Pro tip: you've been missing a few cuts here and there. Next time you know you're going to edit a part out, put a tiny chalk mark on the edge of the board so you can find the restart later. No one will notice ;-) I've been following your videos for a month, part of my quarantine sanity diet, your presentation is awesome, the problems are interesting and unique among UA-cam. I'm inspired to make videos of my own for lower level math with even half the standard of rigor and clarity as in your videos.
I first saw this material in a discrete maths course, and every single time a proof hinged on some constructed sets having (or not having) the same cardinality, the professor would just sort of wave his hands and say "because we can (or cannot) construct a bijection between them." This was always profoundly dissatisfying, so I really like how you've taken the time to show how the necessary functions are constructed. Even when it's intuitively obvious, it is not always trivial to articulate the argument formally.
I just started my first course in Real Analysis and I love these videos! I can tell you're using the same textbook, and these examples and explanations are amazing supplements to the material! Before watching this video I had to do a problem from the textbook for homework that was essentially equivalent to the first example, and I managed to find the exact same function on my own! This stuff is so exciting!
@@kreitlerk We're using a combination of Baby Rudin, Spivak's Calculus on Manifolds, and Taylor's Real Analysis, as well as some supplemental materials on differential forms, topology, and ODEs.
Actually nvm these are what we're using for my current course on real analysis, I took the first course last semester and we used Abbott's Analysis lol
Yup, even though at the end of the day it doesn't really matter. Your function just needs to map all real numbers in (-1,1) to a unique number in R: it just needs to be strictly monotone and continuous in that open interval. But agreed, I also got a bit confused when the graph showed the opposite variation
At 12:55 Michael draws a graph and claims x/(x^2-1) is increasing. Note that the denominator is always negative, so f(x) is positive for x < 0 and negative for x > 0. It can't be increasing. Gnuplot shows x/(x^2-1) to be ~the mirror of Michael's graph. You can show injectivity without monotonicity as follows: Assume f(x) = f(y) then x/(x^2-1) = y/(y^2-1) implying x(y^2-1) = y(x^2-1) implying xy^2 - x - yx^2 + y = 0 implying xy(y - x) = x - y implying xy(y - x) = -(y - x) implying (xy+1)(y-x) = 0 implying y = x or xy = -1 but the latter is impossible if {x, y} is a subset of (-1, 1) since |xy| < 1 = |-1|.
At 13:54 you made a small mistake. Its not enough for your argument to work if the function is only increasing. It must be strictly increasing, which means that f'(x) > 0. But great video though!
Love it! Just a couple of nitpicks, though : @ 10:30 : technically if a = b, the *open* interval from a to b is the empty set -- which is still not equinumerous with |R, mind you ;-) @ 14:30 : technically, the proper condition for f to be injective is f' > 0 (which it is here, thankfully !), for if f' ≥ 0 then in the general case it is possible that there exists an interval where f' = 0 The video is still great, however! Equinumerosity is an equivalence relation (which defines cardinal numbers, so by inclusion natural numbers, actually) so it is necessarily transitive :D
For the argument at the moment 14:40, I believe, despite being mostly correct, one can still raise some suspicions (taking f(x) = -x/(x^2-1)). First of all, f'(x) >= 0, which is correct, implies that f(x) is a non-decreasing function, i.e., it is either increasing or constant for some values of x. As it is non-decreasing, we cannot say f(x) < f(y) but, rather, we can say f(x) = 0 and f'(x) = 0 occurs only at a countably finite number of points, we can generalize the argumentation above.
As an engineer, those formal proofs are my least favorite part mathematics. If I see the graph of these functions, I just see they are bijective - so those proofs kinda annoying for someone that likes practical stuff. But you explain it very well.
Totally agree. I am a physicist, and have used maths all my life in research. I am a practical mathematician though, and, knowing the universe to be quantised don't really understand how the continuum of 'real numbers' can be a real thing. I believe it only to be real in the sense that it has been defined that way, and the axioms allow it. This video, therefore, to me, someone who makes a good living from using maths to design lifesaving drugs, is total gibberish. I am much more an advocate of rational numbers and integers
@@andywright8803 Well, the real numbers are real. Just look at pi. It's not in Q. And it's a continuum. It's not exactly that I've a problem with, it's just the lengthy method of showing that function was a bijection. The first function he used was just a straight line. Of course a straight line is a bijection. I don't think the video is gibberish, the approach is just very formal.
@@kriswillems5661 Yes, pi is a good example. Every value of pi that can ever be written is an approximation, and I am fine with that. Every means of calculating pi is impossible to complete and even if it were possible it could never be expressed exactly other that with its symbol. There is a sense therefore, that pi does not exist in any way other than theoretically and is therefore not real. It's an idealised concept and very useful, but it never actually exists in it's full form. Our universe has imperfections and therefore won't allow it. Calculate it to a trillion decimal places if you wish, but it will still only be approximate. The universe is quantised, and mathematicians insist on a continuum which does not reflect the real world. In my opinion, maths should try to deal with reality first before exploring whims and fancies that do not exist
sin(x) looks bijective if you zoom close enough to the origin. Also, even though the universe might be discrete mathematics does incredibly well to describe it. Most of the interesting (and useful) stuff goes out of the window if you try to be discrete all the time. Calculus and differential equations won't make sense for example.
Yes, I understand you need epsilon delta definitions and limit for calculus to make sense. I was just talking about the proof that function is bijective. You just see at the curve it start near ( -1, - infinity). It's continuous and always rising, and goes to approach (1, + infinity). How can it not be bijective between (-1,1) and the real numbers? It's so obvious.
At 12:55 Michael defines f: (-1, 1) -> R as f(x) = x/(x^2 - 1), then goes on to argue that it's surjective. Here's a non-cheating argument: Note that f(0) = 0, and that f(-x) = (-x) / ((-x)^2 - 1) = (-x) / (x^2 - 1) = - [x/(x^2-1)] = -f(x). Let y be given. If y = 0 then x = 0 is a solution. If y < 0 and there exists an x such that f(x) = -y then f(-x) = -(-y) = y. So assume WLOG that 0 < y. We want an x such that y = f(x) i.e. y = x/(x^2-1) i.e. yx^2 - x - y = 0. This is a second degree polynomial in x, with solutions at [1 +/- sqrt(1 + 4y^2)] / 2y. We want to show that one of these two values is in (-1, 1). Note that sqrt(1+4y^2) > sqrt(4y^2) = sqrt((2y)^2) = 2y. Let sqrt(1+4y^2) = 2y+e where 0 < e. But then [1 + sqrt(1+4y^2)]/2y = (1 + 2y + e)/2y = 1 + (1+e)/2y > 1 so [1 - sqrt(1 + 4y^2)] / 2y is the only possible solution. So we want -1 < [1 - (2y + e)]/2y < 1 i.e. -2y < 1 - 2y - e < 2y i.e. e < 1 < 4y + e Since sqrt(1+4y^2) = 2y+e we have 1 + 4y^2 = (2y+e)^2 = 4y^2 + e^2 + 4ye i.e. 1 = e^2 + 4ye = e(e+4y). But since 0 < y it follows that e < e + 4y. Since 0 < e and e(e+4y) = 1 it follows that e < 1 < e + 4y as desired. (If e >= 1 then 1 = e(e+4y) >= e+4y > e >= 1 and if e < 1 and e+4y
A note: For step 2 the function you chose to show I [-1,1] I = R does not answer the question because: If f (x) = x / (x ^ 2-1) then f´ (x)> 0 and also f (1/2) = - 2/3 0, so just take: F (x) = - x / (x ^ 2-1)
At 12:55 Michael defines a bijection between (-1, 1) and R. Here's a slightly different one, more motivated: As x approaches one endpoint of the interval we want the function to go to infinity, so we should probably have a 1/|x-1| term. As x approaches the other endpoint we want the function to go to negative infinity, so we should probably have a -1/|x - (-1)| term. So let g(x) = 1/|x-1| - 1/|x - (-1)|. Note that 1/|x-1| - 1/|x - (-1)| = 1/(1-x) - 1/|x+1| ;; since x-1 < 0 ergo |x-1| = -(x-1) = 1-x ;; also, x - (-1) = x + 1 = 1/(1-x) - 1/(x+1) ;; since x+1 > 0 and thus |x+1| = x+1 = [x + 1 - (1 - x)] / [ (1-x) * (x+1) ] ;; subtraction of fractions via a common denominator = 2x / [ x + 1 - x^2 - x ] ;; distributive law, twice = 2x / (1 - x^2) ;; x and -x cancel So g(x) = -2f(x), almost the same. I believe Michael drew the graph wrong (mirrored), so maybe he meant f(x) = g(x)/2. More generally, we could use h(x) = 1/(b-x) - 1/(x-a) for any open interval (a, b).
For f : A -> B: For f to be injective then for any a0 and a1 from A we have f(a0) = f(a1) if and only if a0 = a1. That is, two different inputs cannot map to the same output. For f to be surjective then for every b from B there is a value a from A that means f(a) = b. That is, you can map every value from in B using the values in A. With both properties you say that f is bijective. More details and some graphics are on Wikipedia: en.wikipedia.org/wiki/Bijection,_injection_and_surjection
@@Miitchyy Each input mapping to a unique output is true for any function by definition. An injective function must have every output being the image of exactly one input. You had that in the symbolic part of your definition, but not the "that is" part.
Could we also map [a,b] to -infinity,+infinity, using the tan function? We shift the tan function so that its zero coincides with (a+b)/2, and then we scale the range [a,b] to the size of [-pi/2,pi/2]
for the injective part , better to consider cases, f(m) = f(n) >= 0 and then f(m) = f(n) < 0. also 13:46 better to say, if f ' (x) > 0 on (-1,1) , (you might want to change the function to f(x) = -x / (x^2 -1) ) then f is strictly increasing (and therefore f is injective). In general it is not true that if f ' (x) >= 0 then f is injective. since if f(x) = 2, then f '(x) = 0.
He could do that, but he's also proving that the chosen function is a bijection, and that will be messier with tan and arctan. I guess he's just gone with something a bit more amenable to simple arguments.
@@scollyer.tuition If tan(theta1) = tan(theta2), then theta1 = theta2 + kπ for some k in Z. If theta1 and theta2 are both in (-π/2,π/2), then k = 0. Therefore, theta1 = theta2, which proves that tan is injective on (-π/2,π/2). It also follows from the unit circle definition that tan is surjective onto R. Let w be a real number, and suppose w = y/x, with 0 < x < 1. Then x = y/w, and we also require x^2 + y^2 = 1. Therefore, (y/w)^2 + y^2 = 1, and so we choose y = w / sqrt(1+w^2). Therefore, -1 < y < 1, and we have found a point on the unit circle (x,y) with w = y/x = tan(theta) for some theta in (-π/2,π/2). Thus, tan is also surjective onto R. Since tan is bijective, its inverse arctan is also bijective.
The problem you are working on is a form of the continuum hypothesis. The continuum hypothesis has been shown to be undecidable in ZFC (Zermello-Frankel set theory plus the Axiom of Choice). mathworld.wolfram.com/ContinuumHypothesis.html
Very nice topic to develop further (cardinals, ordinals, continuum hypothesis, powersets, etc). Could you make a video about the existance of a bijection from R to R^2 (or any R^n)?
This describes how to construct a continuous surjection from the unit interval [0,1] onto the unit square [0,1] x [0,1]. en.wikipedia.org/wiki/Space-filling_curve#Outline_of_the_construction_of_a_space-filling_curve Since there is a surjection from [0,1] to [0,1] x [0,1], and since [0,1] is homeomorphic with [0,1] x {0} subset of [0,1] x [0,1], it follows that [0,1] has the same cardinality as [0,1] x [0,1]. Finally, since [0,1] has the same cardinality as R, and since [0,1] x [0,1] has the same cardinality as R^2, we have proven that R has the same cardinality as R^2. This construction can probably be extended from R^2 to R^n for natural numbers n > 2.
It would have been a lot easier to prove that (0,1) has the same cardinality as (1,inf) and then showing that (1,inf) has the same cardinality as R. Doing the first part, you could use f(x)=1/x, which is easy to prove that it is bijective from (0,1) to (1, inf). The rest would be fairly easy. At least, this seems easier to me.
Hello Sir. You are too great, as well as your explanations. But, can you help me in this Problem in the next video. Please Sir Find the least composite Number n such that it divides both (2^n)-2 and (3^n)-3. I will be thankful to you
561 is the first composite number that satisfies this equation. And i just for the lolz wrote a python script that gave me the first 10 numbers that satisfy the equation they're listed down below: 561, 1105, 1729, 2465, 2701, 2821, 6601, 8911, 10585, 15841.
@@utsav8981 count = 0 max_count = 10 i = 1 def isprime(n): for i in range(2, int(n**0.5) + 1): if n % i == 0: return False return True while count < max_count: a = (2**i - 2) % i b = (3**i - 3) % i c = isprime(i) if a == 0 and b == 0 and c == False: print(i) count += 1 i += 1 This is a brute force method though, and the numbers we are working with get big real quick so it isn't optimal.
If A is a subset of B, it suffices to show only surjectivity. If A is a subset of B, then |A| ≤ |B|. If there exists a surjection f:A -> B, then |A| ≥ |B|. Therefore, we have |A| ≤ |B| and |A| ≥ |B|. Hence, |A| = |B|. Your map from N -> Z is clearly surjective. Therefore, |N| = |Z|. Furthermore, the map x -> tan(π(x - 1/2)), when restricted to (0,1), maps (0,1) onto R. Therefore, |(0,1)| = |R|.
Here's the homework solution: Show that if we have bijections f: A -> B and g: B -> C there exists a bijection h: A -> C. Proof: let h(x) = g(f(x)) Injectivity: if h(x) = h(y) then g(f(x)) = g(f(y)) implying [as g inj.] f(x) = f(y) implying [as f inj.] x = y Surjectivity: let y be given. There exists z such that g(z) = y. There exists x such that f(x) = z. But then h(x) = g(f(x)) = g(z) = y. But then h is a bijection since it is both injective and surjective.
The natural numbers should be defined to include zero, because then counting cookies in a jar will always be possible with a Natural number. And yes, zero is in, because the jar can get empty! Therefore calling |N the “natural” numbers implies that zero should be in. Excluding zero is as STUPID a definition as dropping one too.
@@utsav8981 Ok ! ^^ Usually when I use the natural numbers they start at 0, but we must have different definitions (Or maybe mine is an abuse of notation)
Wrong. It's a thing - countably infinite, as opposed to uncountably infinite. Surely, you must accept the natural numbers, aka counting numbers, are countably infinite?
@@angelmendez-rivera351 You can't though. Try counting from the first element in a set to the one a googleplex in. Not even talking about infinities here. There are not enough Planck times in the total expected age of the universe, so even some finite sets are uncountable. I know that you mathematicians have defined 'countable' as something like 'potentially listable' if you only had an infinite amount of time. In that case, a journey of 1000 light years is walkable if it's on land, and the moon is definitely a jumpable height. If these examples seem far fetched, remember, none of these values is infinite
@@Roger_Kirk why must I accept that the natural numbers are uncountable? Give me one example of an instance where they have been counted? ( something that started to count it and hasn't finished is not acceptable as I contend that you cannot finish counting it)
Andy Wright countably infinite, even though it practically doesn’t make sense, is usually used in contrast to uncountable sets. a countable set just requires that for every number in that set, you could count up to that number if you just had enough time, no matter how large it is. in the natural numbers, for example, if i wanted you to count up to a million, it’d be easy. 0, 1, 2, 3, 4, ... 999,999, 1,000,000. uncountable sets are sets so unbelievably dense you can’t count them at all like you would with the natural numbers. for example, the real numbers in the interval [0,1]. you could count all the rational numbers: 0, 1, 1/2, 1/3, 2/3, 1/4, 3/4, ... but is there any surefire way to count all the irrational numbers?
Pro tip: you've been missing a few cuts here and there. Next time you know you're going to edit a part out, put a tiny chalk mark on the edge of the board so you can find the restart later. No one will notice ;-) I've been following your videos for a month, part of my quarantine sanity diet, your presentation is awesome, the problems are interesting and unique among UA-cam. I'm inspired to make videos of my own for lower level math with even half the standard of rigor and clarity as in your videos.
This is definitely my favorite math channel. Really interesting topics explained intuitively and fairly rigorously. Keep it up!
ua-cam.com/video/M3eL8njbvkw/v-deo.html
Thanks so much for somehow making your videos relatively short, intuitive, and rigorous all at once!
I first saw this material in a discrete maths course, and every single time a proof hinged on some constructed sets having (or not having) the same cardinality, the professor would just sort of wave his hands and say "because we can (or cannot) construct a bijection between them."
This was always profoundly dissatisfying, so I really like how you've taken the time to show how the necessary functions are constructed. Even when it's intuitively obvious, it is not always trivial to articulate the argument formally.
@12:38 when a=b then the open interval (a,b) is not a point but an empty set :D.
Love the old school chalkboard.
I just started my first course in Real Analysis and I love these videos! I can tell you're using the same textbook, and these examples and explanations are amazing supplements to the material! Before watching this video I had to do a problem from the textbook for homework that was essentially equivalent to the first example, and I managed to find the exact same function on my own! This stuff is so exciting!
Do you know what textbook he is using?
@@kreitlerk We're using a combination of Baby Rudin, Spivak's Calculus on Manifolds, and Taylor's Real Analysis, as well as some supplemental materials on differential forms, topology, and ODEs.
Actually nvm these are what we're using for my current course on real analysis, I took the first course last semester and we used Abbott's Analysis lol
Should have been x / ( 1 - x^2 ), i.e. the opposite sign.
Yup, even though at the end of the day it doesn't really matter. Your function just needs to map all real numbers in (-1,1) to a unique number in R: it just needs to be strictly monotone and continuous in that open interval. But agreed, I also got a bit confused when the graph showed the opposite variation
2/pi* arctan seems good too
@@georgesanxionnat5054 Or artanh. (Inverse hyperbolic tangent.)
At 12:55 Michael draws a graph and claims x/(x^2-1) is increasing.
Note that the denominator is always negative, so f(x) is positive for x < 0 and negative for x > 0. It can't be increasing. Gnuplot shows x/(x^2-1) to be ~the mirror of Michael's graph.
You can show injectivity without monotonicity as follows:
Assume f(x) = f(y)
then x/(x^2-1) = y/(y^2-1)
implying x(y^2-1) = y(x^2-1)
implying xy^2 - x - yx^2 + y = 0
implying xy(y - x) = x - y
implying xy(y - x) = -(y - x)
implying (xy+1)(y-x) = 0
implying y = x or xy = -1 but the latter is impossible if {x, y} is a subset of (-1, 1) since |xy| < 1 = |-1|.
At 13:54 you made a small mistake. Its not enough for your argument to work if the function is only increasing. It must be strictly increasing, which means that f'(x) > 0. But great video though!
Love it! Just a couple of nitpicks, though :
@ 10:30 : technically if a = b, the *open* interval from a to b is the empty set -- which is still not equinumerous with |R, mind you ;-)
@ 14:30 : technically, the proper condition for f to be injective is f' > 0 (which it is here, thankfully !), for if f' ≥ 0 then in the general case it is possible that there exists an interval where f' = 0
The video is still great, however!
Equinumerosity is an equivalence relation (which defines cardinal numbers, so by inclusion natural numbers, actually) so it is necessarily transitive :D
good catch
For the argument at the moment 14:40, I believe, despite being mostly correct, one can still raise some suspicions (taking f(x) = -x/(x^2-1)). First of all, f'(x) >= 0, which is correct, implies that f(x) is a non-decreasing function, i.e., it is either increasing or constant for some values of x. As it is non-decreasing, we cannot say f(x) < f(y) but, rather, we can say f(x) = 0 and f'(x) = 0 occurs only at a countably finite number of points, we can generalize the argumentation above.
If Michael were to change it to f'(x) > 0, I believe there is no need for the cases.
When proving surjectivity of that last function, shouldnt x respectively approach from the right and left in stead of left and right for 1 and 2?
Yeah I think you are correct
I saw that and was hoping there was a correction down here, so am glad to have found somebody else seeing the same thing:)
6:31 Is it enough to prove that the inverse function is injective?
Your example of bijection between (-1,1) and R are so interesting.
As an engineer, those formal proofs are my least favorite part mathematics. If I see the graph of these functions, I just see they are bijective - so those proofs kinda annoying for someone that likes practical stuff. But you explain it very well.
Totally agree. I am a physicist, and have used maths all my life in research. I am a practical mathematician though, and, knowing the universe to be quantised don't really understand how the continuum of 'real numbers' can be a real thing. I believe it only to be real in the sense that it has been defined that way, and the axioms allow it. This video, therefore, to me, someone who makes a good living from using maths to design lifesaving drugs, is total gibberish. I am much more an advocate of rational numbers and integers
@@andywright8803 Well, the real numbers are real. Just look at pi. It's not in Q. And it's a continuum. It's not exactly that I've a problem with, it's just the lengthy method of showing that function was a bijection. The first function he used was just a straight line. Of course a straight line is a bijection. I don't think the video is gibberish, the approach is just very formal.
@@kriswillems5661 Yes, pi is a good example. Every value of pi that can ever be written is an approximation, and I am fine with that. Every means of calculating pi is impossible to complete and even if it were possible it could never be expressed exactly other that with its symbol. There is a sense therefore, that pi does not exist in any way other than theoretically and is therefore not real. It's an idealised concept and very useful, but it never actually exists in it's full form. Our universe has imperfections and therefore won't allow it. Calculate it to a trillion decimal places if you wish, but it will still only be approximate. The universe is quantised, and mathematicians insist on a continuum which does not reflect the real world. In my opinion, maths should try to deal with reality first before exploring whims and fancies that do not exist
sin(x) looks bijective if you zoom close enough to the origin. Also, even though the universe might be discrete mathematics does incredibly well to describe it.
Most of the interesting (and useful) stuff goes out of the window if you try to be discrete all the time. Calculus and differential equations won't make sense for example.
Yes, I understand you need epsilon delta definitions and limit for calculus to make sense. I was just talking about the proof that function is bijective. You just see at the curve it start near ( -1, - infinity). It's continuous and always rising, and goes to approach (1, + infinity). How can it not be bijective between (-1,1) and the real numbers? It's so obvious.
At 12:55 Michael defines f: (-1, 1) -> R as f(x) = x/(x^2 - 1), then goes on to argue that it's surjective. Here's a non-cheating argument:
Note that f(0) = 0, and that f(-x) = (-x) / ((-x)^2 - 1) = (-x) / (x^2 - 1) = - [x/(x^2-1)] = -f(x).
Let y be given. If y = 0 then x = 0 is a solution. If y < 0 and there exists an x such that f(x) = -y then f(-x) = -(-y) = y.
So assume WLOG that 0 < y. We want an x such that y = f(x) i.e. y = x/(x^2-1) i.e. yx^2 - x - y = 0.
This is a second degree polynomial in x, with solutions at [1 +/- sqrt(1 + 4y^2)] / 2y. We want to show that one of these two values is in (-1, 1).
Note that sqrt(1+4y^2) > sqrt(4y^2) = sqrt((2y)^2) = 2y. Let sqrt(1+4y^2) = 2y+e where 0 < e.
But then [1 + sqrt(1+4y^2)]/2y
= (1 + 2y + e)/2y
= 1 + (1+e)/2y
> 1
so [1 - sqrt(1 + 4y^2)] / 2y is the only possible solution.
So we want -1 < [1 - (2y + e)]/2y < 1
i.e. -2y < 1 - 2y - e < 2y
i.e. e < 1 < 4y + e
Since sqrt(1+4y^2) = 2y+e we have 1 + 4y^2 = (2y+e)^2 = 4y^2 + e^2 + 4ye i.e. 1 = e^2 + 4ye = e(e+4y).
But since 0 < y it follows that e < e + 4y. Since 0 < e and e(e+4y) = 1 it follows that e < 1 < e + 4y as desired.
(If e >= 1 then 1 = e(e+4y) >= e+4y > e >= 1 and if e < 1 and e+4y
Wonderful. This reminds me the exams at the university.
ua-cam.com/video/M3eL8njbvkw/v-deo.html
keep up the good work man !
A note: For step 2 the function you chose to show I [-1,1] I = R does not answer the question because:
If f (x) = x / (x ^ 2-1) then f´ (x)> 0 and also f (1/2) = - 2/3 0, so just take:
F (x) = - x / (x ^ 2-1)
dang, your correction of michael penn is in need of correction.
i think you mean , if f(x) = x / (x^2 -1) then f ' (x) < 0.
At 12:55 Michael defines a bijection between (-1, 1) and R. Here's a slightly different one, more motivated:
As x approaches one endpoint of the interval we want the function to go to infinity, so we should probably have a 1/|x-1| term. As x approaches the other endpoint we want the function to go to negative infinity, so we should probably have a -1/|x - (-1)| term.
So let g(x) = 1/|x-1| - 1/|x - (-1)|.
Note that 1/|x-1| - 1/|x - (-1)|
= 1/(1-x) - 1/|x+1| ;; since x-1 < 0 ergo |x-1| = -(x-1) = 1-x ;; also, x - (-1) = x + 1
= 1/(1-x) - 1/(x+1) ;; since x+1 > 0 and thus |x+1| = x+1
= [x + 1 - (1 - x)] / [ (1-x) * (x+1) ] ;; subtraction of fractions via a common denominator
= 2x / [ x + 1 - x^2 - x ] ;; distributive law, twice
= 2x / (1 - x^2) ;; x and -x cancel
So g(x) = -2f(x), almost the same. I believe Michael drew the graph wrong (mirrored), so maybe he meant f(x) = g(x)/2.
More generally, we could use h(x) = 1/(b-x) - 1/(x-a) for any open interval (a, b).
could anyone explain to me what does "injective" and "surjective" mean please?
For f : A -> B:
For f to be injective then for any a0 and a1 from A we have f(a0) = f(a1) if and only if a0 = a1. That is, two different inputs cannot map to the same output.
For f to be surjective then for every b from B there is a value a from A that means f(a) = b. That is, you can map every value from in B using the values in A.
With both properties you say that f is bijective.
More details and some graphics are on Wikipedia: en.wikipedia.org/wiki/Bijection,_injection_and_surjection
@@Miitchyy Each input mapping to a unique output is true for any function by definition. An injective function must have every output being the image of exactly one input. You had that in the symbolic part of your definition, but not the "that is" part.
I see what you mean, Steve. I will rephrase to address the ambiguity. Thank you.
@@Miitchyy You bet! I've done the same thing a few times before, myself.
thank you guys
Could we also map [a,b] to -infinity,+infinity, using the tan function? We shift the tan function so that its zero coincides with (a+b)/2, and then we scale the range [a,b] to the size of [-pi/2,pi/2]
for the injective part , better to consider cases, f(m) = f(n) >= 0 and then f(m) = f(n) < 0.
also 13:46 better to say, if f ' (x) > 0 on (-1,1) , (you might want to change the function to f(x) = -x / (x^2 -1) ) then f is strictly increasing (and therefore f is injective).
In general it is not true that if f ' (x) >= 0 then f is injective. since if f(x) = 2, then f '(x) = 0.
For mapping (a,b) to the real numbers, what if you map x in (a,b) to (-pi/2, pi/2), and then take tan(x)? The surjection would use arctan x.
He could do that, but he's also proving that the chosen function is a bijection, and that will be messier with tan and arctan. I guess he's just gone with something a bit more amenable to simple arguments.
@@scollyer.tuition If tan(theta1) = tan(theta2), then theta1 = theta2 + kπ for some k in Z. If theta1 and theta2 are both in (-π/2,π/2), then k = 0. Therefore, theta1 = theta2, which proves that tan is injective on (-π/2,π/2). It also follows from the unit circle definition that tan is surjective onto R. Let w be a real number, and suppose w = y/x, with 0 < x < 1. Then x = y/w, and we also require x^2 + y^2 = 1. Therefore, (y/w)^2 + y^2 = 1, and so we choose y = w / sqrt(1+w^2). Therefore, -1 < y < 1, and we have found a point on the unit circle (x,y) with w = y/x = tan(theta) for some theta in (-π/2,π/2). Thus, tan is also surjective onto R. Since tan is bijective, its inverse arctan is also bijective.
Approach -1 from the right and 1 from the left
For proof on step 2, could we have also used a function f:(-1,1) to R, f(x)=tan(PI/2* x). The reasoning and graph would still look similar
I love your classes professor
ua-cam.com/video/M3eL8njbvkw/v-deo.html
Skip all the steps and take f:(a,b)->R f(x)=tan[pi(2x-a-b)/(2b-2a)]
You sort of drew its graph.
Thank you Prof!!
Are you teaching a real analysis course?
He is/ (was) at the time of making this to my knowledge
Deserves million subscribes
What if we have closed sets [a,b] ? Can we make function for it?
Yes we can. The proof (to me at least) wasn't obvious though.
Thanks for the great video. Funny coincidence i have just been working on the |Powerset N| = |R| which I thought was surprising.
ua-cam.com/video/M3eL8njbvkw/v-deo.html
Power Set |N is more cleanly written 2^|N.
The problem you are working on is a form of the continuum hypothesis. The continuum hypothesis has been shown to be undecidable in ZFC (Zermello-Frankel set theory plus the Axiom of Choice).
mathworld.wolfram.com/ContinuumHypothesis.html
@@foreachepsilon Thanks.
@@mushroomsteve Thanks for the insight. That might be above my level but i will definitely check it out!
Very nice topic to develop further (cardinals, ordinals, continuum hypothesis, powersets, etc).
Could you make a video about the existance of a bijection from R to R^2 (or any R^n)?
Is it even possible to have a bijection from R to R^n for n>1 ??
This describes how to construct a continuous surjection from the unit interval [0,1] onto the unit square [0,1] x [0,1].
en.wikipedia.org/wiki/Space-filling_curve#Outline_of_the_construction_of_a_space-filling_curve
Since there is a surjection from [0,1] to [0,1] x [0,1], and since [0,1] is homeomorphic with [0,1] x {0} subset of [0,1] x [0,1], it follows that [0,1] has the same cardinality as [0,1] x [0,1]. Finally, since [0,1] has the same cardinality as R, and since [0,1] x [0,1] has the same cardinality as R^2, we have proven that R has the same cardinality as R^2.
This construction can probably be extended from R^2 to R^n for natural numbers n > 2.
@@hoodedR It is probably difficult to explicitly construct a bijection from R to R^n. However, consider the point (x,y), 0
It would have been a lot easier to prove that (0,1) has the same cardinality as (1,inf) and then showing that (1,inf) has the same cardinality as R. Doing the first part, you could use f(x)=1/x, which is easy to prove that it is bijective from (0,1) to (1, inf). The rest would be fairly easy. At least, this seems easier to me.
Hello Sir. You are too great, as well as your explanations. But, can you help me in this Problem in the next video. Please Sir
Find the least composite Number n such that it divides both (2^n)-2 and (3^n)-3.
I will be thankful to you
561 is the first composite number that satisfies this equation.
And i just for the lolz wrote a python script that gave me the first 10 numbers that satisfy the equation they're listed down below:
561, 1105, 1729, 2465, 2701, 2821, 6601, 8911, 10585, 15841.
Can you tell me the python code please
@@utsav8981
count = 0
max_count = 10
i = 1
def isprime(n):
for i in range(2, int(n**0.5) + 1):
if n % i == 0:
return False
return True
while count < max_count:
a = (2**i - 2) % i
b = (3**i - 3) % i
c = isprime(i)
if a == 0 and b == 0 and c == False:
print(i)
count += 1
i += 1
This is a brute force method though, and the numbers we are working with get big real quick so it isn't optimal.
ua-cam.com/video/M3eL8njbvkw/v-deo.html
Good content!
ua-cam.com/video/M3eL8njbvkw/v-deo.html
If A is a subset of B, it suffices to show only surjectivity. If A is a subset of B, then |A| ≤ |B|. If there exists a surjection f:A -> B, then |A| ≥ |B|. Therefore, we have |A| ≤ |B| and |A| ≥ |B|. Hence, |A| = |B|.
Your map from N -> Z is clearly surjective. Therefore, |N| = |Z|.
Furthermore, the map x -> tan(π(x - 1/2)), when restricted to (0,1), maps (0,1) onto R. Therefore, |(0,1)| = |R|.
a bijective function may be just
f(x) = 1/(b-x)+1/(a-x)
f(b-0)=+inf
d(a+0)=-inf
yes
I wish you were a bit more rigorous at the proofs , but that’s ok 👍, still love the videos!
Here's the homework solution:
Show that if we have bijections f: A -> B and g: B -> C there exists a bijection h: A -> C.
Proof: let h(x) = g(f(x))
Injectivity: if h(x) = h(y) then g(f(x)) = g(f(y)) implying [as g inj.] f(x) = f(y) implying [as f inj.] x = y
Surjectivity: let y be given. There exists z such that g(z) = y. There exists x such that f(x) = z. But then h(x) = g(f(x)) = g(z) = y.
But then h is a bijection since it is both injective and surjective.
The natural numbers should be defined to include zero, because then counting cookies in a jar will always be possible with a Natural number. And yes, zero is in, because the jar can get empty! Therefore calling |N the “natural” numbers implies that zero should be in. Excluding zero is as STUPID a definition as dropping one too.
Is there a reason, why you didn't take 0 in N ?:
N is the set of natural numbers, starting from 1, 2, 3, 4...
So, this set does not include 0
@@utsav8981 Ok ! ^^ Usually when I use the natural numbers they start at 0, but we must have different definitions (Or maybe mine is an abuse of notation)
Utsav u must be American
@@georgesanxionnat5054 You are thinking of the whole numbers = {0} union N.
@@georgesanxionnat5054 Actually the set which includes 0 and all the natural number is the set of Whole Numbers. It starts from 0, 1, 2, 3,...
Countably infinite is an oxymoron
Wrong. It's a thing - countably infinite, as opposed to uncountably infinite. Surely, you must accept the natural numbers, aka counting numbers, are countably infinite?
@@angelmendez-rivera351 but if it's infinite, you can never count it
@@angelmendez-rivera351 You can't though. Try counting from the first element in a set to the one a googleplex in. Not even talking about infinities here. There are not enough Planck times in the total expected age of the universe, so even some finite sets are uncountable. I know that you mathematicians have defined 'countable' as something like 'potentially listable' if you only had an infinite amount of time. In that case, a journey of 1000 light years is walkable if it's on land, and the moon is definitely a jumpable height. If these examples seem far fetched, remember, none of these values is infinite
@@Roger_Kirk why must I accept that the natural numbers are uncountable? Give me one example of an instance where they have been counted? ( something that started to count it and hasn't finished is not acceptable as I contend that you cannot finish counting it)
Andy Wright countably infinite, even though it practically doesn’t make sense, is usually used in contrast to uncountable sets. a countable set just requires that for every number in that set, you could count up to that number if you just had enough time, no matter how large it is. in the natural numbers, for example, if i wanted you to count up to a million, it’d be easy. 0, 1, 2, 3, 4, ... 999,999, 1,000,000. uncountable sets are sets so unbelievably dense you can’t count them at all like you would with the natural numbers. for example, the real numbers in the interval [0,1]. you could count all the rational numbers: 0, 1, 1/2, 1/3, 2/3, 1/4, 3/4, ... but is there any surefire way to count all the irrational numbers?