Puzzle time - algebra edition

[Klaus]

We have all learned the following equations about numbers in school.

1. x + y = y + x
2. (x + y) + z = x + (y + z)
3. x · 1 = x
4. x · y = y · x
5. (x · y) · z = x · (y · z)
6. x · (y + z) = x · y + x ·z
7. 1^x = 1
8. x^1 = x
9. x^(y + z) = x^y · x^z
10. (x · y)^z = x^z · y^z
11. (x^y)^z = x^(y · z)
12. 0 + x = x
13. x·0 = 0
14. x^0 = 1

Now let's say the variables stand for sets and not numbers. Can you give an interpretation of "+","·" and "^", "1" and "0" such that the equations hold, when "=" stands for "there exists a bijection between left hand side and right hand side"?

Now let's say the variables stand for logical propositions and not numbers. Can you give an interpretation of "+","·" and "^", "1" and "0" such that the equations hold, when "=" stands for logical equivalence?

In both tasks, can you find a "2" such that "1+1=2"?

[bachophile]

Link to video

[taiwan_girl]

@bachophile I am glad I am not the only one who did not understand. 555

[George K]

@taiwan_girl said in Puzzle time - algebra edition:

@bachophile I am glad I am not the only one who did not understand. 555

That makes at least three of us.

[Klaus]

Hey, what are you all doing in this thread? I was hoping that the title would scare off math-challenged posters

[jon-nyc]

Do the operator mappings have to be unique?

[bachophile]

Jon chimes in to let us know he understands the question.

Right....tell the truth Klaus, u just made up the word bijection.

Sounds filthy to me.

[Mik]

Needs pronouns.

[Klaus]

@jon-nyc said in Puzzle time - algebra edition:

Do the operator mappings have to be unique?

I'm not sure what you have in mind, but I mean that, for instance, "+" is interpreted as an operation on sets (for the first task) or logical propositions (for the second task). For instance, you could map "+" to "intersection" (that choice would satisfy equations 1 and 2, but it would be hard to make it work for the other equations, too).

[Klaus]

Just as a little teaser, this puzzle illustrates the probably most beautiful result of 20th century mathematics/logic. A result that inspired the works of Field medalists and created whole new subfields in mathematics.

[jon-nyc]

Could you map two symbols to the same operator?

[Klaus]

Sure.

[Klaus]

OK, maybe not the right audience here for this kind of puzzle

Anyway, just for closure, here's a solution:

For sets:

the operator + is set union
· is set intersection
A^B is the set of functions from B to A
1 is any one-element set
0 is the empty set
2 is any two-element set

For logical propositions:

the operator + is disjunction
· is conjunction
A^B is the implication B -> A
1 is logical truth
0 is logical falsity
2 is also logical truth (because 2 = 1 + 1 = true OR true = true )

As usual for mathematics, I'll leave the verification of the equations from the original post as an exercise to the reader .

These mappings form the basis the so-called "Curry-Howard correspondence" and, more recently, "homotopy type theory", an influential modern attempt to "reboot" mathematics.

[jon-nyc]

@klaus said in Puzzle time - algebra edition:

A^B is the implication B -> A

I tried that and ruled it out (erroneously) thinking it failed 7