It's not. "Close" refers naturally to a relation between elements of set - metric space, we could say, but let's keep it vague. On the other hand, "dense" qualifies a relation between two sets. Obviously they are related concepts, and one very useful to understand the second, but your wording may be a bit too cowboyish for the standard mathematician.
It’s not immediately clear that there is 00, g:=|g_n-g_m|0 because the sequence is not eventually constant -- otherwise m in G, contradicting the assumption) Therefore, for every ε>0, there is g in G_+ such that: 0
@@JM-us3fr You're right, m just exists so we can take a limit approaching it, using the assumption that it is not attained inside G_+. We're done after getting 0
Yeah, this bothered me too. You can only conclude the existence of a g here with the requisite property by using the group properties of G, and all that was referred to was the mere lack of a minimum. The proof in the video is not sufficient on this point.
Your proof is again not quite sufficient. As @MatthiasGorgens notes below, if G is the trivial group, the lemma is fails, and your proof bakes in the assumption that G is non-trivial. (Specifically, if G is trivial, then G_+ is empty, and so the infimum does not exist. If we alter the proof to allow m to be zero, then the statement that the sequence is not eventually constant fails, but without contradicting the assumption.)
We can generalize Lemma 2 to note that { a + b y | a, b ∈ ℤ and y is irrational } is dense. And likewise any continuous function on a dense set is dense on its codomain. So then you could use that to perhaps find other interesting examples of dense functions on integers. We can also generalize the claim made as part of the proof to say that if G (a subgroup of R) has a least positive element then it's cyclic, which is kind of interesting by itself.
minor nitpick, you probably should have used image, instead of codomain, since we can choose any codomain, as long as it includes the image of the function.
@@fuxpremier yes, it is. The part where we used the periodicity of sin, was to show that the image of the natural numbers under sin is the same as the image of the dense set we constructed.
In 10:58 is asssuming that nd≠0. It needs some justification. Notice if nd=0 then 1=md, with m,d integers, so the unique way it happens is when m,d are units in Z, i.e. m=±1 and d=±1, but then g=m (cause dg=md) and G===Z, which is not true since G contains irrational numbers like 2π when a=0 and b=1.
You can also say since g = m, and it's the least positive element, then g = 1. But we can easily find a positive element of G smaller than 1 (e.g. -6 + 2*pi), so it's a contradiction.
Sweet. I went to a maths specialist high school and one of the coolest things I remember from the year 9 syllabus was properties of sin(n) for positive integers n: things like how it's always irrational, pseudorandom, never repeats (one-to-one), and indeed that it's dense on [-1:1].
2:14 “Since G does not have a least positive element we should be able to find a g in G such that g < E for a given E.” That actually sounds like it needs to be proven. Hypothetically, for example, not knowing anything else about groups or G, it may be that the elements of G form a set that has an infinimum but not a minimum (e.g. looks like numbers of the form 1 + 1/2^n plus 0. That set, if it were a group, would have no least positive element (assuming we’re not defining 0 as positive) but its infimum would be 1, not 0, so of E=1/4 for example there would be no positive element of G less than E. Note that I’m not saying the claim in the video is wrong, I’m just saying it doesn’t immediately follow at face value from the definitions in the video, it’s something that should be proven.
Yeah, but the set { 0, 1+1/2, 1+1/4, ... 1+1/2^n } where n ∈ ℤ, isn't an additive group, as you can see that 1+1/2 + 1+1/2 = 3, which pretty clearly isn't a member of the set.
@@RexxSchneider That’s true of the set I mentioned, but consider the set which is all numbers which are equal to finite sums of elements from that original set. (E.g. 3 = 1.5 + 1.5 so 3 is in the new set). That set is now closed under addition but still has no minimum positive element, and the infimum of its positive elements is 1, not 0. Of course, it’s still not a group yet, but not because of not being closed under addition. It’s because, at least using the standard addition operation for the Reals, it doesn’t include the inverses of its elements. But my point here is that you do have to prove in general that no such group can exist which has an infimum of its positive elements but no minimum and the infimum of the positives is not 0. (Note that one of the commenters above did prove it elsewhere in the comments, it’s worth looking for that if you haven’t done it yourself yet.)
@@BodyknockI did see that proof above, but you were arguing by example, and it was easy to show the flaw in that example simply by lack of closure under addition. I didn't need to prove that no such group exists. I don't _have to prove_ anything in general if you're arguing by example. What I do know is that whatever example you come up with, I can show it's not a group under addition.
one of the best videos on this channel yet, and that's saying a lot. especially nice how the density of G in the second lemma boiled down to pi being irrational
A plug for the hole in the first argument in Lemma 1. Assume that (0, ε) and G are disjoint. Since G is a group, the difference of any two elements of G is an element of G, and hence |g-h| >= ε for all g, h in G. Let N be the minimal positive integer such that there exists positive g in G such that g < Nε. This exists by the axiom of Archimedes as long as we assume that G contains at least one positive element. As G contains no least positive element, there exists h in G such that 0 < h < g. Thus, g-h >= ε or, equivalently, h+ε
There might be another solution: Consider m,n integers and x irrational. Further consider k as a natural number and the list of numbers , , …, and note they all are different, as x is not rational. „“ denotes the fractional part. We then split the interval [0,1) in k parts of equal length, so there are two numbers in our list in one section. Hence 0< - 0< < 1/k => 0lm+lnx is contained in (a,b). Write sin(n) = sin(n+2pim) = sin(x), where x can be in any interval for suiting integers m and n. *because pi is irrational Hence sin(n) takes all values sin(x) takes eventually, which concludes the proof. If there us anything wrong with this, just let me know
We're in the setting of infinite sets so {0} is automatically excluded, but yes, in order to be 100% correct, there needs to be mention of G being infinite
He wrote xor, which means exclusive or. It cannot be simultaneously both. A set G is either a subgroup, or it has a least positive element. It cannot be both. The lemma is correct.
11:00 I think it needs to be pointed out explicitly that g must be smaller than 1 (by showing that G contains at least one element that is between 0 and 1) _and therefore_ n cannot be 0 . Because if n can be 0 , then it may be possible to write 1 as d*g without violating the irrationality of π .
An alternative proof: First use the continued fraction expansion of 2π to construct rational numbers p_i/q_i that converge to 2π. Now |p_i - 2πq_i| converges to 0 (because a theorem about continued fractions says so; it does not already follow from p_i/q_i converging to 2π). To approximate x = sin(theta) in [-1,1], choose integers a_i such that a_i(p_i-2πq_i) is closest to theta. Note that the difference with theta will be smaller than |p_i-2πq_i|. Now take n_i = a_i * p_i, and we will have sin(n_i) = sin(a_i(p_i-2πq_i)) -> sin(theta) = x. Since for any x in [-1,1] we can find a sequence n_i with sin(n_i) -> x, it follows that sin(N) is dense in [-1,1].
He neglected to point out that since G contains elements between 0 and 1 (for example, -6+2π), g must be smaller than 1, and therefore if g = m+2nπ for some integers m and n, then n cannot be 0. Or alternatively, since it was already established that G = , i.e. g is a generator of G, then if n = 0 then g can not generate (for example) 2π, even though 2π is a known element of G; therefore, if g is indeed the smallest positive element of G, then n cannot be 0.
The theorem, his entire proof, and all the lemmas included in this proof are fantastic and really interesting. Nonetheless, there is a passage which is not clear in my humble opinion, and probably should be clarified. At 2:28 you said that for all ε > 0, we can find g in G such that 0 < g < ε because G does not have a least positive element. This is not true, because not every subset S of R admits a least positive element. Take for example S = (1, 2], it has not a least positive element; instead, S has a positive infimum, which is 1. So, the lemma should be rewritten as: "If G is an additive subgroup of R, then either G is dense in R or G has a positive infimum". I don't know, maybe for every additive subgroup G of R a positive infimum of G always belongs to G, but this result, if true, is not trivial for me.
Acutally, I do believe it appears there lacks a precision in the demonstration here. But be reassured that the lemma is true as written as, yes, a subgroup not admitting a least positive value admits a positive (or potentially null) infimum, the possibility to get distinct elements arbitrarly close to this infimum lets you construct differences of elments of G (which are thus still elements of G) arbitrarly close to 0. So the positive or null infimum appears to necessarly be 0.
@@MarcoMate87 Hmm, Yes, I don't feel like it's relevant when proving this point as we only work on (strictly) positive values of the subgroup here. But it can be satisfying to notice thus that the infimum is always an element of the group
Challenge: given y in [-1 ,1] find a *monotonic* sequence a_k = sin ( n_k) (n_k being a sequence of integers) that converges to y. In other words, if the sequence is monotonichally increasing, we would have sin(n_0)
if n = 0 then m is the least positive number in the intgers 1 but If we take the floor of pi we get 3 so 3 < pi < 4 So 0 < pi - 3 < 1 So 1 cannot be the minimum positive constant of a + 2n(pi) and there is a contradiction.
@@fonaimartin98 I actually made a mistake in my workings, it was 2pi so I should have used 6 haha, the logic still works though! And the concept can be used to prove generally that for some integer "k" not equal to 0 the set {kn + mr | n,m are integers} is dense if "r" is any irrational number.
Does anybody have a constructive proof of this? i.e. actually finding n such that |sin(n) - y| < epsilon, given an arbitrary y in [-1, 1] and epsilon > 0.
Use the continued fraction expansion of 2π to construct rational numbers p_i/q_i that converge to 2π. Now |p_i - 2πq_i| will converge to 0 (because a theorem about continued fractions says so; it does not already follow from p_i/q_i converging to 2π). To approximate x = sin(theta), choose integers a_i such that a_i(p_i-2πq_i) is closest to theta. Note that the difference with theta will be smaller than |p_i-2πq_i|. Now take n_i = a_i * p_i, and we will have sin(n_i) = sin(a_i(p_i-2πq_i)) -> sin(theta) = x.
One small issue with the second part of the proof of lemma 2: you are making the assumption that n and d are not equal to zero so that you can divide by them. Of course if d=0, then 1=dg=0 which is clearly false, but if n=0, then we could have g=m and 1=md which would be possible if g=m=d=1. Basically, you need to show that 1 is not the smallest positive number of the form a+2bpi. I suggest noting that for a=-6, b=1, we have -6+2pi = 0.28... which is positive and smaller than 1.
Lemma 2, "inside proof", hypothesis: why would the x be bounded above by a multiple of g? The whole point of being cyclic is that a certain multiple of g is smaller than the previous one. I am puzzled. (And as always, when something seems obvious, there is a non-zero chance it isn't at all)
It's important to note that although the projection of integers through sin onto the real numbers from [-1, 1] is dense, it does not span [-1, 1]. If it did, it would imply that there is a bijective mapping from the real numbers to the integers, which cannot work by pigeonhole principle since the integers are a subset of the real numbers.
The result is fairly easy to prove by counter-example. For instance, sin(n) | n ∈ ℤ, cannot have the value 0.5. That is because if sin(θ) = 0.5, then θ must be of the form π/6+2kπ or 5π/6+2kπ | k ∈ ℤ. In either case, θ is a non-zero rational multiple of π, which cannot be equal to any integer n. The same would hold for any rational non-zero value you chose.
"since the integers are a subset of the real numbers" is not a valid argument. For example, the positive real numbers are also a subset of the real numbers, yet there exists a bijective mapping between the positive real numbers and the real numbers. (For example: y = ln(x) .)
a+xb with x being irrational is also dense in R and this is equivalent to the fractional part of xb being dense in [0,1). There is another nice proof of this in Ebrahimian channel.
Since the set of prime numbers contains arbitrarily long arithmetic progressions, sin(p) is dense in [-1,1]. Edit: hmmm, it might not be as simple as this... 🤔
Great video! Though at first the formulation with XOR was a bit confusing, it would have been easier to understand if it was "subgroup G of the additive group over R is a dense set if and only if it doesn't have the least positive element"
By irrationality of pi, asin(y)+k*pi gets arbitrarily close to an integer. Therefore by continuity of sine, sin(round(asin(y)+k*pi)) gets arbitrarily close to y.
I'm new to this type of math (I've never encountered the mathematical definition of "dense" before), so this video is very enlightening to me. I guess that by a similar argument, we can prove that cos(n) is also dense in [-1,1] . But what about tan(n) ? tan(x) is not continuous, but it is continuous in the (sub)domain(s) where tan(x) is in the interval [-1.1 , +1.1], so it should give no problems getting arbitrarily close to any value between -1 and +1 , right? In fact, can we derive by a similar argument that tan(n) is dense on any finite interval [a,b] ? (i.e., any interval [a,b] where a and be are finite, real numbers; for example [a,b] = [41 , 1700] .)
So, there are *only* a countably infinite number of points in sin(n) but an uncountable in [-1,1]. Can we find one of the real numbers not in sin(n)? For example, is 1/π in sin(n)? If so which integer gives that value?
For example, 1/2 is not in sin(n), because sin(n) = 1/2 requires either n = π/6 + 2kπ = (1/6 + 2k)π , or n = 5π/6 + 2kπ = (5/6 + 2k)π for some integer k , which is not possible since (1/6 + 2k)π and (5/6 + 2k)π cannot be an integer for any integer value of k . For similar reasons, -½ , ±1 , ±½√2 , ±½√3 are not in sin(n) either. I don't know if there exists integer n such that sin(n) = 1/π .
Since each element of the set we're working with is generated by a single integer, this is a countable set. It's not too often that I've come across countable sets that are dense, other than Q of course. Are there some other interesting examples?
algebraic reals (includes Q), Z+pi Z, Z+e Z, Z+sqrt(2)Z, Z+sqrt(3)Z, Z+r Z for any irrational real r, diadic Q, and other forms that aren't coming to mind immediately.
d can't be zero because then 1 = 0 times something. It doesn't matter if n is zero because G is cyclic. (There are some weeds, because G being cyclic is a consequence of the initial assumption we're disproving.) (It's been a few decades since my last algebra class, so I may be misremembering things here.)
This is the proof I came up with: g (the least positive element of G) is equal to m + 2nπ. The number 7 - 2π is in G. This number is strictly between 0 and 1, so we have 0 < g ≤ 7 - 2π < 1. If n were zero, then g would be an integer, but that contradicts the preceding inequality, so n cannot be 0. And, as the comment above points out, d cannot be zero because then 1 = 0.
I don't think it was obvious. Like, sure, I can see it. But I wouldn't say it's obvious enough to leave off. It's also clear that in this video, Michael is doing a more rigorous proof than in some of his other videos
He didn’t prove that it doesn’t have a least positive element. He assumed it didn’t, and proved it was necessarily dense. There’s a bit of a hole in his argument when he assumes we can find g, but other than that it looks fine.
@@Keithfert490 I don't think you quite understood my comment. We're both referring to exactly the same proof. I'm clarifying what the proof was actually proving, and merely mentioning there was a hole. You're just reiterating that there was a proof of something.
@@JM-us3fr I think you are incorrect. He proved that G has no least positive element and then proved that sin(n) is dense on [-1,1]. Two separate proofs.
That's like fifth class mistake when you divide by a variable that may be zero and get no solutions... In that case, the "2*n*d = 0" scenario rules itself out after some argument, but that's not even that obvious. Also, "2nd" is second.
Thank you...i had a problem in sight via phone and your dark blackboard...if possible write in great font or use white blackboard...thank you respect from Morocco
Do you mean that, because an edge-connected square can be completely "covered" by a line with an irrational slope, the result in the video follows? If so, wonderful analogy.
@@cheedozer7391yes that is the same. Glueing together opposite sides of the square will give you a torus. If the continued fraction expansions of two irrational numbers have the same (infinite) ending, then the corresponding irrational flows on the torus are homeomorphic. I believe it is an open question whether this classifies all the irrational flows on the torus or whether maybe they are all homeomorphic, or something in between.
@@NC-hu6xd if you draw a straight line with slope alpha over a torus (or just a 1x1 square and "modulo 1" in both directions), then the line will never return to the origin if alpha is irrational, but it will necessarily get arbitrarily close to the origin, and you can easily see that this means the image of the line on the torus will be dense. I don't know if this really simplifies the approach of the video (which I haven't watched yet). You do need some kind of symmetry argument.
@@ronald3836If you watch the video, and then read the comments, you'll see that there is a trivial case that we have to exclude. Interestingly, your analogy also has a trivial case that fails to cover the square, and that is a line from the centre of the square with slope √2.
Now then Michael - would you see what you have done here! Just have a look at the comments below and above and nested betwixt and between. You have people discussing math in a very sophisticated way (at least it looked like that when I skip read a few comments) and there were no threats or insults too! This is a remarkable achievement in my opinion. Plus! Bringing cyclical subgroups as a subgroup of a subgroup in the reals on closed interval -1 to 1 is pretty extravagant. I hope it was deliberate! Over all impression: you did it again dude! You broke the rigid containers used to teach math and used that to further teach math. I hope we viewers are worthy of the talents on display.
It baffels me how this set G can be dense in R. I understand the proof, it makes sense to me, but I don’t get the intuition… is there some way of thinking about an algorithm to given any x approximate it with a+2b(pi) ? Sorry I am studying CS can’t stop thinking about algorithms…
There isn't any way in general of doing that approximation if a and b are integers, as you can clearly see that x=0.5 has no good approximation, for example. The a and b are not really part of the solution, since sin(a) = sin(a+2bπ) anyway. If you mean finding integer n such that sin(n) is an approximation to within epsilon of some value, where -1 < value < 1, I think that's a tough question to do analytically. Trial and error would get you an answer, eventually. In pseudocode: y = value eps = epsilon limit = 1e6 n = -1 for idx = 1 to limit do: if abs(sin(idx) - y) < eps then: n = idx break end if end for if n > 0 then: print n, sin(n) else print "try a bigger limit" end if That would check a million possibilities. Increase to whatever your computing power will stand if your epsilon is small. Given a target value, we know that an n exists for any epsilon, but the proof of existence isn't an algorithm for finding n as a function of and .
Look at the set S = { n mod 2π : n in Z } as a subset of [0, 2π]. It might be intuitively obvious to you that S is dense in [0, 2π] (even if proving it formally still needs a bit of work). S has infinitely many elements (because pi is irrational), and they are all spaced out more or less equally over the interval [0, 2π].
@@ronald3836 what does mod b mean if b is not an integer? I studied the basic concepts of number theory and „mod“ is there defined like this a mod b = r, where a=k*b+r and r and k are unique integers.
Huh? Eh, no. For example, sin(n) is not dense on the interval [-5, +5] . Because if we pick (let's say) y=3 (which is in that interval) and epsilon = 0.25 , then we cannot find n such that |sin(n) - y| < epsilon .
![I.N "HALLUCINATION" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/n5B5q1Hwt_U/mqdefault.jpg)
"Super duper close" is an excellent explainer for "dense".
I like "as close as we want"
@@MattHudsonAtx The phrasing you offer, or very slight variations on it, is essentially how I've always heard it described in "serious" settings.
It's not. "Close" refers naturally to a relation between elements of set - metric space, we could say, but let's keep it vague. On the other hand, "dense" qualifies a relation between two sets.
Obviously they are related concepts, and one very useful to understand the second, but your wording may be a bit too cowboyish for the standard mathematician.
It’s not immediately clear that there is 00, g:=|g_n-g_m|0 because the sequence is not eventually constant -- otherwise m in G, contradicting the assumption)
Therefore, for every ε>0, there is g in G_+ such that:
0
Why is m
@@JM-us3fr You're right, m just exists so we can take a limit approaching it, using the assumption that it is not attained inside G_+.
We're done after getting 0
@@tiripoulain Sounds good, thanks. And thanks for plugging this hole in Michael's argument
Yeah, this bothered me too. You can only conclude the existence of a g here with the requisite property by using the group properties of G, and all that was referred to was the mere lack of a minimum. The proof in the video is not sufficient on this point.
Your proof is again not quite sufficient. As @MatthiasGorgens notes below, if G is the trivial group, the lemma is fails, and your proof bakes in the assumption that G is non-trivial. (Specifically, if G is trivial, then G_+ is empty, and so the infimum does not exist. If we alter the proof to allow m to be zero, then the statement that the sequence is not eventually constant fails, but without contradicting the assumption.)
We can generalize Lemma 2 to note that { a + b y | a, b ∈ ℤ and y is irrational } is dense. And likewise any continuous function on a dense set is dense on its codomain. So then you could use that to perhaps find other interesting examples of dense functions on integers.
We can also generalize the claim made as part of the proof to say that if G (a subgroup of R) has a least positive element then it's cyclic, which is kind of interesting by itself.
For example e^in is dense on S_1.
We have explicitly made use of the periodicity of sin. Continuity is not a sufficient condition here.
was about to write exactly the same. quite cool facts
minor nitpick, you probably should have used image, instead of codomain, since we can choose any codomain, as long as it includes the image of the function.
@@fuxpremier yes, it is. The part where we used the periodicity of sin, was to show that the image of the natural numbers under sin is the same as the image of the dense set we constructed.
In 10:58 is asssuming that nd≠0. It needs some justification. Notice if nd=0 then 1=md, with m,d integers, so the unique way it happens is when m,d are units in Z, i.e. m=±1 and d=±1, but then g=m (cause dg=md) and G===Z, which is not true since G contains irrational numbers like 2π when a=0 and b=1.
You can also say since g = m, and it's the least positive element, then g = 1. But we can easily find a positive element of G smaller than 1 (e.g. -6 + 2*pi), so it's a contradiction.
First time seeing an XOR lemma in maths, so cool!
It's cool but maybe it's more conventional to write A if and only if notB instead of A xor B
@@angel-ig another way is also just writing the set of subgroups as a partition.
But the A iff not B is also what he proved in the lemma
@@angel-igAnd B if and only if notA.
@@АндрейДенькевич That is implied by my condition. That's why it's iff and not if
From where such people read and connect all the concepts of mathematics.
Sweet. I went to a maths specialist high school and one of the coolest things I remember from the year 9 syllabus was properties of sin(n) for positive integers n: things like how it's always irrational, pseudorandom, never repeats (one-to-one), and indeed that it's dense on [-1:1].
Wow. I wish I'd gone to a school like that.
This is the most confusing thing I have ever witnessed, having never taken math courses beyond linear algebra, d.e., and calc 3
2:14 “Since G does not have a least positive element we should be able to find a g in G such that g < E for a given E.”
That actually sounds like it needs to be proven. Hypothetically, for example, not knowing anything else about groups or G, it may be that the elements of G form a set that has an infinimum but not a minimum (e.g. looks like numbers of the form 1 + 1/2^n plus 0. That set, if it were a group, would have no least positive element (assuming we’re not defining 0 as positive) but its infimum would be 1, not 0, so of E=1/4 for example there would be no positive element of G less than E.
Note that I’m not saying the claim in the video is wrong, I’m just saying it doesn’t immediately follow at face value from the definitions in the video, it’s something that should be proven.
Go read my comment to that effect :)
@@tiripoulain Thanks, I figured someone in the comments would post a proof. 🙂
Yeah, but the set { 0, 1+1/2, 1+1/4, ... 1+1/2^n } where n ∈ ℤ, isn't an additive group, as you can see that 1+1/2 + 1+1/2 = 3, which pretty clearly isn't a member of the set.
@@RexxSchneider That’s true of the set I mentioned, but consider the set which is all numbers which are equal to finite sums of elements from that original set. (E.g. 3 = 1.5 + 1.5 so 3 is in the new set). That set is now closed under addition but still has no minimum positive element, and the infimum of its positive elements is 1, not 0.
Of course, it’s still not a group yet, but not because of not being closed under addition. It’s because, at least using the standard addition operation for the Reals, it doesn’t include the inverses of its elements. But my point here is that you do have to prove in general that no such group can exist which has an infimum of its positive elements but no minimum and the infimum of the positives is not 0. (Note that one of the commenters above did prove it elsewhere in the comments, it’s worth looking for that if you haven’t done it yourself yet.)
@@BodyknockI did see that proof above, but you were arguing by example, and it was easy to show the flaw in that example simply by lack of closure under addition. I didn't need to prove that no such group exists. I don't _have to prove_ anything in general if you're arguing by example.
What I do know is that whatever example you come up with, I can show it's not a group under addition.
one of the best videos on this channel yet, and that's saying a lot. especially nice how the density of G in the second lemma boiled down to pi being irrational
its kind of intuitive too, that you can approximate all possible decimal expansions just by cycling pi and taking of the integer part
A plug for the hole in the first argument in Lemma 1. Assume that (0, ε) and G are disjoint. Since G is a group, the difference of any two elements of G is an element of G, and hence |g-h| >= ε for all g, h in G.
Let N be the minimal positive integer such that there exists positive g in G such that g < Nε. This exists by the axiom of Archimedes as long as we assume that G contains at least one positive element.
As G contains no least positive element, there exists h in G such that 0 < h < g. Thus, g-h >= ε or, equivalently, h+ε
Thanks! Really liked this video and your other recent one about Fermat Little Theorem
1:09 ruclips.net/video/wuT8I97kYMM/видео.htmlm1s
15:42 Good Place To Stop
Did you create this account just to comment to Michael Penn's videos? 🤔
@@deepani.a Yes i did 😂
Very nice explanation of a beautiful result !
There might be another solution:
Consider m,n integers and x irrational.
Further consider k as a natural number and the list of numbers , , …, and note they all are different, as x is not rational. „“ denotes the fractional part.
We then split the interval [0,1) in k parts of equal length, so there are two numbers in our list in one section. Hence
0< - 0< < 1/k => 0lm+lnx is contained in (a,b).
Write sin(n) = sin(n+2pim) = sin(x), where x can be in any interval for suiting integers m and n.
*because pi is irrational
Hence sin(n) takes all values sin(x) takes eventually, which concludes the proof.
If there us anything wrong with this, just let me know
This is easier/more intuitive and avoids the gap in the video's proof.
Lemma 1 seems wrong? {0} is an additive sub-group of R, but it's neither dense nor does it have a least positive element.
0 is the least element of G. So this is not a problem, but Z on the other hand?
We're in the setting of infinite sets so {0} is automatically excluded, but yes, in order to be 100% correct, there needs to be mention of G being infinite
@@louisreinitz5642 Z has a least positive element.
That wasn’t my issue. My issue was that the elements of G could be bounded away from 0, making his g not exist.
He wrote xor, which means exclusive or. It cannot be simultaneously both. A set G is either a subgroup, or it has a least positive element. It cannot be both. The lemma is correct.
11:00 I think it needs to be pointed out explicitly that g must be smaller than 1 (by showing that G contains at least one element that is between 0 and 1) _and therefore_ n cannot be 0 .
Because if n can be 0 , then it may be possible to write 1 as d*g without violating the irrationality of π .
An alternative proof:
First use the continued fraction expansion of 2π to construct rational numbers p_i/q_i that converge to 2π. Now |p_i - 2πq_i| converges to 0 (because a theorem about continued fractions says so; it does not already follow from p_i/q_i converging to 2π).
To approximate x = sin(theta) in [-1,1], choose integers a_i such that a_i(p_i-2πq_i) is closest to theta. Note that the difference with theta will be smaller than |p_i-2πq_i|. Now take n_i = a_i * p_i, and we will have sin(n_i) = sin(a_i(p_i-2πq_i)) -> sin(theta) = x.
Since for any x in [-1,1] we can find a sequence n_i with sin(n_i) -> x, it follows that sin(N) is dense in [-1,1].
Think you also can use Weil’s Equidistribution theorem
10:39 Why not just set md = 1 and nd = 0 ? Possible if you set d = 1, m = 1, n = 0
Then also pi = 0/0 which isn't a rational number anymore
He neglected to point out that since G contains elements between 0 and 1 (for example, -6+2π), g must be smaller than 1, and therefore if g = m+2nπ for some integers m and n, then n cannot be 0.
Or alternatively, since it was already established that G = , i.e. g is a generator of G, then if n = 0 then g can not generate (for example) 2π, even though 2π is a known element of G; therefore, if g is indeed the smallest positive element of G, then n cannot be 0.
@@yurenchuMakes sense, but feels kind of glossed over.
@@geryz7549 I agree, I wrote the same thing two days ago.
but it's not pi either.
Very clean! Happy i’m at the point where this stuff makes sense to me
Super cool! I’ve always wanted to see a proof of this fact
The theorem, his entire proof, and all the lemmas included in this proof are fantastic and really interesting. Nonetheless, there is a passage which is not clear in my humble opinion, and probably should be clarified. At 2:28 you said that for all ε > 0, we can find g in G such that 0 < g < ε because G does not have a least positive element. This is not true, because not every subset S of R admits a least positive element. Take for example S = (1, 2], it has not a least positive element; instead, S has a positive infimum, which is 1. So, the lemma should be rewritten as: "If G is an additive subgroup of R, then either G is dense in R or G has a positive infimum". I don't know, maybe for every additive subgroup G of R a positive infimum of G always belongs to G, but this result, if true, is not trivial for me.
You seem to be mixing up arbitrary subsets of R and additive subgroups of R.
@@canonicaltomAbsolutely not. Every single sentence I wrote is correct, but probably you are not a mathematician.
Acutally, I do believe it appears there lacks a precision in the demonstration here. But be reassured that the lemma is true as written as, yes, a subgroup not admitting a least positive value admits a positive (or potentially null) infimum, the possibility to get distinct elements arbitrarly close to this infimum lets you construct differences of elments of G (which are thus still elements of G) arbitrarly close to 0. So the positive or null infimum appears to necessarly be 0.
@@uitrepafraiche5097 and 0 must be in G, because it is an additive subgroup of R.
@@MarcoMate87 Hmm, Yes, I don't feel like it's relevant when proving this point as we only work on (strictly) positive values of the subgroup here. But it can be satisfying to notice thus that the infimum is always an element of the group
Challenge: given y in [-1 ,1] find a *monotonic* sequence a_k = sin ( n_k) (n_k being a sequence of integers) that converges to y. In other words, if the sequence is monotonichally increasing, we would have sin(n_0)
Projecting a continued fraction logo onto the clouds with a spotlight...
it’s SO dense, every single subinterval has SO many points going on
It’s like poetry; it rhymes.
Also, this proof, well, IT BROKE NEW GROUND
10:50 Can't n be 0? Then that wouldn't be a contradiction.
m and d are also integers, with n=0 you'll be saying that pi is an integer
if n = 0 then m is the least positive number in the intgers 1 but
If we take the floor of pi we get 3 so 3 < pi < 4
So 0 < pi - 3 < 1
So 1 cannot be the minimum positive constant of a + 2n(pi) and there is a contradiction.
@@kerbalprogram3405 that is actually the good place to stop, nicely spotted
@@fonaimartin98 I actually made a mistake in my workings, it was 2pi so I should have used 6 haha, the logic still works though! And the concept can be used to prove generally that for some integer "k" not equal to 0 the set {kn + mr | n,m are integers} is dense if "r" is any irrational number.
This is a great exercise, lots of fun details in there.
Does anybody have a constructive proof of this? i.e. actually finding n such that |sin(n) - y| < epsilon, given an arbitrary y in [-1, 1] and epsilon > 0.
Use the continued fraction expansion of 2π to construct rational numbers p_i/q_i that converge to 2π. Now |p_i - 2πq_i| will converge to 0 (because a theorem about continued fractions says so; it does not already follow from p_i/q_i converging to 2π).
To approximate x = sin(theta), choose integers a_i such that a_i(p_i-2πq_i) is closest to theta. Note that the difference with theta will be smaller than |p_i-2πq_i|. Now take n_i = a_i * p_i, and we will have sin(n_i) = sin(a_i(p_i-2πq_i)) -> sin(theta) = x.
that's quite a nice proof. the proposition seems obvious, but I'm sure that I would have had a hard time formalizing it. Thank you :)
Oh no, the analysis started and I immediately drifted off.
Right back to my undergrad days!
One small issue with the second part of the proof of lemma 2: you are making the assumption that n and d are not equal to zero so that you can divide by them. Of course if d=0, then 1=dg=0 which is clearly false, but if n=0, then we could have g=m and 1=md which would be possible if g=m=d=1. Basically, you need to show that 1 is not the smallest positive number of the form a+2bpi. I suggest noting that for a=-6, b=1, we have -6+2pi = 0.28... which is positive and smaller than 1.
Lemma 2, "inside proof", hypothesis: why would the x be bounded above by a multiple of g? The whole point of being cyclic is that a certain multiple of g is smaller than the previous one.
I am puzzled. (And as always, when something seems obvious, there is a non-zero chance it isn't at all)
11:15 would have be nice to prouve n≠0 , it is by contradiction because then g=1 but 0 < 7-2π < 1=g so g is not the least positive element of g
It's important to note that although the projection of integers through sin onto the real numbers from [-1, 1] is dense, it does not span [-1, 1]. If it did, it would imply that there is a bijective mapping from the real numbers to the integers, which cannot work by pigeonhole principle since the integers are a subset of the real numbers.
Although it's true that there is no bijection from N to R, it's not because of the pigeonhole principle.
For instance there is a bijection from N to Q
The result is fairly easy to prove by counter-example. For instance, sin(n) | n ∈ ℤ, cannot have the value 0.5. That is because if sin(θ) = 0.5, then θ must be of the form π/6+2kπ or 5π/6+2kπ | k ∈ ℤ. In either case, θ is a non-zero rational multiple of π, which cannot be equal to any integer n. The same would hold for any rational non-zero value you chose.
"since the integers are a subset of the real numbers" is not a valid argument. For example, the positive real numbers are also a subset of the real numbers, yet there exists a bijective mapping between the positive real numbers and the real numbers. (For example: y = ln(x) .)
You didn’t prove that n0. If G Is generated by g, you can show that 0
a+xb with x being irrational is also dense in R and this is equivalent to the fractional part of xb being dense in [0,1). There is another nice proof of this in Ebrahimian channel.
Is sin(p) dense in [-1,1] for p prime? What about sin(p^(p^p))?
Very cool question! I found it asked in questions 4637412 and 109782 of Math Stackexchange.
@@angel-ig Thank you for searching this.
Since the set of prime numbers contains arbitrarily long arithmetic progressions, sin(p) is dense in [-1,1]. Edit: hmmm, it might not be as simple as this... 🤔
Great video!
Though at first the formulation with XOR was a bit confusing, it would have been easier to understand if it was "subgroup G of the additive group over R is a dense set if and only if it doesn't have the least positive element"
By irrationality of pi, asin(y)+k*pi gets arbitrarily close to an integer. Therefore by continuity of sine, sin(round(asin(y)+k*pi)) gets arbitrarily close to y.
supeer duppeeer cool!!
I asked it previously as well, but what's going on with the editing? I've noticed the usual intro also isn't present in this video?
Magnificent thumbnail!
I'm dense too, but no need to prove that, as everybody knows! 😂
I thought this would be based on closed covers
Good practise would be to show that sin(1/n) is also dense
I'm new to this type of math (I've never encountered the mathematical definition of "dense" before), so this video is very enlightening to me. I guess that by a similar argument, we can prove that cos(n) is also dense in [-1,1] . But what about tan(n) ?
tan(x) is not continuous, but it is continuous in the (sub)domain(s) where tan(x) is in the interval [-1.1 , +1.1], so it should give no problems getting arbitrarily close to any value between -1 and +1 , right?
In fact, can we derive by a similar argument that tan(n) is dense on any finite interval [a,b] ? (i.e., any interval [a,b] where a and be are finite, real numbers; for example [a,b] = [41 , 1700] .)
I miss the board-tap transitions!
really cool video!
It would be interesting to find the probability density of this points in the interval
Totally irrelevant question: What kind of chalk do you use? Those colors really pop
So, there are *only* a countably infinite number of points in sin(n) but an uncountable in [-1,1]. Can we find one of the real numbers not in sin(n)? For example, is 1/π in sin(n)? If so which integer gives that value?
For example, 1/2 is not in sin(n), because sin(n) = 1/2 requires either n = π/6 + 2kπ = (1/6 + 2k)π , or n = 5π/6 + 2kπ = (5/6 + 2k)π for some integer k , which is not possible since (1/6 + 2k)π and (5/6 + 2k)π cannot be an integer for any integer value of k .
For similar reasons, -½ , ±1 , ±½√2 , ±½√3 are not in sin(n) either.
I don't know if there exists integer n such that sin(n) = 1/π .
Sin(p/q), with p,q elements of Z has many more points but still countably infinite. Besides 0, do we always get an irrational?
continuum hypothesis
Is the decimal part of sqrt(n) dense in (0-1) ?
Since each element of the set we're working with is generated by a single integer, this is a countable set. It's not too often that I've come across countable sets that are dense, other than Q of course. Are there some other interesting examples?
algebraic reals (includes Q), Z+pi Z, Z+e Z, Z+sqrt(2)Z, Z+sqrt(3)Z, Z+r Z for any irrational real r, diadic Q, and other forms that aren't coming to mind immediately.
At 11:10. i Wonder where you claimed that nd is Not zero
d can't be zero because then 1 = 0 times something. It doesn't matter if n is zero because G is cyclic. (There are some weeds, because G being cyclic is a consequence of the initial assumption we're disproving.)
(It's been a few decades since my last algebra class, so I may be misremembering things here.)
This is the proof I came up with:
g (the least positive element of G) is equal to m + 2nπ.
The number 7 - 2π is in G. This number is strictly between 0 and 1, so we have 0 < g ≤ 7 - 2π < 1. If n were zero, then g would be an integer, but that contradicts the preceding inequality, so n cannot be 0.
And, as the comment above points out, d cannot be zero because then 1 = 0.
@@owenbechtel @xizar0rg thank you
It seems to me that "G does not have a least positive element" is quite obvious. Why Michael put so much effort to prove it?
I don't think it was obvious. Like, sure, I can see it. But I wouldn't say it's obvious enough to leave off. It's also clear that in this video, Michael is doing a more rigorous proof than in some of his other videos
He didn’t prove that it doesn’t have a least positive element. He assumed it didn’t, and proved it was necessarily dense. There’s a bit of a hole in his argument when he assumes we can find g, but other than that it looks fine.
@@JM-us3fr not true. He did an entire proof of this before the actual connection to the sin function
@@Keithfert490 I don't think you quite understood my comment. We're both referring to exactly the same proof. I'm clarifying what the proof was actually proving, and merely mentioning there was a hole. You're just reiterating that there was a proof of something.
@@JM-us3fr I think you are incorrect. He proved that G has no least positive element and then proved that sin(n) is dense on [-1,1]. Two separate proofs.
TIL that the trivial group {0} is dense in R
Can this be used to prove that pi is irrational, or even transcendental?
No, because it is *depending on* the fact that pi is irrational (around 10:50).
That's like fifth class mistake when you divide by a variable that may be zero and get no solutions... In that case, the "2*n*d = 0" scenario rules itself out after some argument, but that's not even that obvious. Also, "2nd" is second.
That's really good
so am i
Thank you...i had a problem in sight via phone and your dark blackboard...if possible write in great font or use white blackboard...thank you respect from Morocco
May you post an affiliate link if this chalkboard?
he has it in his merch shop; however it appears to be currently sold out :(
But on amazon I only find the 36“x24“ chalkboard By marble Field. Michael‘s is much greater
How many of your students chose to come to the college just to witness you?
One of my acquaintance came up with a proof of this overnight... He was really a monster.
why wouldn't sin(n) be dense!? You'd expect that right ?
Because you screw up and set your calculator to degrees instead of radians, so now sin(n) is a finite set of 181 distinct values ? 🤔
Use of group theory is superfluous, unfortunately
Simply understanding on a torus with a line with an irrational slope is sufficient.
Do you mean that, because an edge-connected square can be completely "covered" by a line with an irrational slope, the result in the video follows? If so, wonderful analogy.
@@cheedozer7391yes that is the same. Glueing together opposite sides of the square will give you a torus.
If the continued fraction expansions of two irrational numbers have the same (infinite) ending, then the corresponding irrational flows on the torus are homeomorphic. I believe it is an open question whether this classifies all the irrational flows on the torus or whether maybe they are all homeomorphic, or something in between.
Can you explain the reasoning ? I absolutely do not get it
@@NC-hu6xd if you draw a straight line with slope alpha over a torus (or just a 1x1 square and "modulo 1" in both directions), then the line will never return to the origin if alpha is irrational, but it will necessarily get arbitrarily close to the origin, and you can easily see that this means the image of the line on the torus will be dense.
I don't know if this really simplifies the approach of the video (which I haven't watched yet). You do need some kind of symmetry argument.
@@ronald3836If you watch the video, and then read the comments, you'll see that there is a trivial case that we have to exclude. Interestingly, your analogy also has a trivial case that fails to cover the square, and that is a line from the centre of the square with slope √2.
Now then Michael - would you see what you have done here! Just have a look at the comments below and above and nested betwixt and between.
You have people discussing math in a very sophisticated way (at least it looked like that when I skip read a few comments) and there were no threats or insults too!
This is a remarkable achievement in my opinion.
Plus! Bringing cyclical subgroups as a subgroup of a subgroup in the reals on closed interval -1 to 1 is pretty extravagant. I hope it was deliberate!
Over all impression: you did it again dude! You broke the rigid containers used to teach math and used that to further teach math.
I hope we viewers are worthy of the talents on display.
It baffels me how this set G can be dense in R. I understand the proof, it makes sense to me, but I don’t get the intuition… is there some way of thinking about an algorithm to given any x approximate it with a+2b(pi) ? Sorry I am studying CS can’t stop thinking about algorithms…
He didn't show an algorithm, you have to use different metod to find a and b for given x and epsilon.
@@Danielf667 thats why I asked if someone could help me out with this…
There isn't any way in general of doing that approximation if a and b are integers, as you can clearly see that x=0.5 has no good approximation, for example.
The a and b are not really part of the solution, since sin(a) = sin(a+2bπ) anyway.
If you mean finding integer n such that sin(n) is an approximation to within epsilon of some value, where -1 < value < 1, I think that's a tough question to do analytically.
Trial and error would get you an answer, eventually. In pseudocode:
y = value
eps = epsilon
limit = 1e6
n = -1
for idx = 1 to limit do:
if abs(sin(idx) - y) < eps then:
n = idx
break
end if
end for
if n > 0 then:
print n, sin(n)
else
print "try a bigger limit"
end if
That would check a million possibilities. Increase to whatever your computing power will stand if your epsilon is small.
Given a target value, we know that an n exists for any epsilon, but the proof of existence isn't an algorithm for finding n as a function of and .
Look at the set S = { n mod 2π : n in Z } as a subset of [0, 2π]. It might be intuitively obvious to you that S is dense in [0, 2π] (even if proving it formally still needs a bit of work). S has infinitely many elements (because pi is irrational), and they are all spaced out more or less equally over the interval [0, 2π].
@@ronald3836 what does mod b mean if b is not an integer? I studied the basic concepts of number theory and „mod“ is there defined like this a mod b = r, where a=k*b+r and r and k are unique integers.
Since it's dense on [-1,1] and continuous, it's also dense on every closed interval [-a,b] of the real numbers
Huh?
Eh, no. For example, sin(n) is not dense on the interval [-5, +5] . Because if we pick (let's say) y=3 (which is in that interval) and epsilon = 0.25 , then we cannot find n such that |sin(n) - y| < epsilon .
Does the following happen because 11/7 is approximately π/2?
n sin(n)
344 -0.99999033951
355 -0.00003014435
366 0.99999007269
Yes, but also because the graph of y = sin(x) is quite "flat" (horizontal) around the points where y = ±1 .