# Tag Info

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.

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 ...

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.

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.

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.

2

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.

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 $D$.

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.

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 ...

1

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

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 ...

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 ...

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