Puzzle time - algebra edition
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We have all learned the following equations about numbers in school.
- x + y = y + x
- (x + y) + z = x + (y + z)
- x · 1 = x
- x · y = y · x
- (x · y) · z = x · (y · z)
- x · (y + z) = x · y + x ·z
- 1^x = 1
- x^1 = x
- x^(y + z) = x^y · x^z
- (x · y)^z = x^z · y^z
- (x^y)^z = x^(y · z)
- 0 + x = x
- x·0 = 0
- 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"?
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@bachophile
I am glad I am not the only one who did not understand. 555 -
@bachophile
I am glad I am not the only one who did not understand. 555@taiwan_girl said in Puzzle time - algebra edition:
@bachophile
I am glad I am not the only one who did not understand. 555That makes at least three of us.
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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.
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@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).
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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 setFor 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.
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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 setFor 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.
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