we have $n \perp m_1$. So essentially $m$ is parallel to a line which is perpendicular to $n$. $m_1$ is coplanar ...


2

It is no abuse of notation to say that $F(f)$ and $G(f)$ are isomorphic: $F(f)$ is isomorphic to $G(f)$ in the arrow category of $D$.


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I would say that the maps are naturally equivalent. You're right that saying naturally isomorphic is kinda abusing the terminology but it doesn't sound monstrous to me. Whatever you choose you should just make sure to explain it the first time you use it.


0

The reason you want to specify "along $Y$" is that you may have to deal with multiple orthogonal projections at the same time, and then it is useful to have a way to speak of different projections without too much fuss. For example, this occurs in the Gram-Schmidt orthonormalization process, where you have to iteratively take a projection along a ...


1

This denotes the abbreviation of $f$. For a map $f:X\to Y$ and subsets $A \subset X$ and $f(A) \subset B \subset Y$ the abbreviation of $f$ is just the map from $A $ to $B$ defined by $x \mapsto f(x)$. Roughly speaking, it is just $f$ with a more restricted domain and codomain; but as a map also depends on its domain and codomain on should not continue to ...


0

After a rigorous analysis, I am coming up with my answer that is, they ($x$ and $y$) can be said proportional if $scale$ is changed. I found my answer as an outgrowth of study of relation between $Temperature$ and $Volume$. In Celsius scale (theoretically), the temperature at which volume become $Zero$ (however it is not true as gaseous molecules vibrates at ...


-2

The term that the textbook that we use for Algebra 1 is "contradiction" and the term used for an equation that has infinite solutions (4x + 2 =4x + 2) is called "identity"


1

It generally means that $\frac{A}B$ remains in some interval $(c,C)$ for $0<c<C$.


4

$1$ is not a prime number, but it's important to understand that the ancient Greeks saw numbers mainly as geometric constructs, whereas we see them as algebraic constructs. We must also remember that the Greeks thought of the primes as "first numbers," and that survives in our terminology. For example, let's say you have $n > 0$ square tiles all the same ...


1

They have to be able to be represented as $\frac{x}{y}=k$ or $x=ky$ Ratio must be constant. So in your hypothetical, not proportional.


0

A function that has exactly one minimum/maximum (or turning point, if we're speaking of discrete functions) and is otherwise monotonic is called unimodal. If we have two maxima which are also turning points, such as in $1 ,3, 2, 3, 1$, such a function could be called bimodal. This can clearly be generalized to arbitrarily many turning points.


-1

Dollars, employees and months are all units. Dollars per employee per month is also a unit. You can distinguish between these by calling "dollars" a base unit or a fundamental unit. Dollars per employee per month is a composite unit or a derived unit. Turning dollars per employee per month into, for instance, dollars per team per workday is performing a ...


0

Set of all integer is $\mathbb{Z=}${$0,\pm1,\pm2,\pm3, \ldots\ldots,\pm{1000 \ldots}$}. So just remove $0$ from $\mathbb{Z}$. Rest any of these infinite numbers can be used.


1

An integer is any whole number or its negative, e.g. ..., -2, -1, 0, 1, 2, ... A non-zero integer is any of these but 0. Your definition of a rational number is just a mathematically rigorous way of saying that a rational number is any fraction of whole numbers, possibly with negatives, and you can't have 0 in the denominator.


0

This is more of a scientific question than linguistic, but "dollars per employee per month" is a perfectly good unit. It is not fundamentally different from "miles per hour" or "metres per second squared" (a common unit of acceleration). Another similar usage of unit might be "dollars per person" or "dollars per capita", which are commonly used to measure ...


1

Given you have a person "John Doe", then to... his picture he is 'isomorphic' his brother he is 'equivalent' his twin brother he is 'equal' himself he is 'identical'


4

They have different types. "Equal" and "identical" take as input two elements of a set and return a truth value. They both mean the same thing, which is what you think they'd mean. For example, we can consider the set $\{ 1, 2, 3, \dots \}$ of natural numbers, and then $1 = 1$ is true, $1 = 2$ is false, and so forth. "Equivalent" takes as input two ...


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Convention may vary, but the following is, I guess, how most mathematicians would use these notions. Identical and equal are very often used synonymously. However, sometimes identical is meant to say that the two things are not just equal, but actually are syntactically equal. For instance, take $x=2$. The claim that $x^2=4$ is saying that $x^2$ and $4$ are ...


1

Equal means two entities are the same entity; equivalent means that two entities have the same EFFECT, in some sense. That is, when two things are same in some specific way, but not identical, they are said to be equivalent. (Identical is not really a math term, it is an English word used to convey the idea of exact sameness). See ...


1

In general it's a tricky question. For example is the natural number 2 equal to the real number 2? As sets, they are not equal. But as numbers, everyone would consider them equal. But then what's a number? As soon as you try to answer this question, you're down the rabbit hole and into category theory and philosophy. Barry Mazur wrote a famous essay on this ...


1

this mapping is a Riemann mapping, by the Riemann Mapping Theorem, The Riemann Mapping Theorem does not say how the map is called. It says that it exists. How to call it is a matter of conventions and definitions, not theorems. The convention I am familiar with is (as on Riemann mapping theorem wiki page): Riemann mapping = a holomorphic ...


1

If a graph $G$ is dense enough, then we can guarantee the crossing number is at least a certain value. Let $e$ be the number of edges in a graph $G$ of order $n$ with $e\ge 4n$, then $\operatorname{cr}(G)\ge\frac{e^3}{64n^2}$.


1

The crossing number is an invariant of a graph describing how far away from being planar the graph is. A graph is planar if and only if its crossing number is zero. A graph which always has at least one set of edges crossing when drawn in the plane has crossing number at least one, and so on. The crossing number inequality is a lower bound on the crossing ...


0

On p. 50 in Goldblatt's book Lectures on the Hyperreals, hyperreal numbers $b$ which are "limited but not infinitesimal" ($r<|b|<s$ for some $r,s \in \mathbb{R}^+$) are called appreciable.


0

Transfinite numbers are numbers that are "infinite" in the sense that they are larger than all finite numbers, yet not necessarily absolutely infinite. The term transfinite was coined by Georg Cantor, who wished to avoid some of the implications of the word infinite in connection with these objects, which were nevertheless not finite. Few contemporary ...


2

$x$ is in the domain of the function f. Edit: Just realised I'm not really answering your question. I've never had to refer to the relationship between x and y. In my classes, x is either referred to as the input or the argument of the function. However, I do believe that you can say that "$x$ is the pre-image of $y$".


5

You can call it a representative of the equivalence class.


0

I was able to obtain a copy of the paper referenced. This appears to be the original use of the FASS acronym. The authors define the two terms in question as such: Self-avoidance -- the segments of the curve do not touch nor intersect. Simplicity -- the curve can be drawn by a single stroke of the pen, without lifting the pen nor drawing any ...


2

It is usually called the free category on (over) the diagram, but the term 'induced by' is also clear. We can formalize abstract diagrams as directed graphs (i.e. quivers) with a set of commutativity conditions, which are just pairs of parallel paths. Then we have the usual free-forgetful adjunction between these abstract diagrams and categories.


2

I think for the Legendre question it should be "a on p". But I could be wrong.


0

A way to express this is that $n$ is in the annihilator of $A$, or the set containing $A$ to stay completely in line with the linked to resource, when you consider the set of all matrices involved as a $\mathbb{Z}$ module.


1

As was explained by Hurkyl, thin groupoids are known as setoids. They appear in algebraic geometry. Whereas algebraic stacks are categories fibered in groupoids, schemes are categories fibered in setoids. For example, the fiber of $\mathbb{P}^n$ over $S$ is the category of $(\mathcal{L},s_0,\dotsc,s_n)$, where $\mathcal{L}$ is a line bundle on $S$ and ...


2

Search engines these days know a lot. It certainly scares me when I search for "flag manifold" and receive results for "flag variety." However, they don't know everything. If you look at the Wikipedia article for combinatorics, it tells you that the terms that are in use are combinatorialist and combinatorist. These are also the only terms I myself have ...


2

You seem to be describing a setoid. Also, every such category is equivalent to a set (assuming the category is small); typically we only care about categories up to equivalence, so in such situations this notion doesn't give anything new. However, as an aside, I think setoid is a better notion than set for many purposes, even outside the context of ...



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