defining multiplication is trickier than you think
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Without multiplication, arithmetic-Presburger Arithmetic, to be specific-was provably decidable, complete, and consistent. Sadly, by defining multiplication, the resulting arithmetic-Peano Arithmetic-triggered Gödel's incompleteness theorems, and consequently, is neither decidable nor complete. And further, its consistency cannot be proved from within itself (but can be proved from ZFC-thank heaven for small mercies!).
As an engineer, those concerns are irrelevant to me. But I'm given to understand that they cause existential crises among mathematicians-particularly grad students-even today. Luckily, as of now, the earlier existential crises involving Zeno's paradoxes, irrational numbers, imaginary numbers, non-Euclidean geometry, and Russell's Paradox have all been resolved to the satisfaction and relief of mathematicians.
Isn't it possible to prove arithmetic with addition & multiplication decidable, complete, and consistent if one leaves out both subtraction & division? (e.g. leave out their inverses)
@@FrankHarwald Sadly, no 😢. Thanks to Gödel's theorems, any arithmetic (over natural numbers or bigger sets) that has both addition and multiplication immediately becomes incomplete and undecidable, and its consistency cannot be proved internally.
It has nothing to do with subtraction or division. In fact, as you well know, if you apply subtraction or division to two natural numbers, the result often isn't a natural number.
To be clear, when we say that multiplication is _defined_ in Peano arithmetic, it includes an axiom that allows any two numbers to be multiplied. We can certainly define the same multiplication _operation_ in Presburger as well. But without the corresponding axiom, it doesn't guarantee that the result always exists. (And no, you can't prove it via other Presburger axioms.) Consequently, multiplication in Presburger is not as powerful-nor as useful-as in Peano. This handicap makes Presburger decidable, complete, and consistent-which makes sense, when you think about it.
What do you suppose happens if you invert the Presburger case, and define an arithmetic that includes multiplication but no addition? (It's called Skolem arithmetic, BTW.) Clearly, it can't be used for totaling your restaurant tab or grocery purchases, but will it be decidable, complete, and consistent?
One of the things we had in my equivalent class was examples of what you accidentally include if you miss one of the axioms. For example, if you miss induction, you could have N union {alpha}, where S(alpha)=alpha. This fits all of the other axioms, and you would need induction to prove that all natural numbers aren't alpha. The examples were great for understanding why these axioms were necessary, rather than just a set of statements that were obviously true of the natural numbers.
the definition of naturals above lacks the induction axiom
I really like this defining series. Look forward to the next one.
So many eye-openers from this video. Thanks Michael Penn!
Dr Penn, this was one of the most interesting videos I've seen in awhile. I too learned all this in graduate school and still remember it! BTW, I'm thinking of finishing up my degree at SUNY Albany
I love u man, u are genius 💙
took a class like that; was one of my favorites
This really should have more views. What I'm wondering is, I've heard that there are multiple ways to define the natural numbers. How come?
Right distributivity together with commutativity immediately implies left distributivity, i.e. k(m+n) = (m+n)k = mk + nk = km + kn.
I can’t find the set theoretic model of natural numbers video mentioned in the beginning.
to meet the requirements of the square root of the equation he derived it from you only need to take a root of 1 that squares to 1, and can anti commute with the other roots.
Asian math had it best:
The amount of intersection points between two sets of parallel lines with different slopes to each other {({a},{b}) } to turn the points back into lines do a perpendicular to the drawing plane into 3d then flatten back to a 2d plane trough a projection for further crossings and multiplications (be careful to not overlap lines).
This is true because every two lines will intersect at a point and the two sets will construct a rectangle with sides a and b there every line is a continuous but yet discreet set of blocks and each crossing is a square of the rectangle either from a's perspective or b's and that is why a*b=b*a which is a diagonal flip of the drawing plane.
I'm not sure that those three axioms are sufficient to define N as the set of objects that are obtained as an iterated application of the function s to 0.
Take for example the usual set N and the function s defined as s(0)=3, s(1)=2, s(2)=1, s(n)=n+1 for all n>2. We have that 0 belongs to N, we have injective s, we have that s(N)=N-{0}, but the set obtained by iterating the application of s to 0 does not contain 1 and 2.
Indeed, it is not sufficient. You also need the induction axiom: Any set that contains 0, and for each number contains its successor, contains all of N. Your example violates the induction axiom, since the set {0, 3, 4, 5, ...} is a proper subset that fulfills the conditions.
This video is really informative and interesting
bijectivity of the successor function is not enough to prevent cycles next to the inductive set.
What's really interesting is to move up to exponentiation, attempt the same proof of commutativity and associativity, and see where it fails.
The base case doesn't even work so it fails pretty quickly
These are my fav type of math vid, where basic mathematical operations are proved starting with some base assumptions.
The naturals are not uniquely specified by any algorithmic first-order theory. Here is a structure (X,x,S:X->X) where x is an element of X and S:X->X is an injection where S(X) = X\{x}.
Let X be any set containing the natural numbers.
Define S:X->X to be the function where S(x) = x+1 for a natural number x and S(x) = x for all non-natural numbers in X. Now
0 is an element of X and S is an injection. Furthermore, S(X) = X\{0}.
Here is a second-order theory that does specify the naturals:
A tuple (X,x,S:X->X) defines natural numbers, if
x is an element of X
S is an injection
S maps no element to x
If I is a subset of X, where S(I) is a subset of I and x is an element of I, then I = X.
The recursion theorem says that this structure satisfies a universal property:
Let (Y,y,T:Y->Y) be a tuple where y is an element of Y. Then there exists a unique map f:X->Y where f(x) = y and f(S(x)) = T(f(x)).
If one thinks about the set X as the natural numbers and x as the zero element, then f:X->Y is a sequence of elements of Y and one could denote f = (y, T(y), T(T(y)), T(T(T(y))),...).
Because of this universal property, any two structures of natural numbers are isomorphic. The universal property itself implies that the definition above is satisfied. The recursion theorem allows a category theoretical definition of the natural numbers.
but do they work for the dirac equation? :) nice exercise to do
4:26 Actually all you need for your base case is that 0*0 = 0. Once you have that, then you can prove by induction that, for all n, 0n = 0.
Base case: 0*0 = 0 was assumed by definition
Induction: Assume that for a given n that 0n = 0. Then
0s(n)
= 0n + 0 (by definition of multiplication in the video)
= 0n (since 0 is already by definition the additive identity from the previous definition of addition)
= 0 (by the inductive hypothesis).
Therefore if 0*0 = 0 then 0n = 0 for all natural numbers n.
If we consider a non-standard model of the naturals (e.g, via an ultrafilter extension), we will get another models that from the first order logic point of view are the same as the naturals, but from an external viewpoint contains infinite elements, so I don't think we can prove uniqueness (unless we go to second order logic with quantification over arbitrary subsets of the naturals)
can you base the quaternions on a root of 1 which is positive and doesn't commute instead of the regular 1? if so does it make for a good substitute for the gamma matrices in the dirac equation?
I think any root of one will commute with itself and with scalars. So you need at least two roots of one to generate the quarternions.
wait where's the second video?
"the origins of addition" is the title of the video he references at the beginning.
Edit: now i see he put that one on the video, so i don't know if that is the one you're refering to.
Professor, did this careful understanding of addition,...., multiplication precede addition ...etc? Why do I think not!
This definition of the naturals is insufficient. While every such set contains a copy of the natural numbers via 0, s(0), s(s(0)), ..., it can contain additional elements. For instance you can just add some a that is not in the list above and set s(a) = a. In fact you can add any set A you want as long as you make s bijective on A.
s being injective and s(N) = N - {0} doesn't imply that N = {0, s(0), s(s(0)), ... } unless some other conditions are implicit.
if 0 belongs to N then infinity(carry) must belong to N (otherwise logical contradiction) and s(infinity)=0 , no any excluding "s(N)=N-{0}".
Real mathematicians use the disjoint union to define addition, the cartesian product for multiplication and sets of functions for exponentiation
what about complex mathematicians ?
@@charlievane idk i heard they had an imaginary part, sounds pretty made up
How do you define exponentation with set of functions?
@@julos1471 The number of possible functions from A to B is |B| ^ |A|
There are many kinds of real mathematicians and it appears that you’re not one of those.
Quick, find a prime number p such that p^p - p is divisible by 1000.
One guy recorded video in my native language about area of rectangle and it is geometrical interpretation for this if we are limited to real numbers
OK, I pretty much follow everything here. But how do we extend multiplication from the Naturals to the Reals?
basically just test whether they transform nicely and serve the same job as the gammas.
I guess there wasn't a good place to stop for this video!
l, m and k are perfect variables to use for MLK day ;)
all of the homework is already solved by the proof of commutativity being given in the video
Planning to present the proof that the cardinality of a Cartesian product is the product of the cardinalities of the factors?
Why'd you have to go on sabbatical right when I'm ready for Pre Cal 2 😢
neat. now potentiation
I've been searching a true proof of multiplication commutativity for a long time.nobody could answer,they all said "that's by definition, stupid question,or at the best,imagine a rectangle,make a rotation,the rectangle is the same.yes but that's not a proof
After all,4×3=4+4+4 and 3×4=3+3+3+3, the egality is not evident
What about this? We have to prove that (1) (m+m+m+m....)[n times] equals (2) (n+n+n+n...) [m times].
Let us assume n>m. I can write (1) as :
(3) (m+m+m..)[m times] + (m+m...) [n-m times].
I can write (3) as : (4) (m+m+m..)[m times] + m*(n-m).
I can write (4) as : (5) (m+ m*(n-m)/m + m+ m*(n-m)/m +...) [m times].
Given that m+m*(n-m)/m=n I can write (5) as: (n+n+n...) [m times].
QED
If m = { 0, 1, 2, ..., m-1 } and n = { 0, 1, 2, ..., n-1 }, we could define m.n as the cardinality of m x n ( "x", here, is the Cartesian product of sets ). Nah, let's not do that. Let's accept Penn's definition and prove | m x n | = m.n
I think we're going to need s(n) = n U { n }
Hey, this site suggests "carnality" and "cardinal" as corrections for "cardinality"!
10, 11. 12.
It's not tricky at all. Just count rows and columns and totals and you're fine unless you're looking to make a complexity disaster. You're not looking to make a complexity disaster, are you?
Can't wait till you get to the reals and the proof that they are complete
so, you had to admit that 0 belongs to N hehehehe
the base problem is that Set theory is wrong about these things.
you can't add Naturals, because addition only works over objects which have both a magnitude and sign, and Naturals are supposed to only have magnitudes.
this is easily seen by considering the claim that 1+1=2, which is so thoroughly false that most of mathematics exists specifically because it's not true:
1 dog +1 dog = 2 dogs; this is the only case that Set theory is acknowledging
1 dog +1 quail = 2 wings; so 1+1=2 because 0+2=2... that's different
1 dog +1 quail = 6 legs; so 1+1=6 because of a word problem we might give a second grade student
1 foot +1 yard = 48 inches; so 1+1=48 because of unit conversion
1 half +1 third = 5 sixths; so 1+1=5 because of fraction addition
1 frog +1 pond = 1 pond; so 1+1=1 because frogs live in ponds
1 stone +1 mountain = 1 mountain; so 1+1=1 because mountains are made of stones
1 C water +1 C dirt < 2 C mud; so 1+1 is between 1 and 2 because fluids can fill the spaces between granulated solids
1x +1y cannot be simplified; so 1+1 is undefined because of like terms
what this tells me is that the people who invented Set theory could not do mathematics beyond a kindergarten level.
if you want to claim that 1+1=2, and this is referring to the Natural numbers 1, 1 and 2, that's simply impossible. as demonstrated above, addition only behaves this way when all the terms have exactly the same unit. so if you're working with bare numbers, addition cannot be performed by definition. this can be demonstrated another way, as well, by considering the 'commutative property of addition':
4 +3 = 3 +4; so this is said to be commutative because the numbers moved around the + operator but the result remained the same
4 -3 = -3 +4; but what about this one? why does the - stick to the 3? where did the + come from?
the answer to this is exceptionally simple. + and - are not arithmetic operators. the notion that they should be comes from overextending generalizations from Boolean Logic by the Logicists. but the problem is that there's an unrecognized grammar rule in mathematical notation which explains everything in one move. expression-initial + signs are optional. so now look at what we get:
4 +3 = +4 +3 = +3 +4 = 3 +4
4 -3 = +4 -3 = -3 +4
now the commutativity is not a property of addition, because we can now easily see that there is no such thing. all we're doing here is processing an unordered list of 1-dimensional vectors. so {+4, +3} = {+3, +4} because those are the same unordered list. and {+4, -3} = {-3, +4} because they're the same unordered list. but now you can't claim that addition is abelian while subtraction is not, because those claims are now just pure gibberish. further, we can see that putting unsigned Naturals into such an unordered list is also gibberish:
{3, 4} ... this just doesn't even mean anything, because we don't have any way of determining if the 3 and 4 are of commensurable size or direction.
think about how silly it is to say that -7 is a number but 7 giraffes is not. they have the same number component, all that differs between them is which unit they have. yet Set theory says that -7 is a number, despite being composed of a number and a unit.
or put another way, how does it make sense to claim that the Naturals are the positive Integers when the notion of positive is completely foreign to the context of the Naturals? the norm of a vector should be completely unsigned, not positive.
obviously Set theory is built wrong from the start.
as for multiplication, the biggest problem is that you're leaving out the initial condition. this manifests most obviously in the special case of multiplication which is exponentiation.
classically we claim that exponentiation is iterated multiplication, but we also say:
2^3 = 2 *2 *2; this only has 2 multiplications, not 3
2^0 = 1; this has no multiplications, but we have to make up a special rule about where the 1 came from
2^-3 = 1 /(2^3); again, there's no actual explanation of where the 1 in the numerator came from
if instead we realize that the 2^3 example only really makes sense if we include the initial condition, 1, then we get:
2^3 = 1 *2 *2 *2; 1 multiplied by 2, 3 times
2^0 = 1; 1 multiplied by 2, zero times
2^-3 = 1 )/2 )/2 )/2 = 1 multiplied by 2, -3 times
and this also completely explains why 0^0=1, and why i = v(-1) is perpendicular to the Real axis, and halfway between 1 and -1. and note that Euler's discussion of the topic do not actually help you here if you want to defend Set theory, because his claims were directly contradictory. he said that division by 0 is always infinite (despite knowing about l'Hopital's Rule), and that the way you find that 2^0 = 1 is by noticing that 2^2 = (2^3) /2, so 2^0 = (2^1)/2 = 1, which he then notes in his schizophrenic matter cannot explain 0^0. it's also conspicuous that he published this almost immediately after his mentor, who invented l'Hopital's Rule, died.
the real trouble we get into now is that the hyperoperations are all defined explicitly without taking the initial condition into account. apparently because people like Donald Knuth were too dumb to count as high as 0.
the real magic of my account of things come in when you realize that what I've said actually allows unification of all of arithmetic and calculus into a single equation.
given that + and - are not arithmetic operations in the first place, we can pretty much ignore them, and just move on to multiplication, division, exponentiation, logarithms, differentiation and integration.
returning to the idea that summation is generalized by unordered lists of vectors, we can just add in a notation to handle the initial condition:
{0; +3, +4} = 0 +3 +4 = 3 +4; note that the 0 does not make sense in any other position in classical notation, because it lacks a sign
now multiplication is simply reusing the same vector:
{0; +3; +3} = 0 +3 +3 = 3 +3 = 2 *3
how about we introduce notation that allows us to indicate this repetition?
{0; (+3)R.+2} = 0 +3 +3 = 3 +3 = 2 *3
but now, why can't we just have the argument of the loop be a list, so that we can unify everything we have so far?
x = {a, b, c}
{0; xR.+3} = {0; a, b, c} = 0 +a +b +c; some cases of this will be multiplications
we can now simply say that * represents the structure we've established so far. so now:
{1; (*x)R.+3} = 1 *a *b *c; some cases of this will be exponentiations
ok, now let's consider the inverse of multiplication, division. its initial condition is the result of a multiplication, and its output is the number of subtractions it takes to get to the origin:
{n; (*x)R.-3} = n )/a )/b )/c; some cases of this will be exponentiations
now let's notice that this allows us to find the n-dimensional volumes of rectangular solids, and the volumes of all other shapes are just some fraction of those. which means there's an implicit multiplication by 1 occurring on all terms of x. so let's set up a second list to carry these shape coefficients, and call it z. but now we need the terms of x and z to correspond, so we need them both to be ordered lists. and we should then store the output of each step into a similar ordered list, w. the easiest way to specify this is to just make them all base-1 polynomials:
a, b, c, d\ = z; o, p, q, 1\ = x; k = +3
{z(0); (*x)R.k} = d *(c*q) *(b*p) *(a*o) = w
a*o, b*p, c*q, d\ = w
now, if you solve for k you get logarithms, and if your initial condition, z(0), is a function you automatically get calculus. all without the nonsensical contradictions of Set theory.
the successor function doesn't work because it's begging the question.
nothing in the definition of the successor function yields the Naturals. and nothing in the Naturals implies an order, since they cannot be added or subtracted by definition. so the only way to arrive at the conclusion that the successor function relates in any way to the Naturals is to presuppose that conclusion, which is a begged question and thus logically fallacious.
a simple way to tell that what you've done is disingenuous is to notice that you're using unordered lists to imply order. either this is an abuse of notation, or you're following a tradition of intentional deception, but either way, it's obviously incorrect. that is, given the two unordered lists:
{0, S(0), S(S(0)), etc.}
{0, 1, 2, etc.}
it's dishonest to say that there's only one mapping possible between them. if you wanted to sincerely claim this then you need to specify ordered lists. as such, under the standard definition we can't determine whether 7 is the successor to 6 or to -34. as such, the successor function as it's defined unambiguously does not do what it's alleged to do. to achieve the claimed meaning you must use ordered lists:
(0, S(0), S(S(0)), etc.)
(0, 1, 2, etc.)
which neither you, nor anyone else, ever does.
it's also notable that adding multiplication to Set theory is specifically the thing which introduced Incompleteness. this is what Presburger Arithmetic was all about, after all.
which should tell us that there's something wrong with Set theory, since multiplication in Euclidean geometry does not introduce Incompleteness.
The listed axioms do not uniquely define ℕ. They are also satisfied by the disjoint union of ℕ and ℤ with "0" meaning the zero in ℕ and s acting as x ↦ x + 1 on both ℕ and ℤ.
on Z, however, 0 does have a pre-image on S (namely -1), whereas in N it does not. So, for Z we'd have S: Z → Z, while for N we have (as stated) S: N → N/{0}
The counter-example is not ℤ but ℕ ⊔ ℤ (read ℕ disjoint union ℤ). 0 ∊ ℕ and 0 ∊ ℤ are considered distinct elements of ℕ ⊔ ℤ and the former is the one that acts as the "0" element required by the axioms. @@okra_
You are indeed correct, the requirements only guarantee that there exists a subset which is isomorphic to the naturals, you can add refinements to ensure uniqueness or simply construct the aforementioned subset.
The rule(s) I remember being added is basically "induction works", which constrains the set to be equivalent to the natural numbers with nothing extra added.