#bijection

[cauchyfeelings]

Set seeking another set for a bijective relationship.

Must be invertible.

[lesbianslovebts]

I got excited about linguistics again after that last ask, so I want to elaborate on how I view translation as a mathematical mapping between two sets. Observe the following sentences written in Korean, Japanese, and English, respectively.

First, going from Korean to Japanese, you will notice that the function (translation) is bijective (word-for-word) because there exists a map between the two sets (sentences) where each element (word or word-part) in the first set (Korean) is paired with exactly one unique element of the second set (Japanese), each element of the second set is paired with exactly one unique element of the first set, and there are no repeats. Moreover, the mapping preserves the relations between elements in the sets and is, therefore, a homomorphism. In other words, the translation preserves the grammatical part of speech of each word or word-part (i.e., the direct object remains the direct object). Thus, the translation of this sentence from Korean to Japanese is a grammatical isomorphism. Obviously, the same can be said of the Japanese-to-Korean translation.

On the other hand, the translation from Japanese to English is not even a proper function because there exist no elements (word or word-parts) in the codomain (English) for two elements in the domain (Japanese): the topic particle は and the direct object particle を. Likewise, in the translation from English to Japanese, there exist no elements in the codomain for two elements in the domain: the definite article "the" and the possessive adjective "my." Additionally, there are two elements in the domain that map onto two elements each in the codomain: "went" to 行き and ました, and "with" to と and 一緒に. There is no function, let alone isomorphism, between English and Japanese. By extension, there is none between English and Korean, either.

In conclusion, neither English-Japanese-English nor English-Korean-English translation is as direct or complete as Korean-Japanese-Korean translation, which is a grammatical isomorphism, at least in the case of the sentence pictured above.

[isomorphismes]

Galileo observed that the set of positive integers can be put into 1‒1 correspondence with the set of square integers, even though the set of nonsquares seems more numerous than the squares. He deduced from this that "the attributes =, >, < are not applicable to infinite quantities".

Stan Wagon

[math-memes]

[jinglecats]

bijection replied to your post: hey.  why tf is it so difficult to find screen...

OBS (Open Broadcast Software) is free and provides decent quality. The settings are a little tricky to pin down though. It can also be used for streaming though.

ah man, it actually looks pretty promising so far :0  I’ll probably try it out tomorrow since im probly gonna go to bet but tysm for the suggestion

[enchantedsource]

0.9 Cantor Schroeder Berenstein pt.1 Showing that if A injects into B and B injects into A, then there exists a bijection between A and B This one can be a little tricky! Here is a gif illustrating what happens: http://enchantedconsole.tumblr.com/post/149574778527/enchantedconsole-contest-hey-everyone-i-am

[pbsdigitalstudios]

It turns out some infinities ARE bigger than other infinities.  *chuckles*

[matan-matika]

“same-sized” infinities

The idea of “two sets are the same size iff there is a bijection between them” leads to the interesting property that you can extend the idea of size to infinite sets.

Here are illustrations of how two infinite size sets can have the same size, some of which might seem surprising if you are only used to finite sizes.

Some sets I will use as examples are integers Z={...,-2,-1,0,1,2,...}, positive integers Z+={1,2,3...}, non-negative integers Z*={0,1,2,3...}, the non-negative even integers 2Z*={0,2,4,6,...}, the positive rational numbers Q={p/q : p,q are positive integers and q is not 0}, and positive real numbers R+, which include the irrationals.

Z+ and Z* are the same size

The function is f:Z*->Z+, f(x)=x+1

Z* and 2Z* are the same size

The function is f:Z*->2Z*, f(x)=2x

Z* and Z are the same size

The function is one where we split Z* into evens and odds, and then take the evens to the negatives and the odds to the positives

Q on the interval [1,infinity) and Q on the interval (0,1] are the same size

The function is f:Q[1,infinity)->Q(0,1], f(x)=1/x. (We can also use a similar function to show any interval of real numbers has the same size as the entire real number line)

Z* and Q* are the same size

This one is a bit trickier. We will show that Z* is the same size as the cartesian product Z*×Z*, and notice that Q* is a subset of Z*×Z*, so cannot be larger than it, but Q* contains Z*, so it cannot be smaller.

First, we split the cartesian product into lines based on the sum of the two numbers in each pair

Let t(n) be the nth triangular number, or the sum of all non-negative integers below and including it. Notice that t(p+q) is the number of pairs with sum below p+q, because each row where the sum is m has m-1 elements.

We construct a function f:Z*×Z*->Z*, f(p,q)=t(p+q)+p

R and R×R have the same size

First, we restrict R to the interval [0,1), which has the same size as R. Then, we realize that we can represent each real number by its decimal expansion, which has a Z+ sized expansion. We can split every expansion into even and odd number places, and make a new number in R×R restricted to [0,1)×[0,1) using the two expansions to make its two numbers.