yeah lol. and i love that someone pointed out that infinitesimal size and only rational coordinates = 0 chance of you hitting one, as long as you foot is infinitesimal size. but never mind that, the practicality is allready out the window with this XD for starters with a infinit amount of mines you have a 100% chance of them all blowing up by random. That or they dont have such a weak explosion we back at you just walking to the boat with no problems to worry about.
I assume that each mine is infinitely small? Yes, they would need to be in order to fit into a finite space. So, if only the rational coordinates have a mine of infinitesimal size, that would mean that the total area covered by mines is actually zero, so one could walk about with impunity, safe in the knowledge that you'll never be blown up.
Yeah, I can't tell you the number of times this exact scenario comes up. Why, just last week I was trapped on an infinitely small sailboat with infinite mines on board and some mathematician evenly distributed them all over this cartesian plane we were passing by. The whole thing made me late to my meeting with the Natural Numbers; it was a cardinal sin, I tell you.
The smooth curve (x, x+pi/4(x-x^2)) starts at (0,0), ends at (1,1), and has no interior points with both coordinates rational. For, if x is rational and x-x^2 is not zero, obviously y is irrational. Otherwise x is irrational or 0 or 1, and we don't care if y is rational or not. After considerable thought, I see that this argument is valid: since every rational point intersects a line through the origin with rational slope, then every line through the origin with irrational slope cannot intersect any rational points. The worry is that when the irrational slope is say √2/2, then when x=√2/2, y is rational. But the point is still not a rational point.
@@pafnutiytheartist “y = x^a with irrational a” is not sufficient. For example, if a = log₂(3) then (1/2, 1/3) is on the curve. However, if a is algebraic (as well as irrational), then the Gelfond-Schneider theorem applies and any rational x corresponds to a transcendental y.
@fatih3806 Correct. Proof: Assume 0 < x < 1. Obviously 0 < 2^x-1 < 1, since 2^x is strictly increasing, 2^0=1, and 2^1=2. We need (given 0 < x < 1) If 2^x ∈ ℚ, then x ∉ ℚ, which is equivalent to the contrapositive, If x ∈ ℚ, then 2^x ∉ ℚ so proof of one proves the other. Since x ∈ ℚ, x = p/q with p,q ∈ ℕ and q > 1 and p < q and q does not divide p. Then 2^x = 2^(p/q) is a fractional power of 2. But all fractional powers of 2 are irrational. Hence 2^x ∉ ℚ. So in general, either x ∉ ℚ or 2^x ∉ ℚ.
My first thought was "there's infinitely more free spots than mines so just pick a random direction and send it," then I realized you still need to reach the boat so I thought "just slap a second one in that hits your line and the boat and you're golden," was not disappointed with the conclusion
My first thought was similar: "When people walk, they almost never are very precise with their movements. Even if you WANTED to step on a mine, you'd probably fail, because there's infinitely more space that's mineless than mined. Even the probability that your subatomic particles touch an exact point where a mine is, is minuscule. So, just walk. Even try walking in a straight line directly there, because you'd never get it straight enough to ACTUALLY hit a mine."
Morphocular: "But for an idea that seems like the most up in the clouds of theoretical musings it finds its way into some pretty practical fields of study." Also Morphocular: let's talk about how to escape an infinitely dense mine field, with 1-dimensional mines, on our 1-dimensional toes. 😂😂👊🤪
@@reidflemingworldstoughestm1394 Yes but how can it be applied to solving problems if the solution requires the assumption that your minefield is infinite
@@Tonatsi ?? The problem in the video is solved in this manner. Once the relationships and processes are understood, insight is gained, and the problem is solved. The practical value gained from solving the problem is the insight required to solve it -- whether you discovered it yourself, or simply learned it from instruction.
Pick an irrational slope m so that y = mx. Then travel to the intercept with the line y = 1 - x at point x=1/(1+m) y = m/(1+m). You've made it halfway to the ship. By symmetry, you can now switch your slope to 1/m also without running into rational points. The line will now be y = (x-1)/m + 1 For any rational x, y =mx must be irrational, else m = y/x would be a rational number, which contradicts the assumption of irrational m. Thus, x and y cannot both be rational.
I took a similar tact using the same idea of irrational slopes. Note you chose an intercept at y=1-x, so neither x nor y can be rational at the intercept because if either were, they both would be. But, depending how far back we go to first principles, that can create the perhaps slightly awkward requirement there be non-rational (x,y) points on the line y=1-x (because some would argue there is no such intercepting point given the two lines). This can be avoided by not mentioning y=1-x in the first place. Of course, you will end up there regardless when you invert the slope to create the second line running through the point (1, 1).
Indeed. In addition, your proof is constructive, i.e. it tells you where to go, whereas the proof in the video is not. The video and the proof in it are still nice, though.
I love how SoME 1 has led to so many new math channels being created. please make more videos, I just cant wait to say I've been following you since 150 subs when you inevitably get popular.
I like how you can use Fermat's Last Theorem to solve this. x^n + y^n = z^n has no integers solutions with n>2. So let n=3 and z=1 to get x^3 + y^3 = 1, rearrange to get y = cbrt(1 - x^3). Finally flip it vertically with y = 1 - cbrt(1 - x^3), and you have a path that never hits a mine (since if x is rational, y cannot also be rational without finding a solution to x^3 + y^3 = z^3). Obviously there are many other paths that work, but I think this one is quite elegant since it brings in such a famous theorem.
In retrospect, this might not have been the best example of a "practical" application of cardinality, but given just how abstract the concept is, I guess I figured even an infinitely dense minefield of point-like mines might qualify as an application! Though I'm sorry if the introduction was misleading. I did make a follow on video where I tried to present a more serious problem where cardinality plays a pivotal role. Not sure if I succeeded, but you can find it here if you're interested: th-cam.com/video/uLja-yAwuCI/w-d-xo.html
@@TrueBladeSoul in theory, not only is it possible, it is guaranteed that you will hit something with *every* step you take, unless we assume that we touch the ground at a single point. This is precisely what it means that rational numbers are dense in the set of real numbers. On the other hand, when we walk, we don't drag our feet on the ground, i.e. we don't make a continuous path on the ground, but take steps. Therefore, one could simply step over the mines along the diagonal connecting the points (0,0) and (1,1). What I want to say, it's pointless to look for the shortcomings of the analogy of this problem with navigating a minefield, since it breaks immediately if we are to take it literally. The only purpose of the minefield story is that it's a cute way to introduce the problem to general audience, the problem being: find a continuous curve inside the square connecting (0,0) and (1,1) that doesn't cross any point that has both coordinates rational, except the starting and finishing point.
well if I remember correctly, in physics (fundamental) particles are modelled as being point-like / infinitely small, so it's not that bold of an assumption
Another consequence of what you've outlined here: If you pick a random point in your minefield, the chances of hitting a mine are exactly zero! It's actually a pretty safe place. Here we have another paradox. In spite of the infinity of rational numbers between 0 and 1, you can't pick any one of them at random if you're choosing from the set of all real numbers in that interval. Here's a proof. Consider the decimal expansion of the rationals. They all eventually end in an infinitely repeating sequence (even if that sequence is all zeros). But when you pick a real number at random, and look at its decimal expansion, it will just be a random sequence of digits, which will certainly NOT end in an infinitely repeating sequence. You may see a repeating sequence somewhere to start with, but eventually (with probability increasing with each next digit) the repeating sequence will break. In short, your random number was irrational.
Randomly selecting a rational number out of all reals has 0 probability, but it doesn't make it impossible. In fact, fix a real number x. Probability to randomly select x out of all real numbers is 0. This is true for any x. But, if I do randomly select a number, I will in fact get some number, so an event that had 0 probability a priori just happened.
@@Ennar You need to integrate over sets of points to get useful answers. The probability integral over all rationals is zero, the one over the irrationals is one (provided your integral formalism doesn't break with disjoint sets like that)
Seems to me that there's a problem not with your conclusion but with your argument. Yes, at any given point along the decimal expansion the current sequence might break, but that doesn't mean that the number is definitely irrational. If you see 0.1361361368..., you might just be looking at a rational number with a repeated 10 decimal places, rather than the repeated 3 it initially looks like. More generally, no matter how many digits past the decimal point we've "observed", there are still infinitely many rational numbers that start with this sequence, and the same will hold true no matter what the next digit turns out to be. I think part of the problem here lies in reasoning sequentially (i.e. about the "next digit") with something that has been chosen oracularly, with an infinite amount of information pre-determined.
@@hughcaldwell1034 Forget what we "observe" as we look at the decimal expansion. The fact is this: the decimal expansion of every rational number must eventually end in an infinitely repeating sequence. Finding a sequence in a rational number's expansion that repeats many times, but not infinitely many times, means the infinitely repeating sequence has yet to begin. So we ignore that sequence and keep looking. Actually, we don't even need to look. A rational number is defined as the quotient of 2 integers. Knowing those 2 integers it's not difficult to calculate exactly where the infinitely repeating sequence of digits begins. So, if you claim to have picked a rational number at random among the real numbers, I would ask you for the 2 integers that define that rational number, calculate the location of its repeating digits in the decimal expansion, and compare that with the decimal expansion of the number you claim to have picked at random. Eventually, if your number was truly chosen at random, your number's decimal expansion MUST deviate from the predictable repeating sequence of the rational number you claimed to have chosen.
To stay as far as possible from the mines, move along the slope (1+√5)/2 until you hit the line y=1-x, then move directly toward the point (1, 1) along the slope 2/(1+√5).
At first I thought, "Just step 'at random' and you have probably 0 of stepping on a mine," as the irrationals are infinitely denser than the rationals. But the conclusion that there are solid lines along which we may step continuously (which NEVER take a (Q,Q) value) was mind blowing!
I saw this problem for the first time on stackexchange, and you explained it in a really good way! For the curious people, here's the mathematically precise statement of what he proved in this video: Let A be countable a subset of R^2. Then R^2 / A is arc-wise connected.
Whoa, this is amazingly interesting and extremely well made. I always thought it was cool how different sets had the same infinite cardinality but this showed me a lot of really cool stuff I didn't know. Liked and subbed
My first approach was to use a segment of a translated ln(x) function such that it passed through both (1,1) and the origin. That being y=ln(x(e-1)+1) for 0
How to Navigate Infinitely Dense Minefield: shoot one of the mines from a distance, the others will explode in a chain reaction, walk through the giant crater left over. or you know, its only 1km, just swim around the island and get onto the boat.
Found this channel this weekend and I'm enjoying the videos. :) I only have one small quibble with this one, the minefield problem was to "find a solution", but the video only actually proved that an infinite number of solutions exist which isn't quite the same thing. (It's still an interesting topic though!)
@@gregoryford2532 it absolutely doesn't. Infinity doesn't work the same way as finite sets. Imagine randomly picking a number between 0 and 1. The probability that you will pick irrational number is 1, yet you could also pick a rational number (with probability 0). So probability 1 doesn't mean certain, but almost certain, which is a subtle but important distinction.
Travel the curve ( t , (pi/4)^{t-1}*t) Note that the curve starts at (0,0), ends at (1,1), lies within the square for t in [0,1] and when t is rational (so the x-coordinate is rational), then the y coordinate is not because if it was, this would imply pi/4 (and hence pi) is an algebraic number (which it isn't).
Interesting; I definitely wouldn’t have thought of it that way. My approach would’ve probably been to just choose some combination of lines with irrational slopes, so one of the coordinates is always guaranteed to be irrational (and so not on a mine).
An even simpler and safer solution is to stay exactly 1 unit distance from either 1,0 or 0,1. This path results in a 1/4 circle, and since π is irrational, there cannot be any mines along this path. 🤯
This method you used is one way to show that removing countably infinite collection of points from the plane ( or in general R^n ) will give u a space that is still path connected!
Could you also choose your path to be a sin (or cos or tan)? Since for nonzero rational arguments they always result in an irrational number, so they won't go through the mines. It would be cool if you could find an infinitely continuously derivable function like the trig functions to be solutions
This problem is very pretty. I was worried about density of rationals, and indeed, in R, this is impossible: a point in R separates the space. But in R^2, a point does not separate the space (a nice topological difference between R and R^2). Thus it is indeed possible to find such a line that always has irrationals in one-coordinate. Clever!
My first thought was to move along the golden ratio, then upon reaching the halfway line (the line connecting 0,1 to 1,0), travel along the inverse of the golden ratio. Why, well I recall the golden ratio being the most irrational number, meaning I'd be the farthest away from an point described entirely by fractions of natural numbers, and logically so would its inverse, that way I can correct how off course I'd be if I only traveled along one or the other. Well, ok, my absolutely first idea was to toss a stone and set off the infinitely powerful explosion. But that has its own problems.
Your idea is correct, but you can take any irrational number instead of the golden ratio. Golden ratio is not the most irrational number. In fact, the golden ratio is just (1+ sqrt(5))/2 so I can even construct it using ruler and compass if I want. Hardly "the most irrational number" if you ask me. Also, you can never choose a point that will be farthest away from all the rational points, since for any point it's possible to find a rational point arbitrarily close to it. That's what it means for rational numbers to be dense in the set of real numbers.
@@Ennar The OP means 'most irrational' in the sense of Diophantine approximation, where indeed the golden ratio realizes the worst approximability by rationals.
Interestingly, when picking your real number for your slope, you do have to be careful because it's easy to not actually pick a truly random value. For example, any slope with a real number that can be written down with finite digits is rational. Additionally slopes with digits that form a repeating pattern are also able to be expressed as a ratio. You need to pick a number that can't be a ratio. Hmmmm. I wonder if the set of constructible/algebraic numbers is countably infinite
Yes, both sets are countably infinite. Algebraic numbers are roots of polynomials with rational coefficients. There are countably many such polynomials because there are countably many rational numbers, and each polynomial has finitely many roots. Even if each polynomial had countably many roots, we would still end up with countably many algebraic numbers since countable times countable is still countable.
You know, it occurred to me while watching this video that, if both coordinates must be rational to have a mine, then any grid line corresponding to an irrational number must have zero mines on it, creating a grid of mine-free points across the island. I kept waiting for that grid to become relevant, but it never happened
I was thinking the answer was something like “(sqrt(N))/(sqrt(2))” for the bottom, and similar (not exactly the same) for the right side Basically, since there’s always irrational spots connected in a line, you’re safe (since points are at rational spots only)
The problem with this is that it's possible for sqrt(N)/sqrt(2) to be a rational number. N = 2, 8, 32 is all rational. This also applies to when N < 0, like N = 1/2, 1/8, 1/32.
Considering that each mine represents a rational number, you can just walk from A to B since the amount of irrational numbers is so much bigger and so much more... omnipresent .
If you walk diagonally towards the boat, you can make sure you don't hit any mines by ensuring that every step's length is a fraction of 1, which always puts you at an irrational point, but if the distance you walk is sqrt(2)/2, then you'll end up right on [.5,.5]
So essentially If I think I'm not on a mine… I'm not on a mine? I could walk anywhere on this island, even the center, but once I realize "Hey I'm in the exact center of this island" I explode?
I was thinking you could take steps of some fractional distance of pi since you aren't taking infinitely small steps you don't need a continuous path. Then I realized you could just walk to the boat without thinking since the probability of landing exactly on any mine approaches 0.
I’m not sure if Zeno would be angry or proud. Or maybe he would feel as unsatisfied as I am about this solution. I think it’s the mapping from sets to the real world. I think it’s the unintuitive reasoning that reduces to “take all of the possible countably infinite paths and subtract those from the set of all paths then take one of those paths (twice, since the ends are both at rational coordinates).” I get the solution but it’s the mapping of real world objects onto sets that feels like a bit of a bait and switch.
My guess is sqrt(2), which is the length of the main diagonal. I know that line itself is not usable, but 2 other line segments infinitesimally angled away would do, and their combined length would surely have a limit of sqrt(2), if not actually reaching it.
@@CarlSmithNZ There would be an infinite sequence of paths approaching the limit of length sqrt(2) but never reaching it. That is to say, there is no "shortest" path as you can always find a shorter one.
Your video helped me understand something I was having trouble with with the old explanations. You showed the... I don't know the correct term but the thing where you take a diagonal of all the numbers to show a new number. For awhile I never understood why we can't take out the staring 0. And say that all natural/normal numbers are as numerous as the real numbers. By making the distinction at the beginning of calling it "countable" I can see that the countability aspect is important. It is an axiom of infinite math. We have decided that a type of infinity is the kind that we can list from 1 to infinity and while you can find new whole number that you didn't count by using that diagonal proof it is unhelpful because it is changing it from a countable infinity to an uncountable one. Thank you for clearing this up for me and thank you for posting this video it really helps?
Frankly, there is quite a straight solution to this problem. Just pick random point inside the square, which has one and only one irrational coordinate, and take a path from start to this point, following with a path from the point to the finish. You would succeed, because the start and destination points have rational coordinates, and so any given point of these two lines, the coordinates would be computed as a linear combination of the irrational and rational numbers, divided by a linear combination of rational numbers (which would be in fact irrational). So, you would always have a way (and infinitely many others) to complete the journey.
Excellent video! Thank you! I would construct my path with two straight lines, y=kx, and (y-1)=k(x-1). Just pick the slope of my lines to be any irrational numbers, for example, k=sqrt(3) and 1/sqrt(3). Start out on the 60-degree line, and finish the second half of the path on the 30-degree line. Since my slope k is irrational number, y=kx, x and y can't be both rational at the same time.
A lot of people don't consider that depending on the scope of a problem, you can approximate some values to infinitely large or infinitely small instead of measuring. That is the true real world application of infinities and limits. For example, calculating focal length. If you are using light from very far away, you can approximate to infinity.
As hard as the problem seems at first, the same countability argument shows that the following strategy reaches the boat with probability 1. Pick a random amount of time to walk and a random direction to walk in. When the time is up, walk straight toward the boat.
Well, it seems to me that we have basically 0 chance of hitting a mine at all, no matter how hard we try unless we know the exact position of atleast one. Technically, there is a chance but there is a huge BUT. Also, it's worth mentioning humans have legs and not wheels, so our path is not even a line but merely a finite set of close positioned points which makes the chance even less than that.
My guess: You can move across on a diagonal with a slope that's a multiple of an irrational number, like y=(pi)x. Of course, because the optimal path is 1:1, you will have to switch direction to some irrational number that has a different slope to reach the goal.
A method I came up with: Walk on a line with slope pi This ensures that for any distance traveled results in a y coordinate of a multiple of pi. If the multiple forms a rational number y coordinate then the multiple has to also contain pi making it an irrational number, in this case the x coordinate is an irrational number. Continue walking until intersecting the diagonal of the island, at this point start walking at a slope of 1/pi, this has the same effect of irrational coordinates and also will lead to the opposite corner. This can be used with any irrational slope eg. e, pi/2 etc.
I have returned, please keep making videos I love your animations and explanations. My only criticism is when giving the background knowledge/introduction the dense information can be hard to follow.
@@morphocular the way you explained background seems to work well as a refresher though. But If it's the first time you're seeing it then yeah it might a bit hard to follow.
An alternative way to find a path is to look at the part of the curve |x|^3 + |y-1|^3 = 1 that connects zero and 1. (f(t) = (t, cuberoot(t^3-1)+1) Now suppose x and y are rational : then you get, writing x = a/b and y = c/d, (ad)^3 + (bc-bd)^3 = (bd)^3, which contradicts Fermat's last theorem. Therefore, this path is safe :)
Sure sizes of infinities seems like something with no practical applications, but imagine if you were on an island with infinitely many infinitely dense land mines. Now you see just how practical this concept is in a real world situation.
@6:37 "and I can just play this trick again with the new list". The fun thing is that not only is this new number not in the new list, but because the new list contains all numbers in the original list, it also cannot be in the original list. So no matter what the original list was, we can generate not just one, but an infinte number of numbers not in that list. Worthwhile noting that the set of all finite decimals is countable: you can just write 'em backwards and get a unique integer.
wait this kinda makes me think of what if there was a portal on both ends of the line but the person can pass threw the portal traveling the same distance while not being to physicaly touch the mines mind blown cool vid Isubed.
It only took a few moments looking at the minefield puzzle for me to conclude: "Just walk along a horizontal irrational number close to 0 and a vertical irrational number close to 1". Sure, it doesn't explain how to get to those corner points before 0,0 and 1,1, but that's most of the path. Of course, I also scored a 32 (out of 36 max) on my Math SAT, so I have a better grasp of math than most people without being a mathematician. I've actually forgotten how to do derivatives myself.
The diagonal from 0,0 to 1,1 has infinitely many free points, such as sqrt(2),sqrt(2). All you need to do is find the nearest multiple of sqrt(2), for instance 0.01 - any number will suffice - that is comfortable for you to step over to (your stride length, as a function of sqrt(2)). Simply take enough steps (approximately the exact number of steps it would take you to casually walk the diagonal) of that length along the 45 degree diagonal, and you'll have the shortest distance that guarantees no mines.
As soon as you stated the problem I came up with a different solution. We know from Fermat's last theorem that there are no positive integer solutions to A^3 + B^3 = C^3. If we set B=C-A (with C>A) that means there are also no positive integer solutions to A^3 + (C-A)^3 = C^3, or equivalently (A/C)^3 + (1 - A/C)^3 = 1. That also means there are no positive rational solutions to t^3 + (1-t)^3 = 1. Therefore if we walk along the curve x=t^3, y=(1-t)^3 we won't hit a mine. That curve goes from the top left to the bottom right, but we can easily flip it upside-down and walk along the curve x=t^3, y=1-(1-t)^3 instead, which solves the problem. Interestingly, x=t^n, y=1-(1-t)^n will work for any n greater than 2, but if we choose n=2 we will hit an infinite number of mines, corresponding to the Pythagorean triples.
Another way I think of uncountable infinity is how there isn’t really a starting point to even begin counting the real numbers. You can arguably start with 0.000…. but the next number cannot even be described without using infinity. Theoretically it would be 0.000…001 where the dots represent an infinite string of zeros. However that isn’t really even a number because it assumes that the zeros end at some point but they don’t.
Good puzzle. I was a math major and I fell into the trap of figuring out how some monstrosity like the Weierstrass function would help. Only once you insisted the answer was simple did I think “wait, just take a line of slope π.”
while watching this all I could think of is how similar this is to faster than light space travel and how the odds of you hitting a celestial object are practically zero unless you're aiming for it
One way to "smoothly" go from 0,0 to 1,1 would be y = SUM[n=>inf](x^(1/n)/2^n) which does hit 0,0 and 1,1 while guaranteeing that any rational number that is arbitrarily large n-th power of 1/m will produce an irrational number(if it were to produce a rational number, it would mean there are finite prime numbers, which isn't true). And the minefield's density is actually zero because infinity is weird, meaning you have 0 chance to actually hit any of the mines, though this assumes 0 size mines and feet which breaks all known physics. 0 sized things can only be infinitely dense or infinitesimaly dense.
Maybe you could wander through the minefield without hitting a mine. But if you tried to program a robot to do it, they wouldn't be able to, since they could only choose a direction with floating point numbers, and those are rational.
How to win war using mines alone: 1. Plant infinite mines where enemy is 2. Since the mine density is infinite, the minefield turns into a black hole 3. your enemy got swallowed by the black hole
Thanks for the practical advice, I'll try to remember it next time I anger a vengeful god of mathematics and am forced to solve an esoteric problem involving set theory for my survival. It happens far more often than you might think.
The way I think about this (I might be wrong, this is just what I understand) is a countably infinite set of numbers means you can find a number greater than another, and not miss any. Uncountably infinite numbers make it impossible, as there is always something in between.
It's pretty easy to give a constructive solution too. The line (x, pi x) will never hit a mine. Where it meets (1 - x, 1 - pi x), switch over to that line which also doesn't have any coordinate with both rational numbers.
The first thing that come to my mind when you say an infinite amount of mines, is A) This will create an infinitely large universe that is composed of nothing but mines, if the space is infinitely dense with mines, and the can't overlap, like normal matter. B) We basically just have a black hole here.
I'd like to offer my favorite example of "weird math being practical", not that I personally need an argument against "your practical application is infinitely dense mine field" Imagine you are truck driver and you have to transport 2D balls (also known as disks, also known in real world as cylinders, for example PVC tubes or toilet paper). How can you pack cylinders to transport as many as you can? well, square grid is inferior to hexagonal, and it can be proven that (except few outliers based on how big you truck and items are) hexagonal packing is always the best. Let's go to 3D. How can you get the most regular balls into given space? I encountered interesting variation, asking how many same sized spheres can touch one given sphere? One solution is making a hexagon of 6 spheres (all touching center sphere), and then you add 3 spheres above and 3 bellow. What you get is perfectly stable form (resembling dodecahedron), so surely that is the best you can do? Well, yes and no. You got 12 spheres each toughing the center one and that is the best you can get, but try this: Make a pentagon of spheres, place center sphere inside, put another "pentagon" on top. You have 10 spheres toughing, and you can put one more on top and on bottom to get 12. But this time, the structure is unstable, there is "wiggle room". This makes it feel, that if you wiggle just right, you might be able to get one more sphere in there? Well, you can't but it is fascinating, that one solution (hexagon+3+3) is "perfect" and another (2xpentagon+1+1) hase room to improve). Turns out, this is inherent to 3D spheres and in packing, there is no general best way to pack number of spheres. We now get to the point I wanted to make in the beginning. There is interesting result, that in 8D, you can make hexagonal packing, that has so much space in it that another hexagonal grid of 8D-spheres fits in that, and that make perfect best packing, similar to what we can see with cylinders ("2D spheres"). The next perfect packing does not appear until 24D. So, how does it help a truck driver that 8D spheres can be packed perfectly? How does navigating mine field as mathematical point help real people? One way is, it helps you think about situations with given rules in a way that produces solutions. But 8D (and 24D) spheres and their packing HAS real world application. Next time you want to pay for something and use your credit card, you are using cryptography for save transactions and sphere packing is used to find efficient codes that allow for self-repairing code to be transferred through unreliable medium (air, overheating cables, overloaded servers,...) securely. Self-repairing code must be compact (so you don't need GB/s internet to send a text) but also repetitive (when some bits get lost you can still tell what the message was), and sphere packing has incredible relation with exactly that (through the use of matrices of all things). Not gonna say more, you can find more yourself of maybe "someone" makes a video about it? :D
I look at it a slightly different way : for a given line that passes through the origin, consider the x-coordinate of its intersection with the line y=1. The line passes through mines iff that x-coordinate is rational. Since the irrational numbers are infinitely more abundant than rational ones, the probability of such a random line hitting a mine is 0
The cardinality has almost nothing to do with the problem... Take the Cantor set, now since the rational numbers are countable, make a count of them (i.e. list them q_1, q_2,...). Now look at the set that is the union from i=1 to infinity of the Cantor set shifted by q_i (call the set of the countable union S). The new set is dense in R and since the Cantor set is uncountable, it is also uncountable. It is both dense and also uncountable on every interval in R no matter how small, and yet, the measure of this set is 0. This means that if instead you take the set S x S and look at the Island [0,1] x [0,1] where every pair (s, t) from this island has a mine where s, t are in S x S, you will still easily find a path that doesn't touch any mine! And so the cardinality isn't a deciding factor and the thing that really matters is the measure!
![PASULOL รามเกียรติ์ ตอนที่ 4 นนทกพบรัก [Ramakien Ep.4: Nonthok in love]](http://i.ytimg.com/vi/tMUQgmZzf_A/mqdefault.jpg)
As a resident of Bosnia I can indeed confirm that method of traversing an infinitely dense minefield
Finally! A use for what I've learned from this video!
This is actually so good im mad that less people actually get it
hope you guys learned something!
F
as a resident of the afterlife, I can tell you this method doesn't work
Mathematicians be like: "Some abstract math concepts have very real practical uses! For example, imagine a minefield with infinitely many mines..."
yeah lol. and i love that someone pointed out that infinitesimal size and only rational coordinates = 0 chance of you hitting one, as long as you foot is infinitesimal size.
but never mind that, the practicality is allready out the window with this XD for starters with a infinit amount of mines you have a 100% chance of them all blowing up by random.
That or they dont have such a weak explosion we back at you just walking to the boat with no problems to worry about.
@@MouseGoat If you wished to make abundantly clear that you do not understand what the word "infinite" means, you have suceeded.
And an uncountably infinity small foot
@@MouseGoat you're such a clown
I mean it helps with my Friday night drinking games
I really love how the solution was "well I mean just walk lol you'll probably be fine"
Don't try to use your infinitely precise mathematical compass though
@@ЕленаКостина-щ8ыif you’re brave enough to fight the monsters
You have infinitesimal extent, right?
I assume that each mine is infinitely small? Yes, they would need to be in order to fit into a finite space.
So, if only the rational coordinates have a mine of infinitesimal size, that would mean that the total area covered by mines is actually zero, so one could walk about with impunity, safe in the knowledge that you'll never be blown up.
if only this didn't require the formalisation of measure theory...
this is only true if you assume your foot is also an infinitesimal point
@@hyperpsych6483 we do assume that, though. the example in the video assumes it too
Is this really seen as a practical example?
@@kreynolds7544 Is the concept of an infinitely dense minefield practical?
Yeah, I can't tell you the number of times this exact scenario comes up. Why, just last week I was trapped on an infinitely small sailboat with infinite mines on board and some mathematician evenly distributed them all over this cartesian plane we were passing by. The whole thing made me late to my meeting with the Natural Numbers; it was a cardinal sin, I tell you.
I don't know, why this comment isn't popular but it made me chuckle :).
@@cemstrumental thank you, I'm unreasonably proud of it :)
@@MrSpeakerCone And you *should* be. I caught myself laughing out loud, and then I thought, "Wait a minute... this is just math stuff! Nicely done!
the fact that the sea monsters are just an infinite sum symbol (with eyes and teeth) is an incredible detail lol
Here be sumations
∑
N=0
They are all divergent series
@@JorgetePanete scp here be dragons reference?
@@irinaseif9691 I don't know about that one, but here be dragons was seen in really old maps in unknown places.
The smooth curve (x, x+pi/4(x-x^2)) starts at (0,0), ends at (1,1), and has no interior points with both coordinates rational. For, if x is rational and x-x^2 is not zero, obviously y is irrational. Otherwise x is irrational or 0 or 1, and we don't care if y is rational or not.
After considerable thought, I see that this argument is valid: since every rational point intersects a line through the origin with rational slope, then every line through the origin with irrational slope cannot intersect any rational points.
The worry is that when the irrational slope is say √2/2, then when x=√2/2, y is rational. But the point is still not a rational point.
It's a bit complicated. You can achieve the same with
y = x ^ a
Where a is an irrational number, like π or e
@@pafnutiytheartist “y = x^a with irrational a” is not sufficient. For example, if a = log₂(3) then (1/2, 1/3) is on the curve.
However, if a is algebraic (as well as irrational), then the Gelfond-Schneider theorem applies and any rational x corresponds to a transcendental y.
Doesn’t y=2^x-1 also work? Goes from (0,0) , (1,1) and doesn’t pass any rational coordinate
@fatih3806 Correct. Proof:
Assume 0 < x < 1.
Obviously 0 < 2^x-1 < 1, since 2^x is strictly increasing, 2^0=1, and 2^1=2.
We need (given 0 < x < 1)
If 2^x ∈ ℚ, then x ∉ ℚ,
which is equivalent to the contrapositive,
If x ∈ ℚ, then 2^x ∉ ℚ
so proof of one proves the other.
Since x ∈ ℚ, x = p/q with p,q ∈ ℕ and q > 1 and p < q and q does not divide p.
Then 2^x = 2^(p/q) is a fractional power of 2. But all fractional powers of 2 are irrational. Hence 2^x ∉ ℚ.
So in general, either x ∉ ℚ or 2^x ∉ ℚ.
My first thought was "there's infinitely more free spots than mines so just pick a random direction and send it," then I realized you still need to reach the boat so I thought "just slap a second one in that hits your line and the boat and you're golden," was not disappointed with the conclusion
i just saw that the edges of the island had a big enough gap for you to walk along them
My first thought was similar: "When people walk, they almost never are very precise with their movements. Even if you WANTED to step on a mine, you'd probably fail, because there's infinitely more space that's mineless than mined. Even the probability that your subatomic particles touch an exact point where a mine is, is minuscule. So, just walk. Even try walking in a straight line directly there, because you'd never get it straight enough to ACTUALLY hit a mine."
I love seeing these "practical applications" of the different sizes of infinities, especially when it's presented so well. Subscribed!
Morphocular: "But for an idea that seems like the most up in the clouds of theoretical musings it finds its way into some pretty practical fields of study."
Also Morphocular: let's talk about how to escape an infinitely dense mine field, with 1-dimensional mines, on our 1-dimensional toes. 😂😂👊🤪
It's comments like this that get people marooned on islands.
I'm glad I'm not the only one who saw that.
correction: 0 dimensional mines, the mines are points, not lines.
@@dragohammer6937 I'm deeply ashamed about this 🙁😲😲
@@things_leftunsaid I could explain that, but I'll send it to you it in Beth_ω days.
> argues that people find comparing infinities impractical
> uses "infinite minefield" as practical example...
Don't you think learning how infinities work and that that knowledge can be applied to solving problems is practical?
@@reidflemingworldstoughestm1394 Yes but how can it be applied to solving problems if the solution requires the assumption that your minefield is infinite
@@Tonatsi ?? The problem in the video is solved in this manner. Once the relationships and processes are understood, insight is gained, and the problem is solved. The practical value gained from solving the problem is the insight required to solve it -- whether you discovered it yourself, or simply learned it from instruction.
@@reidflemingworldstoughestm1394 I am still convinced this only exists to torture university students
@@Tonatsi 13th graders, possibly -- students, not so much.
Pick an irrational slope m so that y = mx. Then travel to the intercept with the line y = 1 - x at point x=1/(1+m) y = m/(1+m). You've made it halfway to the ship.
By symmetry, you can now switch your slope to 1/m also without running into rational points. The line will now be
y = (x-1)/m + 1
For any rational x, y =mx must be irrational, else m = y/x would be a rational number, which contradicts the assumption of irrational m. Thus, x and y cannot both be rational.
I took a similar tact using the same idea of irrational slopes. Note you chose an intercept at y=1-x, so neither x nor y can be rational at the intercept because if either were, they both would be.
But, depending how far back we go to first principles, that can create the perhaps slightly awkward requirement there be non-rational (x,y) points on the line y=1-x (because some would argue there is no such intercepting point given the two lines).
This can be avoided by not mentioning y=1-x in the first place. Of course, you will end up there regardless when you invert the slope to create the second line running through the point (1, 1).
Agree, this problem is absolutely obvious and doesn't require any infinities
Actually x and y have to be rationals at two points, the start and the end but I get what you mean
You complicated this so much. The simple function y=a^x - (a-1)^x when a is irrational and greater than 1 does the trick.
Indeed. In addition, your proof is constructive, i.e. it tells you where to go, whereas the proof in the video is not. The video and the proof in it are still nice, though.
I love how SoME 1 has led to so many new math channels being created.
please make more videos, I just cant wait to say I've been following you since 150 subs when you inevitably get popular.
can you explain more ?
@@secunsecun6313 cam YOU explain more?
@@Schattenhall ??
Well they did make more videos lol
I like how you can use Fermat's Last Theorem to solve this. x^n + y^n = z^n has no integers solutions with n>2. So let n=3 and z=1 to get x^3 + y^3 = 1, rearrange to get y = cbrt(1 - x^3). Finally flip it vertically with y = 1 - cbrt(1 - x^3), and you have a path that never hits a mine (since if x is rational, y cannot also be rational without finding a solution to x^3 + y^3 = z^3).
Obviously there are many other paths that work, but I think this one is quite elegant since it brings in such a famous theorem.
Awesome!
I'm confused as to how this demonstrates the practicality of infinity when the puzzle assumes there can be infinitely small points.
In retrospect, this might not have been the best example of a "practical" application of cardinality, but given just how abstract the concept is, I guess I figured even an infinitely dense minefield of point-like mines might qualify as an application! Though I'm sorry if the introduction was misleading.
I did make a follow on video where I tried to present a more serious problem where cardinality plays a pivotal role. Not sure if I succeeded, but you can find it here if you're interested:
th-cam.com/video/uLja-yAwuCI/w-d-xo.html
Yeah well in theory it’s still possible the size of our feet would always hit something dead on
@@TrueBladeSoul in theory, not only is it possible, it is guaranteed that you will hit something with *every* step you take, unless we assume that we touch the ground at a single point. This is precisely what it means that rational numbers are dense in the set of real numbers. On the other hand, when we walk, we don't drag our feet on the ground, i.e. we don't make a continuous path on the ground, but take steps. Therefore, one could simply step over the mines along the diagonal connecting the points (0,0) and (1,1). What I want to say, it's pointless to look for the shortcomings of the analogy of this problem with navigating a minefield, since it breaks immediately if we are to take it literally. The only purpose of the minefield story is that it's a cute way to introduce the problem to general audience, the problem being: find a continuous curve inside the square connecting (0,0) and (1,1) that doesn't cross any point that has both coordinates rational, except the starting and finishing point.
well if I remember correctly, in physics (fundamental) particles are modelled as being point-like / infinitely small, so it's not that bold of an assumption
@@phee4174 In physics a Planck length is seen as the shortest possible distance, and it’s not a distance of 0
Another consequence of what you've outlined here: If you pick a random point in your minefield, the chances of hitting a mine are exactly zero! It's actually a pretty safe place. Here we have another paradox. In spite of the infinity of rational numbers between 0 and 1, you can't pick any one of them at random if you're choosing from the set of all real numbers in that interval. Here's a proof. Consider the decimal expansion of the rationals. They all eventually end in an infinitely repeating sequence (even if that sequence is all zeros). But when you pick a real number at random, and look at its decimal expansion, it will just be a random sequence of digits, which will certainly NOT end in an infinitely repeating sequence. You may see a repeating sequence somewhere to start with, but eventually (with probability increasing with each next digit) the repeating sequence will break. In short, your random number was irrational.
Randomly selecting a rational number out of all reals has 0 probability, but it doesn't make it impossible. In fact, fix a real number x. Probability to randomly select x out of all real numbers is 0. This is true for any x. But, if I do randomly select a number, I will in fact get some number, so an event that had 0 probability a priori just happened.
@@Ennar You need to integrate over sets of points to get useful answers. The probability integral over all rationals is zero, the one over the irrationals is one (provided your integral formalism doesn't break with disjoint sets like that)
@@sh4dow666 I'm familiar with Lebesgue integral.
Seems to me that there's a problem not with your conclusion but with your argument. Yes, at any given point along the decimal expansion the current sequence might break, but that doesn't mean that the number is definitely irrational. If you see 0.1361361368..., you might just be looking at a rational number with a repeated 10 decimal places, rather than the repeated 3 it initially looks like.
More generally, no matter how many digits past the decimal point we've "observed", there are still infinitely many rational numbers that start with this sequence, and the same will hold true no matter what the next digit turns out to be. I think part of the problem here lies in reasoning sequentially (i.e. about the "next digit") with something that has been chosen oracularly, with an infinite amount of information pre-determined.
@@hughcaldwell1034 Forget what we "observe" as we look at the decimal expansion. The fact is this: the decimal expansion of every rational number must eventually end in an infinitely repeating sequence. Finding a sequence in a rational number's expansion that repeats many times, but not infinitely many times, means the infinitely repeating sequence has yet to begin. So we ignore that sequence and keep looking. Actually, we don't even need to look. A rational number is defined as the quotient of 2 integers. Knowing those 2 integers it's not difficult to calculate exactly where the infinitely repeating sequence of digits begins.
So, if you claim to have picked a rational number at random among the real numbers, I would ask you for the 2 integers that define that rational number, calculate the location of its repeating digits in the decimal expansion, and compare that with the decimal expansion of the number you claim to have picked at random. Eventually, if your number was truly chosen at random, your number's decimal expansion MUST deviate from the predictable repeating sequence of the rational number you claimed to have chosen.
To stay as far as possible from the mines, move along the slope (1+√5)/2 until you hit the line y=1-x, then move directly toward the point (1, 1) along the slope 2/(1+√5).
At first I thought, "Just step 'at random' and you have probably 0 of stepping on a mine," as the irrationals are infinitely denser than the rationals. But the conclusion that there are solid lines along which we may step continuously (which NEVER take a (Q,Q) value) was mind blowing!
I think the problem, and the strength of the conclusion, would be more clear if you said you specified that your are actually riding a bicycle!
What an interesting video about avoiding infinitely many mines that showcases the real-world applications of considering infinity.
Your focus on the specific example as opposed to what it implies is an indicator of low IQ.
Thank you! This is incredibly useful for people with infinitely small feet.
I saw this problem for the first time on stackexchange, and you explained it in a really good way! For the curious people, here's the mathematically precise statement of what he proved in this video: Let A be countable a subset of R^2. Then R^2 / A is arc-wise connected.
You have released only two videos so far but you are already among my top math channels, with Numberphile and 3B1B. I pledge for more!!!
Whoa, this is amazingly interesting and extremely well made. I always thought it was cool how different sets had the same infinite cardinality but this showed me a lot of really cool stuff I didn't know.
Liked and subbed
Thank you! I'm glad you got so much out of this video.
My first approach was to use a segment of a translated ln(x) function such that it passed through both (1,1) and the origin. That being y=ln(x(e-1)+1) for 0
How to Navigate Infinitely Dense Minefield: shoot one of the mines from a distance, the others will explode in a chain reaction, walk through the giant crater left over.
or you know, its only 1km, just swim around the island and get onto the boat.
Said there are water monsters, swimming gets you eaten.
You will probably die if you try shooting them
That's a risk I'm willing to take
@@starburst98 swim through the ground itself but under the mines
Found this channel this weekend and I'm enjoying the videos. :) I only have one small quibble with this one, the minefield problem was to "find a solution", but the video only actually proved that an infinite number of solutions exist which isn't quite the same thing. (It's still an interesting topic though!)
For finding a way out you could just go randomly and reach (1,1) with probability 1.
True. But having an event of probability 1 doesn't guarantee that it will necessarily happen.
@@Ennar ill take my chances
@@HUEHUEUHEPony good for you.
@@gregoryford2532 it absolutely doesn't. Infinity doesn't work the same way as finite sets. Imagine randomly picking a number between 0 and 1. The probability that you will pick irrational number is 1, yet you could also pick a rational number (with probability 0). So probability 1 doesn't mean certain, but almost certain, which is a subtle but important distinction.
Travel the curve
( t , (pi/4)^{t-1}*t)
Note that the curve starts at (0,0), ends at (1,1), lies within the square for t in [0,1] and when t is rational (so the x-coordinate is rational), then the y coordinate is not because if it was, this would imply pi/4 (and hence pi) is an algebraic number (which it isn't).
Interesting; I definitely wouldn’t have thought of it that way.
My approach would’ve probably been to just choose some combination of lines with irrational slopes, so one of the coordinates is always guaranteed to be irrational (and so not on a mine).
An even simpler and safer solution is to stay exactly 1 unit distance from either 1,0 or 0,1. This path results in a 1/4 circle, and since π is irrational, there cannot be any mines along this path. 🤯
Assuming you an step onto the complex plane, you can navigate the minefield easily.
This method you used is one way to show that removing countably infinite collection of points from the plane ( or in general R^n ) will give u a space that is still path connected!
Couldnt you also follow a path of an altered sine wave. Think about it, the only pair of rational numbers for the input and output of sinx is (0,0)
For bonus safety, once you've selected your line out from 0,0 you can simply pick it's mirror going out from 1,1 to be the line which intersects it.
Could you also choose your path to be a sin (or cos or tan)? Since for nonzero rational arguments they always result in an irrational number, so they won't go through the mines. It would be cool if you could find an infinitely continuously derivable function like the trig functions to be solutions
thank you for teaching me to navigate any minefield flawlessly.
I figured you could travel in an arc instead, there wouldn't be any rational coordinates that fall on the path, right?
There are rational points if you just follow the unit circle, but if you picked an arc with an irrational radius then you should be fine
This problem is very pretty. I was worried about density of rationals, and indeed, in R, this is impossible: a point in R separates the space. But in R^2, a point does not separate the space (a nice topological difference between R and R^2). Thus it is indeed possible to find such a line that always has irrationals in one-coordinate. Clever!
My first thought was to move along the golden ratio, then upon reaching the halfway line (the line connecting 0,1 to 1,0), travel along the inverse of the golden ratio.
Why, well I recall the golden ratio being the most irrational number, meaning I'd be the farthest away from an point described entirely by fractions of natural numbers, and logically so would its inverse, that way I can correct how off course I'd be if I only traveled along one or the other.
Well, ok, my absolutely first idea was to toss a stone and set off the infinitely powerful explosion. But that has its own problems.
Lesson Five. The shortest path... was a detour.
Your idea is correct, but you can take any irrational number instead of the golden ratio. Golden ratio is not the most irrational number. In fact, the golden ratio is just (1+ sqrt(5))/2 so I can even construct it using ruler and compass if I want. Hardly "the most irrational number" if you ask me. Also, you can never choose a point that will be farthest away from all the rational points, since for any point it's possible to find a rational point arbitrarily close to it. That's what it means for rational numbers to be dense in the set of real numbers.
@@Ennar
Well then I'm apparently misinterpreting these videos:
th-cam.com/video/sj8Sg8qnjOg/w-d-xo.html
th-cam.com/video/p-xa-3V5KO8/w-d-xo.html
@@Ennar The OP means 'most irrational' in the sense of Diophantine approximation, where indeed the golden ratio realizes the worst approximability by rationals.
@@proxy9321 ok, that makes sense. Thank you for clarifying.
What a cool proof! It almost feels like you sidestepped the entire problem and walked right into the solution.
Interestingly, when picking your real number for your slope, you do have to be careful because it's easy to not actually pick a truly random value. For example, any slope with a real number that can be written down with finite digits is rational. Additionally slopes with digits that form a repeating pattern are also able to be expressed as a ratio. You need to pick a number that can't be a ratio.
Hmmmm. I wonder if the set of constructible/algebraic numbers is countably infinite
Yes, both sets are countably infinite. Algebraic numbers are roots of polynomials with rational coefficients. There are countably many such polynomials because there are countably many rational numbers, and each polynomial has finitely many roots. Even if each polynomial had countably many roots, we would still end up with countably many algebraic numbers since countable times countable is still countable.
You know, it occurred to me while watching this video that, if both coordinates must be rational to have a mine, then any grid line corresponding to an irrational number must have zero mines on it, creating a grid of mine-free points across the island. I kept waiting for that grid to become relevant, but it never happened
Because how do you get onto one of those grid lines from 0,0? That's exactly equivalent to the original problem.
First game in Squid Game 2?
8:07 i love silent shills. you've earned a like for that
I was thinking the answer was something like “(sqrt(N))/(sqrt(2))” for the bottom, and similar (not exactly the same) for the right side
Basically, since there’s always irrational spots connected in a line, you’re safe (since points are at rational spots only)
The problem with this is that it's possible for sqrt(N)/sqrt(2) to be a rational number. N = 2, 8, 32 is all rational. This also applies to when N < 0, like N = 1/2, 1/8, 1/32.
Considering that each mine represents a rational number, you can just walk from A to B since the amount of irrational numbers is so much bigger and so much more... omnipresent .
If you walk diagonally towards the boat, you can make sure you don't hit any mines by ensuring that every step's length is a fraction of 1, which always puts you at an irrational point, but if the distance you walk is sqrt(2)/2, then you'll end up right on [.5,.5]
So essentially
If I think I'm not on a mine…
I'm not on a mine?
I could walk anywhere on this island, even the center, but once I realize "Hey I'm in the exact center of this island" I explode?
I was thinking you could take steps of some fractional distance of pi since you aren't taking infinitely small steps you don't need a continuous path. Then I realized you could just walk to the boat without thinking since the probability of landing exactly on any mine approaches 0.
I’m not sure if Zeno would be angry or proud. Or maybe he would feel as unsatisfied as I am about this solution. I think it’s the mapping from sets to the real world. I think it’s the unintuitive reasoning that reduces to “take all of the possible countably infinite paths and subtract those from the set of all paths then take one of those paths (twice, since the ends are both at rational coordinates).” I get the solution but it’s the mapping of real world objects onto sets that feels like a bit of a bait and switch.
What's the shortest of the oncountably many paths to freedom? Can this be answered at all?
My guess is sqrt(2), which is the length of the main diagonal. I know that line itself is not usable, but 2 other line segments infinitesimally angled away would do, and their combined length would surely have a limit of sqrt(2), if not actually reaching it.
@@CarlSmithNZ There would be an infinite sequence of paths approaching the limit of length sqrt(2) but never reaching it. That is to say, there is no "shortest" path as you can always find a shorter one.
Man, where was this video about navigating infinitely dense minefields back when I still lived with my ex
If they were real Sigma monsters they would have jump on land, dying to a mine under their own terms
Quite so. I just assumed they are divergent.
Your video helped me understand something I was having trouble with with the old explanations.
You showed the... I don't know the correct term but the thing where you take a diagonal of all the numbers to show a new number.
For awhile I never understood why we can't take out the staring 0. And say that all natural/normal numbers are as numerous as the real numbers.
By making the distinction at the beginning of calling it "countable" I can see that the countability aspect is important.
It is an axiom of infinite math. We have decided that a type of infinity is the kind that we can list from 1 to infinity and while you can find new whole number that you didn't count by using that diagonal proof it is unhelpful because it is changing it from a countable infinity to an uncountable one.
Thank you for clearing this up for me and thank you for posting this video it really helps?
finally a strategy to get to the end of the minefield I Skylanders spyro adventure! I've been stuck on it for 2 years
The weirdness of infinite and concrete conclusions we can draw if we carefully work with such an abstract concept is absolutely fascinating.
Yes, yes it is. Nobody really understands why mathematics works, but it does.
Frankly, there is quite a straight solution to this problem. Just pick random point inside the square, which has one and only one irrational coordinate, and take a path from start to this point, following with a path from the point to the finish.
You would succeed, because the start and destination points have rational coordinates, and so any given point of these two lines, the coordinates would be computed as a linear combination of the irrational and rational numbers, divided by a linear combination of rational numbers (which would be in fact irrational). So, you would always have a way (and infinitely many others) to complete the journey.
Excellent video! Thank you!
I would construct my path with two straight lines, y=kx, and (y-1)=k(x-1). Just pick the slope of my lines to be any irrational numbers, for example, k=sqrt(3) and 1/sqrt(3). Start out on the 60-degree line, and finish the second half of the path on the 30-degree line. Since my slope k is irrational number, y=kx, x and y can't be both rational at the same time.
Infinity is very interesting. All mathematicians should think its interesting when you do something like integrate sin(x), 0
A lot of people don't consider that depending on the scope of a problem, you can approximate some values to infinitely large or infinitely small instead of measuring. That is the true real world application of infinities and limits.
For example, calculating focal length. If you are using light from very far away, you can approximate to infinity.
4:20 The Stern-Brocot sequence is a delightfully unexpected way to make such a list! ^_^
As hard as the problem seems at first, the same countability argument shows that the following strategy reaches the boat with probability 1. Pick a random amount of time to walk and a random direction to walk in. When the time is up, walk straight toward the boat.
Well, it seems to me that we have basically 0 chance of hitting a mine at all, no matter how hard we try unless we know the exact position of atleast one.
Technically, there is a chance but there is a huge BUT.
Also, it's worth mentioning humans have legs and not wheels, so our path is not even a line but merely a finite set of close positioned points which makes the chance even less than that.
Thanks for the tip! Next time I'm in a minefield, I'll be sure to use this strategy!
My guess: You can move across on a diagonal with a slope that's a multiple of an irrational number, like y=(pi)x. Of course, because the optimal path is 1:1, you will have to switch direction to some irrational number that has a different slope to reach the goal.
A method I came up with:
Walk on a line with slope pi
This ensures that for any distance traveled results in a y coordinate of a multiple of pi.
If the multiple forms a rational number y coordinate then the multiple has to also contain pi making it an irrational number, in this case the x coordinate is an irrational number.
Continue walking until intersecting the diagonal of the island, at this point start walking at a slope of 1/pi, this has the same effect of irrational coordinates and also will lead to the opposite corner.
This can be used with any irrational slope eg. e, pi/2 etc.
If I lay down on my stomach and crawl with my arms I'm technically not stepping on the mines
This will take a while
I have returned, please keep making videos I love your animations and explanations. My only criticism is when giving the background knowledge/introduction the dense information can be hard to follow.
Thanks for your feedback. I'll keep that in mind.
@@morphocular the way you explained background seems to work well as a refresher though. But If it's the first time you're seeing it then yeah it might a bit hard to follow.
An alternative way to find a path is to look at the part of the curve |x|^3 + |y-1|^3 = 1 that connects zero and 1. (f(t) = (t, cuberoot(t^3-1)+1)
Now suppose x and y are rational : then you get, writing x = a/b and y = c/d, (ad)^3 + (bc-bd)^3 = (bd)^3, which contradicts Fermat's last theorem. Therefore, this path is safe :)
Sure sizes of infinities seems like something with no practical applications, but imagine if you were on an island with infinitely many infinitely dense land mines.
Now you see just how practical this concept is in a real world situation.
I love how every mathematician will explain why their study isn’t just really problems being made up by bringing up problems they made up
@6:37 "and I can just play this trick again with the new list". The fun thing is that not only is this new number not in the new list, but because the new list contains all numbers in the original list, it also cannot be in the original list. So no matter what the original list was, we can generate not just one, but an infinte number of numbers not in that list.
Worthwhile noting that the set of all finite decimals is countable: you can just write 'em backwards and get a unique integer.
I didn't look at subscriber count before watching, and the entire time I absentmindedly assumed I'm watching a 1 million subscriber channel
wait this kinda makes me think of what if there was a portal on both ends of the line but the person can pass threw the portal traveling the same distance while not being to physicaly touch the mines mind blown cool vid Isubed.
"practical problems" :-D. Well, maybe not, but I really loved your style. Here's a new subscriber, that's for sure.
It only took a few moments looking at the minefield puzzle for me to conclude: "Just walk along a horizontal irrational number close to 0 and a vertical irrational number close to 1". Sure, it doesn't explain how to get to those corner points before 0,0 and 1,1, but that's most of the path.
Of course, I also scored a 32 (out of 36 max) on my Math SAT, so I have a better grasp of math than most people without being a mathematician. I've actually forgotten how to do derivatives myself.
The diagonal from 0,0 to 1,1 has infinitely many free points, such as sqrt(2),sqrt(2). All you need to do is find the nearest multiple of sqrt(2), for instance 0.01 - any number will suffice - that is comfortable for you to step over to (your stride length, as a function of sqrt(2)). Simply take enough steps (approximately the exact number of steps it would take you to casually walk the diagonal) of that length along the 45 degree diagonal, and you'll have the shortest distance that guarantees no mines.
Never knew this was one of the obstacles my parents had to go through when going to school.
As soon as you stated the problem I came up with a different solution. We know from Fermat's last theorem that there are no positive integer solutions to A^3 + B^3 = C^3. If we set B=C-A (with C>A) that means there are also no positive integer solutions to A^3 + (C-A)^3 = C^3, or equivalently (A/C)^3 + (1 - A/C)^3 = 1. That also means there are no positive rational solutions to t^3 + (1-t)^3 = 1. Therefore if we walk along the curve x=t^3, y=(1-t)^3 we won't hit a mine. That curve goes from the top left to the bottom right, but we can easily flip it upside-down and walk along the curve x=t^3, y=1-(1-t)^3 instead, which solves the problem. Interestingly, x=t^n, y=1-(1-t)^n will work for any n greater than 2, but if we choose n=2 we will hit an infinite number of mines, corresponding to the Pythagorean triples.
Also had to think of this family of solutions!
If non linear paths are allowed then its as simple as just x^irationall and the solution will always irational exept at x=1.
Today we learned :
Dont make this man angry
Another way I think of uncountable infinity is how there isn’t really a starting point to even begin counting the real numbers. You can arguably start with 0.000…. but the next number cannot even be described without using infinity. Theoretically it would be 0.000…001 where the dots represent an infinite string of zeros. However that isn’t really even a number because it assumes that the zeros end at some point but they don’t.
It's really nice that this solution doesn't require the exact locations of the infinitely dense minefield.
Love the video! Keep up the outstanding work!
Good puzzle. I was a math major and I fell into the trap of figuring out how some monstrosity like the Weierstrass function would help. Only once you insisted the answer was simple did I think “wait, just take a line of slope π.”
while watching this all I could think of is how similar this is to faster than light space travel and how the odds of you hitting a celestial object are practically zero unless you're aiming for it
Luckily you made this tutorial , I wouldve never escaped this island... could you explain how to afford countably infinite mines next?
One way to "smoothly" go from 0,0 to 1,1 would be
y = SUM[n=>inf](x^(1/n)/2^n)
which does hit 0,0 and 1,1 while guaranteeing that any rational number that is arbitrarily large n-th power of 1/m will produce an irrational number(if it were to produce a rational number, it would mean there are finite prime numbers, which isn't true).
And the minefield's density is actually zero because infinity is weird, meaning you have 0 chance to actually hit any of the mines, though this assumes 0 size mines and feet which breaks all known physics. 0 sized things can only be infinitely dense or infinitesimaly dense.
Maybe you could wander through the minefield without hitting a mine. But if you tried to program a robot to do it, they wouldn't be able to, since they could only choose a direction with floating point numbers, and those are rational.
How to win war using mines alone:
1. Plant infinite mines where enemy is
2. Since the mine density is infinite, the minefield turns into a black hole
3. your enemy got swallowed by the black hole
Thanks for the practical advice, I'll try to remember it next time I anger a vengeful god of mathematics and am forced to solve an esoteric problem involving set theory for my survival. It happens far more often than you might think.
The way I think about this (I might be wrong, this is just what I understand) is a countably infinite set of numbers means you can find a number greater than another, and not miss any. Uncountably infinite numbers make it impossible, as there is always something in between.
It's pretty easy to give a constructive solution too.
The line (x, pi x) will never hit a mine.
Where it meets (1 - x, 1 - pi x), switch over to that line which also doesn't have any coordinate with both rational numbers.
take a line with slope phi half the way then with slope 1/phi
The first thing that come to my mind when you say an infinite amount of mines, is
A) This will create an infinitely large universe that is composed of nothing but mines, if the space is infinitely dense with mines, and the can't overlap, like normal matter.
B) We basically just have a black hole here.
This will be useful for the future.
I'd like to offer my favorite example of "weird math being practical", not that I personally need an argument against "your practical application is infinitely dense mine field"
Imagine you are truck driver and you have to transport 2D balls (also known as disks, also known in real world as cylinders, for example PVC tubes or toilet paper). How can you pack cylinders to transport as many as you can? well, square grid is inferior to hexagonal, and it can be proven that (except few outliers based on how big you truck and items are) hexagonal packing is always the best.
Let's go to 3D. How can you get the most regular balls into given space? I encountered interesting variation, asking how many same sized spheres can touch one given sphere? One solution is making a hexagon of 6 spheres (all touching center sphere), and then you add 3 spheres above and 3 bellow. What you get is perfectly stable form (resembling dodecahedron), so surely that is the best you can do? Well, yes and no. You got 12 spheres each toughing the center one and that is the best you can get, but try this: Make a pentagon of spheres, place center sphere inside, put another "pentagon" on top. You have 10 spheres toughing, and you can put one more on top and on bottom to get 12. But this time, the structure is unstable, there is "wiggle room". This makes it feel, that if you wiggle just right, you might be able to get one more sphere in there? Well, you can't but it is fascinating, that one solution (hexagon+3+3) is "perfect" and another (2xpentagon+1+1) hase room to improve). Turns out, this is inherent to 3D spheres and in packing, there is no general best way to pack number of spheres.
We now get to the point I wanted to make in the beginning. There is interesting result, that in 8D, you can make hexagonal packing, that has so much space in it that another hexagonal grid of 8D-spheres fits in that, and that make perfect best packing, similar to what we can see with cylinders ("2D spheres"). The next perfect packing does not appear until 24D.
So, how does it help a truck driver that 8D spheres can be packed perfectly? How does navigating mine field as mathematical point help real people? One way is, it helps you think about situations with given rules in a way that produces solutions. But 8D (and 24D) spheres and their packing HAS real world application. Next time you want to pay for something and use your credit card, you are using cryptography for save transactions and sphere packing is used to find efficient codes that allow for self-repairing code to be transferred through unreliable medium (air, overheating cables, overloaded servers,...) securely.
Self-repairing code must be compact (so you don't need GB/s internet to send a text) but also repetitive (when some bits get lost you can still tell what the message was), and sphere packing has incredible relation with exactly that (through the use of matrices of all things). Not gonna say more, you can find more yourself of maybe "someone" makes a video about it? :D
Have we just found 3blue1brown's aprentice?!?
Great video, gonna show it to all of my friends
infinitely small naval mines don’t exist, they can’t hurt you.
infinitely densely packed mines from this video:
I look at it a slightly different way : for a given line that passes through the origin, consider the x-coordinate of its intersection with the line y=1. The line passes through mines iff that x-coordinate is rational. Since the irrational numbers are infinitely more abundant than rational ones, the probability of such a random line hitting a mine is 0
The cardinality has almost nothing to do with the problem...
Take the Cantor set, now since the rational numbers are countable, make a count of them (i.e. list them q_1, q_2,...).
Now look at the set that is the union from i=1 to infinity of the Cantor set shifted by q_i (call the set
of the countable union S).
The new set is dense in R and since the Cantor set is uncountable, it is also uncountable. It is both dense and also uncountable on every interval in R no matter how small, and yet, the measure of this set is 0. This means that if instead you take the set S x S and look at the Island [0,1] x [0,1] where every pair (s, t) from this island has a mine where s, t are in S x S, you will still easily find a path that doesn't touch any mine!
And so the cardinality isn't a deciding factor and the thing that really matters is the measure!