it's nice to hear "we write this in this way, I have no idea why"; it eases my anxiety that I might have missed some important feature of the mathematics when I see something written a certain way that I am unfamiliar with, knowing that sometimes the reason something is described with a certain notation is rather arbitrary and just needs to be known rather than understood.
Tbh, a lot of things were created by some mathematicians many years ago and have just become standard notation. Sometimes, you won’t know who created the notation but it is just commonly used.
Great lecture! For anyone wondering: There is a small error in the equation at 08:18. If x is an n-th root of unity we have 1+x+x^2+..+x^{n-1} = 0 and not 1. (To see this multiply 1+x+x^2+..+x^{n-1} with x and observe what happens.)
Peculiarly enough, it is corrected in the next line, where the equation is divided by ζ³. This operation would by no means turn a 1 into a 0. It is good practice in lectures to scatter some minor errors across the material so as to test whether the audience is still attentive.
Loving this series so far. It has set off at a really nice pace At 26:28, I think it’s worth mentioning, that although an intuitionist would reject the argument as a proof that one of pi+e and pi*e are transcendental, they are not claiming that the mathematics being done is somehow incorrect. The point of view is simply that there is a meaningful difference between the statements: “It is not the case that both of pi+e and pi*e are algebraic”, and “Either pi+e is transcendental or pi*e is transcendental”. The argument given is intuitionistically valid as a proof for the former statement, but not the latter.
but surely if both pi+e and pi*e are not algebraic then it would be logical to conclude that AT LEAST one of them is transcendental (since a number is either algebraic or transcendental). Can you explain what the meaningful difference between the two are please?
@@20doctorwho09 one of the major ideas of "intuitionistic logic" is removing the "law of the excluded middle" and "law of double negation" as axioms (or even a couple of "Di'Morgan's Laws"), instead requiring there to be a certain proof of that. Apparently this is very useful for questions of constructability and computability when doing proofs. But if you have not proven the law of the excluded middle and law of double negation applies to the situation, NOT( NOT(pi+e is algebraic) AND NOT(pi*e is algebraic)) does not (yet) imply (NOT(pi+e is algebraic) OR NOT(pi*e is algebraic))
It is a bit strange and unintuitive and I kind of blinked when he said that too, but it comes from treating Q[X] as a distinct ring unto itself. (Q[X] being the ring of polynomials with coefficients in Q and in a single variable X). Q is contained as a subring of Q[X] by considering only constant polynomials of Q[X]. One can also construct rings as polynomials over Q in an arbitrary number of variables such as Q[X, Y]. I think the core of what he's saying is that X neither exists in Q, nor can you conjure a polynomial with coefficients in Q with X as a solution. Thus, I'm not really sure what the nature of Q[X] is in relation to Q. It's certainly not a traditional field extension, and in fact I realize after I've been writing that for it to be a field extension, the field extended to would have to be the field of rational functions in a variable X. It seems to exist as a non-algebraic but finite extension induced by adding a variable called X. It is perhaps not particularly meaningful, and perhaps incorrect to call X transcendental over Q, I suppose. It's a tough call because X is inherently a variable, as opposed to adjoining pi or some other transcendental number to Q. The resulting field would look essentially the same if we had no idea ahead of time of the numerical properties of the adjoined number if our extension were embedded into the complex numbers. Conversely, the only property we require of X is compatibility with the ring operations of Q. Edit: Hopefully this is my last edit, just realized I made a major mistake. Q[X] is an infinite dimension vector space over Q. I.e. I was wrong about it being a finite extension. Its dimension is infinite because no power of X can be written as a linear combination of other powers of X with coefficients in Q. I'm still not convinced that it's proper to say X is transcendental over Q, though.
@@MathOnNapkins Forgive me for replying to a 3W old comment. x is transcendental over Q because x is not a solution to a polynomial over Q. One cannot solve for a 0 of a polynomial and obtain x. The polynomial x has solution 0. I believe the following definition makes things easier to understand (Paolo Aluffi Algebra Chapter 0): Consider [F(a) : F] (the dimension of the vector space F(a) over F. "a" is algebraic if the dimension is finite. Transcendental otherwise. F[x] is an infinite dimensional vector space with basis {1, x, x^2, ...} as you've pointed out. Saying the same thing again but closer to the definition, [F[x] : F] is infinite hence "x" is transcendental. How this relates to polynomials is the following, suppose F(a) is a finite extension (i.e. the dimension of the vector space is finite) then there is a unique monic polynomial of minimal degree for "a" such that p(a)=0 and F(a) = F[t] / (p(t)).
@@Luxaray2000 No worries. I appreciate your reply. That is a nice, helpful definition of transcendental. If we are being pedantic, then yes I would have to cede that X is in fact transcendental over Q. My objection was more due to it being unintuitive to apply this label to X, given that it is an indeterminate / variable. So it's not *really* a number, per se, but rather a stand-in for a number. It doesn't exist in C or R, does it? I guess that's not a very rigorous argument either, so I'll just stop arguing about it xD. I don't yet know enough topology or enough about closures to remark further.
Very well explained thank you sir and also astonishing easy to see nowadays why Euclid could not construct the 7-gon because of the cubic polynom - we are lucky to live after Galois
I just noticed, thanks to your comment, that if you treat L as an additive group (L,+) or a multiplicative group (L\{0},*), then the index of the subgroups (K,+) and (K\{0},*) do not match the degree of the field extension. So [L:K] looks just like a subgroup index, but does not correspond to any actual subgroup index, even though there is a very natural subgroup structure to L and K.
Actually below is proved in later lecture, so can be ignored. I feel any polynomial expression of elements in algebraic extension L/K is algebraic like ξ1^2 + ξ2 is algebraic if ξ1, ξ2 are algebraic , since you can use resultant to get minimal poly by computing det of Sylvester matrix. It certainly holds for cos(2\pi/7)
The proof is only "rigorous" in a particular interpretation of the meaning of the logical connectives: one where "P or Q" is equivalent to "not (not P and not Q)". However, this isn't necessarily what those words should mean. "Intuitionists" reject that interpretation. Instead, the intuitionists have a different meaning in mind. In present terminology this intended meaning is called the "Brouwer-Heyting-Kolmogorov" (or BHK) interpretation after L.E.J. Brouwer (of fixed point theorem fame and a strident intutionist...ironic given the aforementioned fixed-point theorem AFAIK isn't intuitionistically valid), Arend Heyting (his student who put intuitionism on a rigorous footing from the mainstream perspective) and Andrey Kolmogorov (of the "Russian school" of intuitionism, and the founder of modern probability theory). It works as follows: to each statement we associate a type of object which constitutes its evidence. Logical connectives can then be associated with type formers. So, the conjunction "A and B" has as evidence "evidence of A and evidence of B" which means it must consist of pairs. OTOH, evidence of implication "A => B" is interpreted as a "intuitive process" (or maybe a function, or in more modern interpretations a computable function or better still a concrete computer program) that turns "evidence of A" into "evidence of B." Finally the disjunction "A or B" has as evidence either evidence of A or evidence of B, which means it must correspond to the disjoint union. The BHK interpretation is just saying "what the words mean", but it turns out to be surprisingly useful since it allows logical connectives to be reinterpreted as the types of a programming language with a proof of a proposition being a program of that type. This idea, called the "Curry-Howard correspondence" underlies computer aided proof systems such as Coq and Lean and has been the inspiration of a substantial part of the research in Programming Language Theory over the last 40 or so years with evident impacts on mainstream programming languages (e.g. the "generics" in java come from thinking about the computational meaning of second order quantification). Specifically, in the intuitionistic setting a proof of "exists x.P" must involve finding a particular "x" which satisfies "P" and a proof of "forall (x in A), exists (y in B).R x y" yields a function (and in the case of Coq say this is a program we can actually run!) from f from A to B such that forall x, R x (f x). That is pretty cool! However, I should admit, that my PhD dissertation was in the area of "classical Curry-Howard" so maybe this connection shouldn't be oversold as a justification for intuitionism--BHK/Curry Howard does lead to favoring intuitionistic logics, but the arguments need to be more subtle. Still though, non classical logics are useful, and to see why perhaps it is better to start with a different model. A simple one that is Kripke semantics of intuitionistic logic and works as follows. Let (F, Q if for every w' in F, w
Here's how to explain in a way that avoids pointless arguments about philosophy. When you construct a proof, and if you use formal logic to write out the proof in tedious detail, the result looks like a computer program. The program has inputs, and produces outputs. The arithmetic proof of a statement like "For all integers n, there exists an integer m such that m equals n times n time n" is simple enough that you can look at the induction steps, and use them to convert the proof to an algorithm that can be executed on a computer, and will actually compute cubes (extremely inefficiently). But other proofs that classically look similar can't be turned into programs and run in this way, at least not to produce the naive output you think they should give. For example, the proof that e+pi and e*pi can't both be algebraic, when converted to an algorithm, doesn't give you what you might think. While it is a correct proof, it will not properly convert to a program which will produce an example of one of the two numbers which will be transcendental. You might say "Who cares? We know one must be!" But the next program might run this program as a subroutine, and when it feeds the pair "e+pi" and "e *pi" as inputs to this program and demands a transcendental output which is one or the other, the program will not be able to comply with the demand. So this proof does not produce a computer program which outputs one of e+pi or e*pi, and therefore, it can't be used as a subroutine systematically to build bigger programs. But this proof still produces SOMETHING! What this proof DOES produce is a different type of computer program, one which takes proofs as inputs, and produces proofs as outputs. This program will convert a proof of "both e+pi and e*pi are algebraic" into a contradiction! This also has an interpretation as a computer program. But since it doesn't output the name of a real number identified as transcendental, it is a different kind of program, it's a program that produces a different and weaker type of output. So EVEN THOUGH BOTH KINDS OF PROOF CORRECTLY PROVE STATEMENTS, it pays to pedantically distinguish between the two types of proof, even if your philosophy thinks that both proofs are equally valid. The distinction is important regardless of philosophy. A weak proof, when converted to an algorithm will not produce what you naively think it's going to produce, it will produce a complicated mess that takes some proofs, produces contradictions, or takes proofs that produce contradictions and produce other contradictions from those proofs, and so on, nesting into a complicated forest of types described by "Heyting algebra", or equivalently, by function-pointer types, the kind of functions that take arbitrary lists of unions and unions of lists of pointers to functions to other unions of pointers to functions. A horrendously complicated collection of functions. This is what proof looks like "under the hood", and it is how you must implement proof on a computer. The intuitionist logic keeps track of those different types of proofs. The classical logic ignores these distinctions. Classical logic embeds inside intuitionistic logic, you just have to put some "not not"s in the statement of a theorem, and a classical proof automatically becomes an intuitionist proof. But an intuitionist proof can be better, it can explicitly be run as a computer program to produce an object of some kind. Classical proofs can't be run, they don't correspond to a model of computation. So it pays to be more nitpicky about the structure of the proof, and "count the contradictions". To create these ideas, Brouwer used inflammatory philosophical language, but the program is nothing to do with philosophy, it's about computer programs and the relation to proofs.
@@philipjohnson-freyd1314 How can you interpret proof as program without the full structure of intuitionistic logic? Is there a copy of your thesis available? I wrote my comment before unrolling yours, sorry.
@@definitelynotofficial7350 "Long before"? Ridiculous. Logicians were clarifying the notion of algorithm throughout the late 19th century and early 20th century, so as to make sense of analysis and set theory. The start of the idea of computation as a concept wasn't with Turing (who gave the final modern form of the idea), but likely with Kronecker and his generation who defined primitive recursive functions, and distinguished between "arithmetic proofs" (using induction), "analytic proofs" (using real numbers) and "set theoretic proofs" (using transfinite induction and uncountable sets). They were aware that this is the heirarchy of abstraction in the 19th century, and Hilbert's goal was to justify the latter methods (analytic and set theoretic) using an arithmetic foundation, and that means 'define the procedures on a computer' in modern language, just the modern language wasn't developed until 1936. This trichotomy of proofs is why the goal of an arithmetic proof of the prime-number theorem was considered so important in the 1920s, they didn't know how exactly how to think of analytic proofs arithmetically precisly. We more or less know today, as we can think of the proof in a formal system. The role of "intuition" in "intuitionism" isn't all that abstract. Brouwer's idea was to clarify that the natural numbers are available as a model directly to our intuition. We know that the model of natural numbers includes all finite digit sequences, and adding and mutliplying are by the grade-school operations. This makes the model of natural numbers 'inutitiive' in a way that the model of real numbers is not, because the real numbers require you to formalize the notion of an infinite list. Why are real numbers not "intuitive" to Brouwer? If you ask children, they generally find real numbers intuitive (after they learn infinite decimals the Stevens way). Obviously Brouwer knows that you can specify and integer finitely on a computer, and you can't do that for an arbitrary real number. He just lacks the standard vocabulary to say it. When you have rapid fundamental advancement like in logic in the early half of the 20th century, you have to be aware that the final form of the ideas are floating around in half-baked versions much earlier than the time during which they are set in stone. The notion of intuitionism is concerned with computation, full stop. That is all it is concerned with, the 'intuition' is used because the only computation model available before Turing was the human brain, and the 'intuition' being described is our intuitive understanding that the model of natural numbers with addition and mulitplication is a uniquely defined thing, even though we could imagine a million different models of the Peano Axioms, only one is the 'right model', and we know that because only one is defined by computable functions. What that means is that when Intuitionism is defined, it immediately gives a formal definition of part of a programming language in the Heyting algebra. The logic is designed to convert proofs automatically into algorithms, and the algorithms are precise, so they turn into computer programs. Heyting Arithmetic, in fact, might be arguably considered the first (accidental) universal computer, as could Hilbert deduction (sort of, it's harder to see there), and Godel recursive functions (which explicitly try to be universal). Turing's universality is the culmination of 50 years of hard work which was trying to get at this main idea.
So are you saying that the intuitionists reject proofs by contradiction? You have shown that it cannot be the case that both e + pi and e*pi are algebraic, undeniably. How would the intuitionist reconcile the obvious conclusion that at least one of them is not algebraic?
it's nice to hear "we write this in this way, I have no idea why"; it eases my anxiety that I might have missed some important feature of the mathematics when I see something written a certain way that I am unfamiliar with, knowing that sometimes the reason something is described with a certain notation is rather arbitrary and just needs to be known rather than understood.
Tbh, a lot of things were created by some mathematicians many years ago and have just become standard notation. Sometimes, you won’t know who created the notation but it is just commonly used.
Great lecture!
For anyone wondering: There is a small error in the equation at 08:18. If x is an n-th root of unity we have 1+x+x^2+..+x^{n-1} = 0 and not 1. (To see this multiply 1+x+x^2+..+x^{n-1} with x and observe what happens.)
I lost here for quite a while before realized that I should check the comments.
Peculiarly enough, it is corrected in the next line, where the equation is divided by ζ³. This operation would by no means turn a 1 into a 0.
It is good practice in lectures to scatter some minor errors across the material so as to test whether the audience is still attentive.
Agree. The sum of all nth roots of unity is equal to zero. i.e.: 1 + [(-1 + √3 i ) /2] + [(-1 - √3 i ) /2] = 0
Thank you!! This is a relief; I was befuddled by this bit for half an hour
I am not a mathematician but was blown away by the argument that led to the cubic polynomial satisfied by cos(2*pi/7).
👍
There is a reverse pascal triangle in that construction which makes it more amazing xD
At 23:57 he says "with roots in this field". Shouldn't it be "with coefficients in the field K(a0, ..., an-1)"?
Loving this series so far. It has set off at a really nice pace
At 26:28, I think it’s worth mentioning, that although an intuitionist would reject the argument as a proof that one of pi+e and pi*e are transcendental, they are not claiming that the mathematics being done is somehow incorrect. The point of view is simply that there is a meaningful difference between the statements:
“It is not the case that both of pi+e and pi*e are algebraic”, and
“Either pi+e is transcendental or pi*e is transcendental”.
The argument given is intuitionistically valid as a proof for the former statement, but not the latter.
but surely if both pi+e and pi*e are not algebraic then it would be logical to conclude that AT LEAST one of them is transcendental (since a number is either algebraic or transcendental). Can you explain what the meaningful difference between the two are please?
@@20doctorwho09 one of the major ideas of "intuitionistic logic" is removing the "law of the excluded middle" and "law of double negation" as axioms (or even a couple of "Di'Morgan's Laws"), instead requiring there to be a certain proof of that. Apparently this is very useful for questions of constructability and computability when doing proofs.
But if you have not proven the law of the excluded middle and law of double negation applies to the situation, NOT( NOT(pi+e is algebraic) AND NOT(pi*e is algebraic)) does not (yet) imply (NOT(pi+e is algebraic) OR NOT(pi*e is algebraic))
At 8:16, shouldn't the right hand side be 0 and not 1?
Yep
At 6:27, why is x transcendental over the rational numbers field?
It is a bit strange and unintuitive and I kind of blinked when he said that too, but it comes from treating Q[X] as a distinct ring unto itself. (Q[X] being the ring of polynomials with coefficients in Q and in a single variable X). Q is contained as a subring of Q[X] by considering only constant polynomials of Q[X]. One can also construct rings as polynomials over Q in an arbitrary number of variables such as Q[X, Y]. I think the core of what he's saying is that X neither exists in Q, nor can you conjure a polynomial with coefficients in Q with X as a solution. Thus, I'm not really sure what the nature of Q[X] is in relation to Q. It's certainly not a traditional field extension, and in fact I realize after I've been writing that for it to be a field extension, the field extended to would have to be the field of rational functions in a variable X. It seems to exist as a non-algebraic but finite extension induced by adding a variable called X. It is perhaps not particularly meaningful, and perhaps incorrect to call X transcendental over Q, I suppose.
It's a tough call because X is inherently a variable, as opposed to adjoining pi or some other transcendental number to Q. The resulting field would look essentially the same if we had no idea ahead of time of the numerical properties of the adjoined number if our extension were embedded into the complex numbers. Conversely, the only property we require of X is compatibility with the ring operations of Q.
Edit: Hopefully this is my last edit, just realized I made a major mistake. Q[X] is an infinite dimension vector space over Q. I.e. I was wrong about it being a finite extension. Its dimension is infinite because no power of X can be written as a linear combination of other powers of X with coefficients in Q. I'm still not convinced that it's proper to say X is transcendental over Q, though.
@@MathOnNapkins Forgive me for replying to a 3W old comment.
x is transcendental over Q because x is not a solution to a polynomial over Q. One cannot solve for a 0 of a polynomial and obtain x. The polynomial x has solution 0.
I believe the following definition makes things easier to understand (Paolo Aluffi Algebra Chapter 0):
Consider [F(a) : F] (the dimension of the vector space F(a) over F. "a" is algebraic if the dimension is finite. Transcendental otherwise.
F[x] is an infinite dimensional vector space with basis {1, x, x^2, ...} as you've pointed out.
Saying the same thing again but closer to the definition, [F[x] : F] is infinite hence "x" is transcendental.
How this relates to polynomials is the following, suppose F(a) is a finite extension (i.e. the dimension of the vector space is finite) then there is a unique monic polynomial of minimal degree for "a" such that p(a)=0 and F(a) = F[t] / (p(t)).
@@Luxaray2000 No worries. I appreciate your reply. That is a nice, helpful definition of transcendental.
If we are being pedantic, then yes I would have to cede that X is in fact transcendental over Q.
My objection was more due to it being unintuitive to apply this label to X, given that it is an indeterminate / variable. So it's not *really* a number, per se, but rather a stand-in for a number. It doesn't exist in C or R, does it? I guess that's not a very rigorous argument either, so I'll just stop arguing about it xD. I don't yet know enough topology or enough about closures to remark further.
Thank you for your effort Richard! May The Universe reward you!
Very well explained thank you sir and also astonishing easy to see nowadays why Euclid could not construct the 7-gon because of the cubic polynom - we are lucky to live after Galois
Thanks for helping us to field more information and give our fund of knowledge extensions!
I think that the L in 16:06 should be an M in terms of the notation previously introduced.
Thanks Mr.Borcherds! You saved my deadline!
Isn't the notation [L:K] like that because it matches that of the subgroup index?
Yeah, the context is too similar for it to be a coincidence. Wonder what notation came first?
I just noticed, thanks to your comment, that if you treat L as an additive group (L,+) or a multiplicative group (L\{0},*), then the index of the subgroups (K,+) and (K\{0},*) do not match the degree of the field extension. So [L:K] looks just like a subgroup index, but does not correspond to any actual subgroup index, even though there is a very natural subgroup structure to L and K.
18:45 why the three notations for a field extension
Sir,please,recommend texts for gallois theory.whuch book would be of best use for first reading ?
He recommends Galois Theory by Emil Artin in the description of the first video
Do we assume the field has characteristic 0?
Actually below is proved in later lecture, so can be ignored.
I feel any polynomial expression of elements in algebraic extension L/K is algebraic like ξ1^2 + ξ2 is algebraic if ξ1, ξ2 are algebraic , since you can use resultant to get minimal poly by computing det of Sylvester matrix. It certainly holds for cos(2\pi/7)
Did we just skip over why x is transcendental at 6:00 but spent an entire 3 minutes explaining the trivial case of cos 2pi/7? :D
Excellent resource
Какой это курс?
Good job
Thankyou. Gentleman
Thank you.
mer ye christmas
Ye rry christmas
What does "intuitionist" mean? Your proof is rigorous that they can't both be algebraic, I don't understand where there would be room to argue that.
The proof is only "rigorous" in a particular interpretation of the meaning of the logical connectives: one where "P or Q" is equivalent to "not (not P and not Q)". However, this isn't necessarily what those words should mean. "Intuitionists" reject that interpretation.
Instead, the intuitionists have a different meaning in mind. In present terminology this intended meaning is called the "Brouwer-Heyting-Kolmogorov" (or BHK) interpretation after L.E.J. Brouwer (of fixed point theorem fame and a strident intutionist...ironic given the aforementioned fixed-point theorem AFAIK isn't intuitionistically valid), Arend Heyting (his student who put intuitionism on a rigorous footing from the mainstream perspective) and Andrey Kolmogorov (of the "Russian school" of intuitionism, and the founder of modern probability theory). It works as follows: to each statement we associate a type of object which constitutes its evidence. Logical connectives can then be associated with type formers. So, the conjunction "A and B" has as evidence "evidence of A and evidence of B" which means it must consist of pairs. OTOH, evidence of implication "A => B" is interpreted as a "intuitive process" (or maybe a function, or in more modern interpretations a computable function or better still a concrete computer program) that turns "evidence of A" into "evidence of B." Finally the disjunction "A or B" has as evidence either evidence of A or evidence of B, which means it must correspond to the disjoint union.
The BHK interpretation is just saying "what the words mean", but it turns out to be surprisingly useful since it allows logical connectives to be reinterpreted as the types of a programming language with a proof of a proposition being a program of that type. This idea, called the "Curry-Howard correspondence" underlies computer aided proof systems such as Coq and Lean and has been the inspiration of a substantial part of the research in Programming Language Theory over the last 40 or so years with evident impacts on mainstream programming languages (e.g. the "generics" in java come from thinking about the computational meaning of second order quantification). Specifically, in the intuitionistic setting a proof of "exists x.P" must involve finding a particular "x" which satisfies "P" and a proof of "forall (x in A), exists (y in B).R x y" yields a function (and in the case of Coq say this is a program we can actually run!) from f from A to B such that forall x, R x (f x). That is pretty cool!
However, I should admit, that my PhD dissertation was in the area of "classical Curry-Howard" so maybe this connection shouldn't be oversold as a justification for intuitionism--BHK/Curry Howard does lead to favoring intuitionistic logics, but the arguments need to be more subtle.
Still though, non classical logics are useful, and to see why perhaps it is better to start with a different model. A simple one that is Kripke semantics of intuitionistic logic and works as follows. Let (F, Q if for every w' in F, w
Here's how to explain in a way that avoids pointless arguments about philosophy. When you construct a proof, and if you use formal logic to write out the proof in tedious detail, the result looks like a computer program. The program has inputs, and produces outputs. The arithmetic proof of a statement like "For all integers n, there exists an integer m such that m equals n times n time n" is simple enough that you can look at the induction steps, and use them to convert the proof to an algorithm that can be executed on a computer, and will actually compute cubes (extremely inefficiently).
But other proofs that classically look similar can't be turned into programs and run in this way, at least not to produce the naive output you think they should give. For example, the proof that e+pi and e*pi can't both be algebraic, when converted to an algorithm, doesn't give you what you might think. While it is a correct proof, it will not properly convert to a program which will produce an example of one of the two numbers which will be transcendental. You might say "Who cares? We know one must be!" But the next program might run this program as a subroutine, and when it feeds the pair "e+pi" and "e *pi" as inputs to this program and demands a transcendental output which is one or the other, the program will not be able to comply with the demand.
So this proof does not produce a computer program which outputs one of e+pi or e*pi, and therefore, it can't be used as a subroutine systematically to build bigger programs.
But this proof still produces SOMETHING! What this proof DOES produce is a different type of computer program, one which takes proofs as inputs, and produces proofs as outputs. This program will convert a proof of "both e+pi and e*pi are algebraic" into a contradiction!
This also has an interpretation as a computer program. But since it doesn't output the name of a real number identified as transcendental, it is a different kind of program, it's a program that produces a different and weaker type of output.
So EVEN THOUGH BOTH KINDS OF PROOF CORRECTLY PROVE STATEMENTS, it pays to pedantically distinguish between the two types of proof, even if your philosophy thinks that both proofs are equally valid. The distinction is important regardless of philosophy.
A weak proof, when converted to an algorithm will not produce what you naively think it's going to produce, it will produce a complicated mess that takes some proofs, produces contradictions, or takes proofs that produce contradictions and produce other contradictions from those proofs, and so on, nesting into a complicated forest of types described by "Heyting algebra", or equivalently, by function-pointer types, the kind of functions that take arbitrary lists of unions and unions of lists of pointers to functions to other unions of pointers to functions. A horrendously complicated collection of functions. This is what proof looks like "under the hood", and it is how you must implement proof on a computer.
The intuitionist logic keeps track of those different types of proofs. The classical logic ignores these distinctions. Classical logic embeds inside intuitionistic logic, you just have to put some "not not"s in the statement of a theorem, and a classical proof automatically becomes an intuitionist proof. But an intuitionist proof can be better, it can explicitly be run as a computer program to produce an object of some kind.
Classical proofs can't be run, they don't correspond to a model of computation.
So it pays to be more nitpicky about the structure of the proof, and "count the contradictions". To create these ideas, Brouwer used inflammatory philosophical language, but the program is nothing to do with philosophy, it's about computer programs and the relation to proofs.
@@philipjohnson-freyd1314 How can you interpret proof as program without the full structure of intuitionistic logic? Is there a copy of your thesis available? I wrote my comment before unrolling yours, sorry.
@@annaclarafenyo8185 It has everything to do with philosophy. Brouwer and intuitionist mathematics came long before computers.
@@definitelynotofficial7350 "Long before"? Ridiculous. Logicians were clarifying the notion of algorithm throughout the late 19th century and early 20th century, so as to make sense of analysis and set theory. The start of the idea of computation as a concept wasn't with Turing (who gave the final modern form of the idea), but likely with Kronecker and his generation who defined primitive recursive functions, and distinguished between "arithmetic proofs" (using induction), "analytic proofs" (using real numbers) and "set theoretic proofs" (using transfinite induction and uncountable sets). They were aware that this is the heirarchy of abstraction in the 19th century, and Hilbert's goal was to justify the latter methods (analytic and set theoretic) using an arithmetic foundation, and that means 'define the procedures on a computer' in modern language, just the modern language wasn't developed until 1936.
This trichotomy of proofs is why the goal of an arithmetic proof of the prime-number theorem was considered so important in the 1920s, they didn't know how exactly how to think of analytic proofs arithmetically precisly. We more or less know today, as we can think of the proof in a formal system.
The role of "intuition" in "intuitionism" isn't all that abstract. Brouwer's idea was to clarify that the natural numbers are available as a model directly to our intuition. We know that the model of natural numbers includes all finite digit sequences, and adding and mutliplying are by the grade-school operations. This makes the model of natural numbers 'inutitiive' in a way that the model of real numbers is not, because the real numbers require you to formalize the notion of an infinite list.
Why are real numbers not "intuitive" to Brouwer? If you ask children, they generally find real numbers intuitive (after they learn infinite decimals the Stevens way). Obviously Brouwer knows that you can specify and integer finitely on a computer, and you can't do that for an arbitrary real number. He just lacks the standard vocabulary to say it.
When you have rapid fundamental advancement like in logic in the early half of the 20th century, you have to be aware that the final form of the ideas are floating around in half-baked versions much earlier than the time during which they are set in stone.
The notion of intuitionism is concerned with computation, full stop. That is all it is concerned with, the 'intuition' is used because the only computation model available before Turing was the human brain, and the 'intuition' being described is our intuitive understanding that the model of natural numbers with addition and mulitplication is a uniquely defined thing, even though we could imagine a million different models of the Peano Axioms, only
one is the 'right model', and we know that because only one is defined by computable functions.
What that means is that when Intuitionism is defined, it immediately gives a formal definition of part of a programming language in the Heyting algebra. The logic is designed to convert proofs automatically into algorithms, and the algorithms are precise, so they turn into computer programs.
Heyting Arithmetic, in fact, might be arguably considered the first (accidental) universal computer, as could Hilbert deduction (sort of, it's harder to see there), and Godel recursive functions (which explicitly try to be universal). Turing's universality is the culmination of 50 years of hard work which was trying to get at this main idea.
So are you saying that the intuitionists reject proofs by contradiction? You have shown that it cannot be the case that both e + pi and e*pi are algebraic, undeniably. How would the intuitionist reconcile the obvious conclusion that at least one of them is not algebraic?