time span proof by contradiction 2:30 prove √2 is irrational by contradiction 4:31 False proof example 13:42 induction proofs 20:03 example 1 on induction proof 24:30 example 2 on induction proof 36:37 example on false induction proof 43:34 Example 3 on induction proof 59:00
Providing more clarity on the Horse Problem. For the Induction Axiom to work, there are two requirements. 1. Base case must be true - P(1) is trivially true, because the 1 horse will always be the same color as itself. 2. P(n) => P(n+1) - So, when we write n horses, we typically write h(1), h(2), h(3) ....h(n) & for n+1 horses h(1), h(2), h(3) ....h(n), h(n+1) In the above statements, we sub-consciously assume that n is a bigger number greater than 1. Here lies the mistake. Since n can be any number, n can also be 1. So, the list can be: h(n) -----> for n horses h(n), h(n+1) -----> for n+1 horses More specifically: h(1) -----> for n horses, where n being 1 h(1), h(2) -----> for n+1 horses, where n being 1 Now, as per inductive step, predicate is assumed to be true. Assumption: Set of any n horses is of the same color. => h(1) is the same color as itself, say color c1 The assumption also implies that h(2) is the same color of itself, say color c2 Now does h(1) => h(2)? i.e. does c1 = c2? We can't determine that, can we? There is no way we say c1 is always same as c2. So, the inductive step breaks. The takeaway from this problem is to remember the "..." bug that we sub-consciously tend to make. Hope it helps someone :)
Although its been 3 years since I studied induction in high school, only today I understood why it is true. All thanks to the Indian education system for teaching induction without teaching mathematical logic.
The sound of that chalk pen hitting the blackboard came as pure music to my ears... One who has a love for computers will find this lecture to be a melody. Loved the horse proof...and the tiling problem.
Damn....have to use a lot of brain power for this class. I like the approach of this professor where he engages the class, compared to other lazy professors who just read from slides.
I just realized that the last tile problem was a nicer way of asking to prove that (2^(2*n))-1 is divisible by 3. Love this more verbose way and as well as the story behind it :) Kudos to MIT
With similarity of triangles you can disprove the statement that says "90>92" mentioned at minute 14:43. In the big triangle before splitting in halves, the size of the vertical side is 10 and the horizontal's is 9. After splitting it, the sizes of the small triangles are 2 for the vertical side and unknown for the horizontal, let's call it x. Their angles continue to be consistent, so we can apply similarity: 10/2 = 9/x x = 1.8 So x > 2 is false.
This man made induction understandable in the space of 5 mins, where my own prof took 2 hours and left only a classroom full of blank stares. Strong knowledge base != Good teaching!!!
It's really an amazing course! I'm going to study in university next year and i'm coming to look for some inspires for the courses i will take. This outstanding professor do solve a lot of questions for me in advance.
proof by contradiction 2:30 i.e. root(2) is irrational false proof 13:42 i.e. 90>92 induction 21:00 28:30 prove sum of 1st n natural number = n(n+1)/2 induction q: 37:00 41:45 STEPS OF induction 57:24 ....
basis: 2x2 can be tile with a single tile that leaves the NE corner empty. Assume this holds for P(n) now show P(n+1). Split the corresponding 2^nx2^n into 2 regions of size 2^n, apply the induction hypothesis, it gives you that all of these 4 regions can be tiled leaving the NE corner empty. Now simply rotate the NW and SE 2^nx2^n regions 90 and -90 , respectively and you get a L-shaped tile missing in the middle - put a tile here. QED. Now to finish it off, when you have the required size you simply rotate the NE sub square by 180 to get its NE corner to the middle.
so are you saying dragons exist? orcs exist? all the other crazy things in movies and tv shows exist? CGI is just computer generated graphics - it doesnt have to be based of any real objects.
Again, flawed. He can be a combination of people if its CGI cuz atleast someone or some people should have thought about making him. Dragons are fictional, yes. But to make a CGI of a real person for one whole lecture with this much perfection surely means that there are people who thought of it. Meaning there are people who know how to explain it. Meaning there is atleast one math professor like him. Please try to think on it for longer. I am not saying he can't be CGI I'm saying his explaining and intellect must be thought by someone else he can't be CGI.
🎯 Key Takeaways for quick navigation: 00:26 🧠 *The course emphasizes the importance of proof components: propositions, axioms, and logical deductions. Reasonable use of axioms is accepted, with a focus on consistency.* 01:49 🔄 *Introduction to proof by contradiction; assuming the opposite of the proposition, deriving a contradiction, and concluding that the proposition is true.* 03:01 🤔 *Steps for a proof by contradiction: Assume proposition P is false, use the assumption to derive a falsehood, leading to the conclusion that P is true.* 04:45 ➗ *Proof by contradiction example: Demonstrating that the square root of 2 is irrational using indirect reasoning.* 09:55 📚 *Historical context: The proof that the square root of 2 is irrational was discovered by the Pythagoreans, leading to challenges to their axioms and the suppression of the result.* 13:42 🚫 *Introduction to false proofs: Highlighting the pitfalls of relying on visual representations and assumptions in proofs.* 20:44 🔄 *Induction as a powerful proof technique: The course emphasizes the importance of understanding and applying proofs by induction in computer science.* 23:33 🎯 *Mathematical induction is like a series of dominoes, where proving a base case and the induction step ensures that a property holds for all values.* 24:30 🧮 *Induction proof example: Proving the sum of natural numbers from 1 to n equals n times n plus 1 over 2, using base case and inductive step.* 28:46 🔄 *In a proof by induction, you establish a predicate (p) representing the property you want to prove, check the base case, and show the inductive step.* 34:41 🛑 *Proof by induction doesn't necessarily provide an intuitive understanding of why a theorem is true; it's more about establishing the truth of a statement.* 36:35 ➗ *Induction example: Proving that for all natural numbers n, 3 divides n cubed minus n, using base case and inductive step.* 43:25 ❌ *False proof by induction: Attempting to prove that all horses are the same color, revealing the importance of correctly formulating the induction hypothesis.* 51:20 🐴 *Inductive proof using "pn" to show all horses in a set of n are the same color, addressing a mistake in the proof for n=1.* 55:35 🤔 *Careful use of the "dot, dot, dot" notation in mathematical proofs, highlighting a flaw in the case n=1 of the horse color proof.* 56:30 🧩 *Importance of proving all steps in an inductive proof, emphasizing the missing link in the proof chain (p1 implies p2) for horse color.* 59:21 🧠 *Application of mathematical induction to prove the existence of a solution and find the solution for a courtyard tiling problem.* 01:03:07 🧩 *Definition of the induction hypothesis ("pn") in the context of proving the tiling of a courtyard with a center square missing.* 01:08:06 🔄 *Iterative refinement of the induction hypothesis to successfully prove the tiling of a 2^n by 2^n region with a corner square missing.* 01:13:18 🔄 *Iterative refinement of the induction hypothesis by considering proving the problem with Bill in the corner first and then using it as a lemma for the center.* 01:14:09 🤔 *Discussion on alternative approaches, including dividing the courtyard into 2 by 2 squares, highlighting potential challenges.* 01:15:37 🛠️ *Stronger induction hypothesis (pn) can simplify the proof, emphasizing the advantage of assuming more inductive power for a more complex problem.* 01:16:34 🔄 *Application of a stronger pn (any square missing) to prove pn implies pn+1 more easily, showcasing the power of a robust induction hypothesis.* 01:18:28 🧠 *General rule: If you struggle with induction, try assuming something stronger as your induction hypothesis; a more powerful pn can make proving complex problems easier.* Made with HARPA AI
I thought I lost hope with mathematics. I was kinda slow in my thinking ability. Now I am working towards understanding maths for computer science so that I can go on studying master and eventually PhD. Hope you all enjoy this series of lectures.
For the tile problem, If you have any square missing, the shape of the tiles will not be L For the corner piece missing in 2^n * 2^n, try finding the last piece to be L in shape while leaving a corner empty.
I'm gonna be honest here. I've never really liked giving my time and my efforts to Mathematics. But damn, the way this professor teaches, it's very engaging and very very interesting. Thank you for these materials! These are really helpful.
I always thought this proof by induction stuff is like magic. NEVER understood it. But now I do (I think). Great video and thank you for providing it to us for free
For the one at the end make an 2^n+1 square out of a 2^n square (one fourth of the field) and the remainder, a 2^(n+1)x2^(n+1) L-shape out of 2^n L-shapes, meaning that each non-two-by-two square can resolve to a smaller square and L-shape where the square is in any quadrant, until you get to the two-by-two which is self-explanatory.
Professor Leighton, thank you for a solid introduction to proof by induction. In the early 1990's , I took Advanced Calculus at the University of Maryland College Park and proof by induction was introduced in a slightly different way.
@19:00 The mistake right up-front was not that it was drawn wrong. That is the secondary mistake. The original mistake is that the Height-measure of the original triangles was altered to 8 (with the remainder of 2 added to the sub-triangles) PRIOR to estimating the new base-measure of the sub-triangles. With the correct Height-Measure of 10 and Base-Measure of 9, we don't assume the base is anything-plus. The base is obviously something-minus. But if you first alter the height to 8 with a base of 9, we then assume the base is something-plus, because we forget about the 2 we just removed. So the idea of perception is correct in that it covers up mistakes. But the "right up-front" issue is actually the order of operations, not the sizes of the triangles. The sizes of the triangles are a secondary issue. When I saw the 2+, I wondered why the base of a sub-triangle was different in proportion to the base of the parent-triangle whose base is smaller than its height. The image wasn't the cause of the issue, the order of operations was.
On the court yard proof... the base case isn't correct because the statue isn't in the center. It's off to the side in one of the quadrants. It should be partially in all four quadrants to be considered centered...
The requirement is more relaxed, but it might seem harder to prove because a solution to the more relaxed problem would automatically prove the more constrained problem. In this sense, the more relaxed problem yields a "stronger" result because proving the strong result automatically proves the "weak" result.
I don't understand why he said 1:09:25 "We don't want put Bill in the corner, but in the center.". In his initial example with n=2 (1:01:40) he put Bill in the corner as well.
The tiles are L shaped , the minimum size of area would’ve 4x4 , four L shaped tiles . Then Bill would be in the corner of each L shaped tile but in the center of the 4 L shaped tiles. To me 2 to the n has to be divisible by 4 .
In the Horse Problem, does it then mean if we take the predicate to be: P(n): In any set of n>=2 horses, all horses in that set are the same color, then we can establish the truth of the theorem (that all horses are the same color), because then depending on the truth of the base case, we can always establish that P(2)->P(3), P(3)->P(4), ..., P(n)->P(n+1) are true. So that is a bit circumventing around what we actually sought to prove, given that we're assuming that in the base case. Really good problem.
Just this second class completely destroyed my ego. One of the most difficult but stimulating and satisfying classes I've ever engaged myself in; for once I'm truly haunted by what lies ahead.
sir Time 15:45 , trying to prove 90>92 there in the doted line of a small triangle whose length is 2 is counted in both (small deducted triangle) square length whose length was supposed to be 9-2=7 but counted as 9 by8 AND In that small Deducted triangle whose length and breadth is 2 either you count in (small deducted triangle)real length 9-2=7by8 Or In that small deducted triangle whose length and breadth is 2 by 2
Horse Problem Clarification: The inductive step doesn’t work when n=1 So P(n) = H1, H2, …, Hn And P(n+1) = H1, H2, …, Hn, Hn+1 1. We assume P(n) is true by induction 2. In P(n+1), we know that the first n horses are the same color because we assumed P(n) is true 3. Now, remove H1 from P(n+1), you’re left with the set {H2, …, Hn, Hn+1}, which now has P(n) horses 4. So now we know that {H2, …, Hn, Hn+1} are all the same color because again P(n) is true 5.(Now here’s the error) The next step WOULD be to say that H1 = {H2, …, Hn} because of step 1 and {H2, …, Hn} = Hn+1 because of step 4, thus implying that P(n+1) are all the same color, BUT look closely at case n = 1: P(1)= H1 P(2) = H1, H2 Remember step 5: H1 = {H2, …, Hn} = Hn+1 But wait, look at the middle set, {H2, …, Hn}. It’s P(n) without H1. Now what is P(1) without H1? *NOTHING, IT’S EMPTY!* Thus saying that H1 is the same color as no horses at all is false, which means step 5 in our induction is false.
For those who are confused with the tile problem, I think the center square means that Bill must be inside a square which is in the center. Like imagine a smaller square for n = 2 which is inside the original square. Then you can cover the 4 corners with the L-shaped tiles and it'll leave a square in the center and bill will lie somewhere in that square. For more details check out the readings of the chapter-3 induction of the same problem
bill problem can be done this way- well we have a area thats 2^n+1 by 2^n+1 so if we break that up we have 4X (2^n by 2^n) thats the total area since bill had a nice center tile in each of those 4 2^n by 2^n areas hence bill has a center tile in the bigger area made up of 2^n+1 by 2^n+1 and now lets say in case n is 2 this center area is like 2X2 space but we put in another L shaped tile and we have 1 sq left for bill in the center
hypothesis: any region of " ((2 power n x 2 power n ) - 1) / 3 " is true. Bill will never end up in the center. number of tiles extending away from bill towards opposite ends of any region will always be different ! to center bill and prove L shaped tiles will fit you have to use the hypothesis "((2 power n x 2 power n)-4)/3" is true, base greater than or equal to 1.
That 90 > 92 thing is kinda sounds like when you divide a 6 by 4 (24 chunks) chocolate and generates 1 more chocolate chunk and rearrange the chocolate back to 6 by 4 and you get 1 extra chunk.
To be more clear, +n - n = 0, so he basically just added 0, which doesn't change the value of the polynomial (which I just incorrectly referred to as an equation). Also, in case you missed it, he didn't just throw in the "-n", he also incremented the 2n (see the change to 3n), which is the +n I was referring to that cancels out the -n, leaving it unchanged.
The Egyptian were the first mathematician not the Greeks. The Greek took the knowledge from the ancient Egyptia. Hence the triangular structure of the pyramids.
Horses are not numbers or centimeters. They don't have any standarts about colors. That is why h(1) can be a set of itself with one specific color. Then h(2) can be a set of itself again with a different specific color. These two set can represent the "n=1"set alone. and These two cases are separately true. Then for a "n+1" set of (h(1), h(2)) you can not imply either of these base case scenarios because they have different results. This breaks the implication from p1 to p2, which breaks all significance of the proof.
*How does induction work?* This is what i understood : Let's take the example of this formula S(n) = 1 + 2 + 3 + 4 + -------- + n = n(n + 1) / 2 Now suppose I asked you the value of S(100) and you solved it for me S(100) = 1 + 2 + 3 + -------- + 100 = 5050 Similarly , You also solved for me f(100) = 100(100 +1) / 2 = 5050 Now, assuming you weren't lying to me about S(100) and I'm required to find S(101), I can do that as follows : S(101) = [1 + 2 + -------100] + 101 = [5050] + 101 = 5151 f(101) = 101(101 + 1) / 2 = 5151 Similarly If i have to find S(102) and assuming you did'nt lie to me about S(100), I can do S(102) = [1 + 2 + 3 +--------- + 101] + 102 = [5151] + 102 = 5253 f(102) = 102(102 + 1) / 2 = 5253 | | | | Similarly I can keep on calculating it for values greater than 100 (means 100 is the base case here) based upon the answer you told me... So here, I'm not verifying the hypothesis, but assuming you verified it for me and using that I proved a conditional statement that If p(n) holds true then p(n+1) also holds true. i.e. p(n) => p(n + 1) And when you check base case you are giving yourself that initial start...(just like you told me that S(100) is true.)..but here solving for the basecase you know its true cuz you verified it yourself... *Sorry for my English. I'm not a native speaker*
I didn't watch the video, but I wanted to point out that there is a really nice proof for 3|(n^3 - n) that involves modular arithmetic. Simply note that n^3 - n = n(n+1)(n-1) which is clearly 0 mod 3.
Induction can be pretty hard to understand at first. I didn't really understand it for a few years after first studying it. Many people struggle with induction, but with some persistence you can understand it. My advice is to watch some more videos about it until you understsnd how to use it. If it still doesn't make sense to you after watching one video, come back the next day and watch another one. Sleeping on something is really helpful. Once you have learned how to use induction, do some practice problems. Finally, after doing practice problems, watch this video again. Your practical experience will make it much easier to understand the theory behind induction proofs. You can't always learn a difficult concept like induction all at once. But, by chipping away at it over time, you can come to truly understand it.
You need the base case to prove everything. The base is, literally, the base. You build from the ground up. When you’ve proved that the ground floor is perfectly done, and you have got the fact that a solid floor would enable you to build a floor on top of it, then you can theoretically construct a building with infinitely many floors. The physical world says such building is not possible, but such thinking is reasonable in the realm of the induction axiom. That’s why the method of induction is an axiom, because it doesn’t apply to everything in the universe, and is chosen according to our purpose.
This is my third lecture I've watched with him and I kept thinking that this MIT room looks like a 70 year old community college in the middle of downtown Detroit, but then Tom drops a bomb saying that building is new, and furnished, was supposed to be under $100 million but skyrocketed to over $300 million.
The "all horses are the same color" proof fails because the variable of interest is "color" which is not represented numerically in the proof. The proof presented refers only to the number of horses, there is no mathematical predicate for the color.
In 2^n + 1 square why prof. Has put the bill in the center in the upper left square...Bill wants to be in the center and the center of the square 2^n+1 is where bill earlier was... Could anyone explain?
The centre was not fixed meaning Bill just didn't want his statue in the corners. And that is why the Professor has changed his theorem from a centre to any centre because we now have more room to play around and figure out the answer.
He didn't say it wasn't divisible by three, he said it didn't "look" like it was divisible by three. Subtracting n, then adding it back to an equation doesn't change the truth of the equation, so the fact that the rewritten equation was divisible by three proved that (n^3 + 3n^2 + 2n) is divisible by three as well.
You can think of 2|a (2 divides a) as an equation: 2q=a, where q is an integer. Squaring both sides: 4q^2=a^2. Now q^2 is just any integer, so you might as well write 4q=a^2. Convert back to other notation: 4|a^2.
okay i have a doubt in the tile question if we are changing the cases for eg from center to corner and corner to any tile then the base case must be change as well and in the base case there will be 16 cases in which we have to proove the question for each case now thats make our life harder but in the last teacher said that this is easy how man how can one totally forget about the base case
by 64x4 he actually meant 4x4 (as he corrects later). In a 4x4 square we have 16 square tiles capacity. Even when there is no square missing(in the 4x4 square), and you try filling it with L shaped tiles (which have size 3), its not possible because there are 16 squares and you cant fill it completely with 3 tile sized squares. It is because 16 is not divisible by 3. The whole bigger square has 64 square tiles and on dividing it into 4 parts we have four 16 tiled(4x4) squares. Hope this helps
He skipped a few steps but, rearrange it so that its in the form. n^3 +3n^2 + 2n = (n^3 + 2n) + 3n^2 . then consider that (-n is just (2n - 3n). add zero, +(3n -3n) to the equation. which helps us rewrite it as (n^3 + (2n-3n) + 3n^2 + 3n. = (n^3 -n) + 3n^2 + 3n
Sid Arta if a - b = c - b and a = e, don't you think we can substitute a with e to get e - b = c - b? Replace a with (n + 1)^3 and b with (n + 1) and it should make sense
Thought the problem was the base case since if one compares the color of the horses, there must be 2 or more horses. It turned out to be the implication from P(1) to P(2)!
In the end, i don't follow. In the Bill's last case, he said that Bill could be anywhere, but the example shows that it specifically is put in the corner--not anywhere, is there a fault? Can it be proved?
Sorry there are no recitation videos available for this course but there are recitation notes available. Hopefully they will be of some help. See the course on MIT OpenCourseWare for the materials at ocw.mit.edu/6-042JF10.
On the horse problem my understanding is this: We cant prove p(2) i.e 2 horses are of same colour using the base case p(1). A set of 1 horse is of same colour but that doesnt prove that in set of 2 horses both of them are same colour. We cant use p(1) to prove p(2)
Yes thats true because if we assume p(1) to be true, then p(2) has to be false since so the implication is false. Because of this, the inductive "Domino" is not started since the base case itself fails.
Can anyone explain why @ 40:30, he throws in a (-n) to make the proof work. I understand that it does work but what rule allowed him to throw it in after the initial proof was false. i.e (n^3 + 3n^2 + 2n) did not divide three, so he changed it to (n^3 - n +3n^2 + 3n) which proved the theorem true.
He simply restated +2n as -n +3n. The rest didn't change. Effectively, nothing changed. It's like restating 5 as 3+2. You can do it if you like, the value doesn't change.
For the set which contains all sets of length n, where all sets of length n+1 has only horses of the same colour, it was not shown that the set of length 1 is a member.
Did this in school when I was 16 or 17 years old. Did a ton of math in my engineering degree. Worked as a software engineer for many years and worked in IT for 30 years. Never used this topic in math once in all that time. Some math is helpful for computer scientists, but not this.
2/(triangleHeight -2) :: (suppposed2+Width/triangleWidth) triangleHeight = 10; traingleWidth = 9 2/8 == supposed2+Width/9 9/4 = suppposed2+Width== 2.25 (triangleHeight -2)x(triangleWidth + supposed2+Width) == actual area 8x(9+2.25) = 90 how does that person @ 18:00 say 1.8 something and also the professor.. can anyone explain?? (a -2)(b + c) == ab (conservation of area) c/b == 2/(a - 2)
Around 34:00, he shows that (n+1)/2+(n+2)=n^2+2n+n+2/2. ... What algebra method is used to determine this? I can see how the third and final result of (n+1)(n+2)/2 is derived from the reverse of the FOIL method, but how did he get from step 1 to 2?
regarding the horse question (I havent seen the solution): p(1) itself cant be used as the base case because p(1) recursively proves itself True. since there is only 1 horse, if we define these variables: p == "H1 is Black" and q == "all horses in any set is Black", we conclude that p => q == 1 (always), because if H1 is in fact Black then: p == 1, and if we check q we see that "all horses" consists of only H1 thus q should be 1 too. now if H1 is NOT Black, then p == 0 and q should be 0 too because, again, there is only H1 so "all horses" means that we're just checking H1 let's see if I got if right! I didnt :(
He added n and subtracted n at the same time, so: (n^3 + 3n^2+2n) = (n^3 + 3n^2+2n) - n + n = (n^3 -n + 3n^2+2n+n) = (n^3 -n + 3n^2+3n) =( n^3 -n) + (3n^2+3n)
For the statue proof, how would one define in mathematical terms the L shape figure of squares used? I proved this is possible using modulo of 3 but it doesn’t specify the shape. Matrice notations maybe?
I thought that the bug was in the predicate itself as color is not a function of numbers so how can we make a predicate that you enter a number and it conclude something about colors. I do not know whether this is right or wrong but I guess it makes sense.
and hey don't Natural Numbers start from 1 and it is Whole Numbers that start from 0 (zero) ?? " 21:36 www.virtualnerd.com/algebra-2/equations-inequalities/real-numbers/number-types/natural-and-whole-numbers-definition
I'm a business informatics student so I am taking introduction to computer science and computer programming, and so far I hate everything that has to do with this major aside from these 2 courses which I am really enjoying so far, so I am considering changing to a computer science and engineering major, but I want to make sure this is what I will enjoy before I make this change so I am researching some stuff relating to computer science online, one of which being discrete mathematics. Do you think discrete maths gives good insight on how CS is like? How about 'Theory of computation' and 'Intro to Artificial Intelligence'? Do you have any other recommendations for courses or subjects that DO give good insight on CS and do not have any course prerequisites so I can look into them and be able to follow? Thanks in advance to those who take the time to answer and help :) Apologies if this has spammed you on more than one video. I'm just trying to get as much information as possible.
Kalernor Good question. As I'm just learning discrete math myself, I can't say if it gives good insight, but I've been involved in Comp Sci in one way or another for 4-5 years now. I think a good place to start would be to learn to program, if you haven't already. Programming is just a part of Computer Science, but I think it's going to pop up a lot along your journey towards the mastery of this topic. I'd personally recommend learning to program in Java or Python to start. Then, let your curiosity guide you to other languages from there (maybe a LISP, since you mentioned Artificial Intelligence and Theory of Computation). If you do start getting into a LISP, I'd recommended reading 'Structure and Interpretation of Computer Programs'. You can find the text for free online.
thank you both of you. I read both your comments and took it into consideration and have benefited from you guys. I made the decision to switch majors and thankfully I got accepted and now I believe this is the right move for me. (btw I already am learning Java and am really enjoying it as well, but I think the real meat for me would be the theories underlying programming and stuff like the courses Benjamin mentioned :D if not then I already know I enjoy programming and writing code )
+“miltonius” miguel juan adjudicator He added +n and -n (which is zero and has no affect on result, so you are not changing anythink by doing it) to have the fact (n^3-n) . For the +n, its added to 2n resulted 3n. Mathematicaly shown; n^3 + 3n^2 + 2n + n - n = (n^3 - n) + 3n^2 + 2n + n = (n^3 - n) + 3n^2 + 3n
The last but one theorem for the induction hypothesis was proven false for p(1) with one of the reasons contributing is 'In the inductive step, there are no horses for n+1=2'. How he has proven the base case of the last theorem as we are not tiling for p(0). I mean how could both of the following phrases are true at the same time for p(0)? 1. 'a way to tile' and 2. 'any square missing' Also any square missing meant a 2*2 square only(for Bill)? If so how could he have a 2*2 square for p(0)? And as clearly not, does it mean the length of the square be the length of the 'L' tile. (That's too assuming the base case taken at p(1)).
time span
proof by contradiction 2:30
prove √2 is irrational by contradiction
4:31
False proof example 13:42
induction proofs 20:03
example 1 on induction proof 24:30
example 2 on induction proof 36:37
example on false induction proof 43:34
Example 3 on induction proof 59:00
thank u !
IDK about you, but this is better than Netflix
Can you prove that
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Thanks to all the mathematic professors who helped me hate mathematics over the past years... I'm loving mathematics again :)
i didn't hate maths but I didn't liked it either now I'm starting to like it hope I will love it when this course is over
Oòlp
True story
I feel like a wizard when successfully doing a proof by induction!
Providing more clarity on the Horse Problem.
For the Induction Axiom to work, there are two requirements.
1. Base case must be true
- P(1) is trivially true, because the 1 horse will always be the same color as itself.
2. P(n) => P(n+1)
- So, when we write n horses, we typically write
h(1), h(2), h(3) ....h(n)
& for n+1 horses
h(1), h(2), h(3) ....h(n), h(n+1)
In the above statements, we sub-consciously assume that n is a bigger number greater than 1. Here lies the mistake. Since n can be any number, n can also be 1. So, the list can be:
h(n) -----> for n horses
h(n), h(n+1) -----> for n+1 horses
More specifically:
h(1) -----> for n horses, where n being 1
h(1), h(2) -----> for n+1 horses, where n being 1
Now, as per inductive step, predicate is assumed to be true.
Assumption: Set of any n horses is of the same color.
=> h(1) is the same color as itself, say color c1
The assumption also implies that h(2) is the same color of itself, say color c2
Now does h(1) => h(2)? i.e. does c1 = c2?
We can't determine that, can we? There is no way we say c1 is always same as c2.
So, the inductive step breaks.
The takeaway from this problem is to remember the "..." bug that we sub-consciously tend to make. Hope it helps someone :)
thank you !!
Thank you so much!
thanks :)
No bruh I think this is incorrect
The implication is false when true implies false.
Since h(1) is true, and h(2) is true, the implication h(1) => h(2) is true.
Although its been 3 years since I studied induction in high school, only today I understood why it is true. All thanks to the Indian education system for teaching induction without teaching mathematical logic.
Yes Bro sams thing happened with me
Teacher aya induction ko use kar k math kar dia 2 homework ko de dia kaam khatam 😟😟😟
yeaaa ritt
IFKR
69th like
We solve problems without even knowing why the hell we are solving it
The sound of that chalk pen hitting the blackboard came as pure music to my ears... One who has a love for computers will find this lecture to be a melody.
Loved the horse proof...and the tiling problem.
58:20 "So always check the base case. You could prove some great stuff if you don't check the base case."
I love this sentence XDDD
This is Awesome!!!
Louis Stuck till the end
I love it when the professor writes on the blackboard. It’s like they’re learning it with you. Almost all of my professors are using PowerPoint. :-/
Lies again? MRT MIT
Always be vary of the powerpoint
Power Point is the lazy man's choice
Damn....have to use a lot of brain power for this class. I like the approach of this professor where he engages the class, compared to other lazy professors who just read from slides.
I love it. The better part of the internet. Challenge us humans.
@@benisrood for sure
"Brain power"- well seems like you don't enjoy this 😊
proof by induction starts here :
28:31
I just realized that the last tile problem was a nicer way of asking to prove that (2^(2*n))-1 is divisible by 3. Love this more verbose way and as well as the story behind it :) Kudos to MIT
Damn i just had the realization too
With similarity of triangles you can disprove the statement that says "90>92" mentioned at minute 14:43.
In the big triangle before splitting in halves, the size of the vertical side is 10 and the horizontal's is 9. After splitting it, the sizes of the small triangles are 2 for the vertical side and unknown for the horizontal, let's call it x.
Their angles continue to be consistent, so we can apply similarity:
10/2 = 9/x
x = 1.8
So x > 2 is false.
this is great
This man made induction understandable in the space of 5 mins, where my own prof took 2 hours and left only a classroom full of blank stares. Strong knowledge base != Good teaching!!!
It's really an amazing course! I'm going to study in university next year and i'm coming to look for some inspires for the courses i will take. This outstanding professor do solve a lot of questions for me in advance.
proof by contradiction 2:30
i.e. root(2) is irrational
false proof 13:42
i.e. 90>92
induction 21:00
28:30 prove sum of 1st n natural number = n(n+1)/2
induction q: 37:00
41:45 STEPS OF induction
57:24 ....
killing that pythogarian was quite irrational... :P
get out
Dude 😬😬
I seeee what you did there
Pun intended
basis: 2x2 can be tile with a single tile that leaves the NE corner empty. Assume this holds for P(n) now show P(n+1). Split the corresponding 2^nx2^n into 2 regions of size 2^n, apply the induction hypothesis, it gives you that all of these 4 regions can be tiled leaving the NE corner empty. Now simply rotate the NW and SE 2^nx2^n regions 90 and -90 , respectively and you get a L-shaped tile missing in the middle - put a tile here. QED. Now to finish it off, when you have the required size you simply rotate the NE sub square by 180 to get its NE corner to the middle.
If only my maths lecturer taught like this guy when I was at university...!
I need proof that such Math profs exist. I think this is CGI....
Your assumption that this is CGI proves that there is at least one math teacher like him. If not, how would the CGI be made?
Boom! Mind = Blown
so are you saying dragons exist? orcs exist? all the other crazy things in movies and tv shows exist? CGI is just computer generated graphics - it doesnt have to be based of any real objects.
Again, flawed. He can be a combination of people if its CGI cuz atleast someone or some people should have thought about making him. Dragons are fictional, yes. But to make a CGI of a real person for one whole lecture with this much perfection surely means that there are people who thought of it. Meaning there are people who know how to explain it. Meaning there is atleast one math professor like him. Please try to think on it for longer. I am not saying he can't be CGI I'm saying his explaining and intellect must be thought by someone else he can't be CGI.
Yeah, you're totally right about that, I misunderstood what you meant initially.
🎯 Key Takeaways for quick navigation:
00:26 🧠 *The course emphasizes the importance of proof components: propositions, axioms, and logical deductions. Reasonable use of axioms is accepted, with a focus on consistency.*
01:49 🔄 *Introduction to proof by contradiction; assuming the opposite of the proposition, deriving a contradiction, and concluding that the proposition is true.*
03:01 🤔 *Steps for a proof by contradiction: Assume proposition P is false, use the assumption to derive a falsehood, leading to the conclusion that P is true.*
04:45 ➗ *Proof by contradiction example: Demonstrating that the square root of 2 is irrational using indirect reasoning.*
09:55 📚 *Historical context: The proof that the square root of 2 is irrational was discovered by the Pythagoreans, leading to challenges to their axioms and the suppression of the result.*
13:42 🚫 *Introduction to false proofs: Highlighting the pitfalls of relying on visual representations and assumptions in proofs.*
20:44 🔄 *Induction as a powerful proof technique: The course emphasizes the importance of understanding and applying proofs by induction in computer science.*
23:33 🎯 *Mathematical induction is like a series of dominoes, where proving a base case and the induction step ensures that a property holds for all values.*
24:30 🧮 *Induction proof example: Proving the sum of natural numbers from 1 to n equals n times n plus 1 over 2, using base case and inductive step.*
28:46 🔄 *In a proof by induction, you establish a predicate (p) representing the property you want to prove, check the base case, and show the inductive step.*
34:41 🛑 *Proof by induction doesn't necessarily provide an intuitive understanding of why a theorem is true; it's more about establishing the truth of a statement.*
36:35 ➗ *Induction example: Proving that for all natural numbers n, 3 divides n cubed minus n, using base case and inductive step.*
43:25 ❌ *False proof by induction: Attempting to prove that all horses are the same color, revealing the importance of correctly formulating the induction hypothesis.*
51:20 🐴 *Inductive proof using "pn" to show all horses in a set of n are the same color, addressing a mistake in the proof for n=1.*
55:35 🤔 *Careful use of the "dot, dot, dot" notation in mathematical proofs, highlighting a flaw in the case n=1 of the horse color proof.*
56:30 🧩 *Importance of proving all steps in an inductive proof, emphasizing the missing link in the proof chain (p1 implies p2) for horse color.*
59:21 🧠 *Application of mathematical induction to prove the existence of a solution and find the solution for a courtyard tiling problem.*
01:03:07 🧩 *Definition of the induction hypothesis ("pn") in the context of proving the tiling of a courtyard with a center square missing.*
01:08:06 🔄 *Iterative refinement of the induction hypothesis to successfully prove the tiling of a 2^n by 2^n region with a corner square missing.*
01:13:18 🔄 *Iterative refinement of the induction hypothesis by considering proving the problem with Bill in the corner first and then using it as a lemma for the center.*
01:14:09 🤔 *Discussion on alternative approaches, including dividing the courtyard into 2 by 2 squares, highlighting potential challenges.*
01:15:37 🛠️ *Stronger induction hypothesis (pn) can simplify the proof, emphasizing the advantage of assuming more inductive power for a more complex problem.*
01:16:34 🔄 *Application of a stronger pn (any square missing) to prove pn implies pn+1 more easily, showcasing the power of a robust induction hypothesis.*
01:18:28 🧠 *General rule: If you struggle with induction, try assuming something stronger as your induction hypothesis; a more powerful pn can make proving complex problems easier.*
Made with HARPA AI
So, i'm learning from MIT...Thank you ❤
this proves to me, it would not have been a good idea for me to go to MIT : (
@@donsurlylyte why not?
I thought I lost hope with mathematics. I was kinda slow in my thinking ability. Now I am working towards understanding maths for computer science so that I can go on studying master and eventually PhD. Hope you all enjoy this series of lectures.
You got this bro
I think it is the consensus among all CS students around the world, that this is the class that's kills us all.
For the tile problem,
If you have any square missing, the shape of the tiles will not be L
For the corner piece missing in 2^n * 2^n, try finding the last piece to be L in shape while leaving a corner empty.
I'm gonna be honest here. I've never really liked giving my time and my efforts to Mathematics. But damn, the way this professor teaches, it's very engaging and very very interesting. Thank you for these materials! These are really helpful.
I always thought this proof by induction stuff is like magic. NEVER understood it. But now I do (I think).
Great video and thank you for providing it to us for free
same here but when he said p(0)->p(1) all came up
For the one at the end make an 2^n+1 square out of a 2^n square (one fourth of the field) and the remainder, a 2^(n+1)x2^(n+1) L-shape out of 2^n L-shapes, meaning that each non-two-by-two square can resolve to a smaller square and L-shape where the square is in any quadrant, until you get to the two-by-two which is self-explanatory.
Professor Leighton, thank you for a solid introduction to proof by induction. In the early 1990's , I took Advanced Calculus at the University of Maryland College Park and proof by induction was introduced in a slightly different way.
Great lecture! I am absolutely in love with MIT open course ware.
@19:00
The mistake right up-front was not that it was drawn wrong. That is the secondary mistake. The original mistake is that the Height-measure of the original triangles was altered to 8 (with the remainder of 2 added to the sub-triangles) PRIOR to estimating the new base-measure of the sub-triangles.
With the correct Height-Measure of 10 and Base-Measure of 9, we don't assume the base is anything-plus. The base is obviously something-minus. But if you first alter the height to 8 with a base of 9, we then assume the base is something-plus, because we forget about the 2 we just removed.
So the idea of perception is correct in that it covers up mistakes. But the "right up-front" issue is actually the order of operations, not the sizes of the triangles. The sizes of the triangles are a secondary issue.
When I saw the 2+, I wondered why the base of a sub-triangle was different in proportion to the base of the parent-triangle whose base is smaller than its height. The image wasn't the cause of the issue, the order of operations was.
22:16 is the best explanation of induction in my life.
Indeed!
Man, that horse proof was a mindfuck...
Have you ever followed the horse proof while *high*, man
zooscientist1 you get get
Please call
@@zooscientist1 hahahahahahahahahahahahahahahahahahahaha
finally i hav no word to say its realy greate lecture thaks sir and thanks MIT
Thank you MIT, for your education.
On the court yard proof... the base case isn't correct because the statue isn't in the center. It's off to the side in one of the quadrants. It should be partially in all four quadrants to be considered centered...
It depends on what 'center' means according to them.
@1:15:30
I don't understand how "any" makes it "stronger". [Do whatever you want] seems much more relaxed of a requirement.
The requirement is more relaxed, but it might seem harder to prove because a solution to the more relaxed problem would automatically prove the more constrained problem.
In this sense, the more relaxed problem yields a "stronger" result because proving the strong result automatically proves the "weak" result.
I don't understand why he said 1:09:25 "We don't want put Bill in the corner, but in the center.". In his initial example with n=2 (1:01:40) he put Bill in the corner as well.
The tiles are L shaped , the minimum size of area would’ve 4x4 , four L shaped tiles . Then Bill would be in the corner of each L shaped tile but in the center of the 4 L shaped tiles. To me 2 to the n has to be divisible by 4 .
Amazing professor most people just learn math and know nothing of the history definitely added to my rewatch list
In the Horse Problem, does it then mean if we take the predicate to be: P(n): In any set of n>=2 horses, all horses in that set are the same color, then we can establish the truth of the theorem (that all horses are the same color), because then depending on the truth of the base case, we can always establish that P(2)->P(3), P(3)->P(4), ..., P(n)->P(n+1) are true. So that is a bit circumventing around what we actually sought to prove, given that we're assuming that in the base case. Really good problem.
Very informative. Teachers in my school never explained why we take p(n) => p(n+1) at inductive step.
Which school!
I'm Indiab Too
Just this second class completely destroyed my ego. One of the most difficult but stimulating and satisfying classes I've ever engaged myself in; for once I'm truly haunted by what lies ahead.
here a 2018 learner pleaseeee keep these videos alive!!! excellent job thank!!!! so
learning in 2022 and yes it's still alive and very useful to me
sir Time 15:45 , trying to prove 90>92
there in the doted line of a small triangle whose length is 2 is counted in both (small deducted triangle) square length whose length was supposed to be 9-2=7 but counted as 9 by8
AND
In that small Deducted triangle whose length and breadth is 2
either you count in (small deducted triangle)real length 9-2=7by8
Or
In that small deducted triangle whose length and breadth is 2 by 2
Horse Problem Clarification:
The inductive step doesn’t work when n=1
So P(n) = H1, H2, …, Hn
And P(n+1) = H1, H2, …, Hn, Hn+1
1. We assume P(n) is true by induction
2. In P(n+1), we know that the first n horses are the same color because we assumed P(n) is true
3. Now, remove H1 from P(n+1), you’re left with the set {H2, …, Hn, Hn+1}, which now has P(n) horses
4. So now we know that {H2, …, Hn, Hn+1} are all the same color because again P(n) is true
5.(Now here’s the error) The next step WOULD be to say that H1 = {H2, …, Hn} because of step 1 and {H2, …, Hn} = Hn+1 because of step 4, thus implying that P(n+1) are all the same color,
BUT look closely at case n = 1:
P(1)= H1
P(2) = H1, H2
Remember step 5: H1 = {H2, …, Hn} = Hn+1
But wait, look at the middle set, {H2, …, Hn}. It’s P(n) without H1.
Now what is P(1) without H1?
*NOTHING, IT’S EMPTY!*
Thus saying that H1 is the same color as no horses at all is false, which means step 5 in our induction is false.
but then for n>2 suddenly all horses are the same colour?
For those who are confused with the tile problem, I think the center square means that Bill must be inside a square which is in the center. Like imagine a smaller square for n = 2 which is inside the original square. Then you can cover the 4 corners with the L-shaped tiles and it'll leave a square in the center and bill will lie somewhere in that square. For more details check out the readings of the chapter-3 induction of the same problem
Motivation : read the comment before watching lecture
This will engage you till end of the courses
Teaching is art ! Great lecture By Tom Leighton .
bill problem can be done this way- well we have a area thats 2^n+1 by 2^n+1 so if we break that up we have 4X (2^n by 2^n) thats the total area since bill had a nice center tile in each of those 4 2^n by 2^n areas hence bill has a center tile in the bigger area made up of 2^n+1 by 2^n+1 and now lets say in case n is 2 this center area is like 2X2 space but we put in another L shaped tile and we have 1 sq left for bill in the center
This is why I find discrete math is more challenging to grasp than any other branch of math.
Really? I prefer discrete math over lets say calculus
Nah man calculus is fun
professor explained things simply
True. Made it easy to understand 🙂
hypothesis: any region of " ((2 power n x 2 power n ) - 1) / 3 " is true. Bill will never end up in the center. number of tiles extending away from bill towards opposite ends of any region will always be different ! to center bill and prove L shaped tiles will fit you have to use the hypothesis "((2 power n x 2 power n)-4)/3" is true, base greater than or equal to 1.
That 90 > 92 thing is kinda sounds like when you divide a 6 by 4 (24 chunks) chocolate and generates 1 more chocolate chunk and rearrange the chocolate back to 6 by 4 and you get 1 extra chunk.
To be more clear, +n - n = 0, so he basically just added 0, which doesn't change the value of the polynomial (which I just incorrectly referred to as an equation). Also, in case you missed it, he didn't just throw in the "-n", he also incremented the 2n (see the change to 3n), which is the +n I was referring to that cancels out the -n, leaving it unchanged.
that 90 > 92 thing really got me... be hold, the splash zone!
I really admit that I am having an interesting experience with these lessons you offer
Great introduction of inductive proofs. The use of invalid inductive processes is very instructive.
The Egyptian were the first mathematician not the Greeks. The Greek took the knowledge from the ancient Egyptia. Hence the triangular structure of the pyramids.
Exactly
Indeed, fabulous. Learned all the induction from here. Thank you so much!
Horses are not numbers or centimeters. They don't have any standarts about colors. That is why h(1) can be a set of itself with one specific color. Then h(2) can be a set of itself again with a different specific color. These two set can represent the "n=1"set alone. and These two cases are separately true.
Then for a "n+1" set of (h(1), h(2)) you can not imply either of these base case scenarios because they have different results. This breaks the implication from p1 to p2, which breaks all significance of the proof.
*How does induction work?*
This is what i understood :
Let's take the example of this formula
S(n) = 1 + 2 + 3 + 4 + -------- + n = n(n + 1) / 2
Now suppose I asked you the value of S(100) and you solved it for me
S(100) = 1 + 2 + 3 + -------- + 100 = 5050
Similarly , You also solved for me f(100) = 100(100 +1) / 2
= 5050
Now, assuming you weren't lying to me about S(100) and I'm required to find S(101), I can do that as follows :
S(101) = [1 + 2 + -------100] + 101 = [5050] + 101 = 5151
f(101) = 101(101 + 1) / 2 = 5151
Similarly If i have to find S(102) and assuming you did'nt lie to me about S(100), I can do
S(102) = [1 + 2 + 3 +--------- + 101] + 102 = [5151] + 102 = 5253
f(102) = 102(102 + 1) / 2 = 5253
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Similarly I can keep on calculating it for values greater than 100 (means 100 is the base case here) based upon the answer you told me...
So here, I'm not verifying the hypothesis, but assuming you verified it for me and using that I proved a conditional statement that
If p(n) holds true then p(n+1) also holds true.
i.e. p(n) => p(n + 1)
And when you check base case you are giving yourself that initial start...(just like you told me that S(100) is true.)..but here solving for the basecase you know its true cuz you verified it yourself...
*Sorry for my English. I'm not a native speaker*
@1:01:05 "I'm not supposed to reveal his name so let's call him Bill" LOL
And its Bill Gates
@58:30 you can prove some great stuff if you don't check the base case.
at 56:16, how is that we have proved that proposition for all n>=2?, the equality bridge broke a few minutes earlier....
"If you don't succeed at first, try something harder". Hahaha great quote.
This actually works believe it or not.
thought that too :D
I didn't watch the video, but I wanted to point out that there is a really nice proof for 3|(n^3 - n) that involves modular arithmetic. Simply note that n^3 - n = n(n+1)(n-1) which is clearly 0 mod 3.
"beat me, it doesn't look like a multiple of 3"
Ali Abdul-Kareem loves his response to that as well haha
Is it okay if I didn't understand proof by induction?
I mean am I dumb or it's a complex math. Or does it require a previous knowledge of math.
Even I didnt understand that much , all I did was that, if you can assume p(n) and show that p(n+1) is also true, then you have proved it
Induction can be pretty hard to understand at first. I didn't really understand it for a few years after first studying it. Many people struggle with induction, but with some persistence you can understand it.
My advice is to watch some more videos about it until you understsnd how to use it. If it still doesn't make sense to you after watching one video, come back the next day and watch another one. Sleeping on something is really helpful.
Once you have learned how to use induction, do some practice problems. Finally, after doing practice problems, watch this video again. Your practical experience will make it much easier to understand the theory behind induction proofs.
You can't always learn a difficult concept like induction all at once. But, by chipping away at it over time, you can come to truly understand it.
You need the base case to prove everything. The base is, literally, the base. You build from the ground up. When you’ve proved that the ground floor is perfectly done, and you have got the fact that a solid floor would enable you to build a floor on top of it, then you can theoretically construct a building with infinitely many floors.
The physical world says such building is not possible, but such thinking is reasonable in the realm of the induction axiom. That’s why the method of induction is an axiom, because it doesn’t apply to everything in the universe, and is chosen according to our purpose.
This is my third lecture I've watched with him and I kept thinking that this MIT room looks like a 70 year old community college in the middle of downtown Detroit, but then Tom drops a bomb saying that building is new, and furnished, was supposed to be under $100 million but skyrocketed to over $300 million.
well done, thanks for mit offer me such high quality lesson.
The "all horses are the same color" proof fails because the variable of interest is "color" which is not represented numerically in the proof. The proof presented refers only to the number of horses, there is no mathematical predicate for the color.
In 2^n + 1 square why prof. Has put the bill in the center in the upper left square...Bill wants to be in the center and the center of the square 2^n+1 is where bill earlier was...
Could anyone explain?
The centre was not fixed meaning Bill just didn't want his statue in the corners. And that is why the Professor has changed his theorem from a centre to any centre because we now have more room to play around and figure out the answer.
He didn't say it wasn't divisible by three, he said it didn't "look" like it was divisible by three. Subtracting n, then adding it back to an equation doesn't change the truth of the equation, so the fact that the rewritten equation was divisible by three proved that (n^3 + 3n^2 + 2n) is divisible by three as well.
7:43 how did he know "a squared" was a multiple of 4 ?
You can think of 2|a (2 divides a) as an equation: 2q=a, where q is an integer. Squaring both sides: 4q^2=a^2. Now q^2 is just any integer, so you might as well write 4q=a^2. Convert back to other notation: 4|a^2.
okay i have a doubt in the tile question if we are changing the cases for eg from center to corner and corner to any tile then the base case must be change as well and in the base case there will be 16 cases in which we have to proove the question for each case now thats make our life harder but in the last teacher said that this is easy how man how can one totally forget about the base case
can someone explain why he said"even when there's a square missing, we're in trouble. say this 64........" at 01:07:45?
by 64x4 he actually meant 4x4 (as he corrects later). In a 4x4 square we have 16 square tiles capacity. Even when there is no square missing(in the 4x4 square), and you try filling it with L shaped tiles (which have size 3), its not possible because there are 16 squares and you cant fill it completely with 3 tile sized squares. It is because 16 is not divisible by 3.
The whole bigger square has 64 square tiles and on dividing it into 4 parts we have four 16 tiled(4x4) squares. Hope this helps
6 divides [(2^2n)-4] for n>1 Is a more elegant proof for the last question.
Whenever i cant sleep, i watch this video so i can go asleep, thanks math video
in minute 40:37 , how does adding a -n result in that equation? n^3 + 3n^2 + 2n goes on to be n^3-n+3n^2+3n... after adding -n?
He skipped a few steps but, rearrange it so that its in the form.
n^3 +3n^2 + 2n = (n^3 + 2n) + 3n^2 .
then consider that (-n is just (2n - 3n).
add zero, +(3n -3n) to the equation.
which helps us rewrite it as (n^3 + (2n-3n) + 3n^2 + 3n.
= (n^3 -n) + 3n^2 + 3n
About 41:00.
I cannot understand why he wrote :
(n + 1)^3 - (n + 1) = n^3 + 3n^2 + 3n + 1 - (n + 1)
(n + 1)^3 is really equal to n^3 + 3n^2 + 3n + 1?
Why? Which rule(s) he applied to find this?
Sid Arta if a - b = c - b and a = e, don't you think we can substitute a with e to get e - b = c - b? Replace a with (n + 1)^3 and b with (n + 1) and it should make sense
Sid Arta (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
That's basic algebra, mate.
Mohit Shrestha Yes, identity... I need revisions. Thanks a lot to you and Rasaq Otunba.
Thought the problem was the base case since if one compares the color of the horses, there must be 2 or more horses. It turned out to be the implication from P(1) to P(2)!
In the end, i don't follow. In the Bill's last case, he said that Bill could be anywhere, but the example shows that it specifically is put in the corner--not anywhere, is there a fault? Can it be proved?
No.he just prove a stronger proposition that bill can be anywhere.
The original question asked him to prove bill can be in the center. He proved that bill can be anywhere, and therefore can be in the center.
Hello, thank you for all your efforts. I would like to know if the Recitation videos for this class are available to the general public.
Sorry there are no recitation videos available for this course but there are recitation notes available. Hopefully they will be of some help. See the course on MIT OpenCourseWare for the materials at ocw.mit.edu/6-042JF10.
@@mitocw it is really wonderful how much mit does so that quality education is available to everyone for free.
On the horse problem my understanding is this:
We cant prove p(2) i.e 2 horses are of same colour using the base case p(1).
A set of 1 horse is of same colour but that doesnt prove that in set of 2 horses both of them are same colour.
We cant use p(1) to prove p(2)
Yes thats true because if we assume p(1) to be true, then p(2) has to be false since so the implication is false. Because of this, the inductive "Domino" is not started since the base case itself fails.
Can anyone explain why @ 40:30, he throws in a (-n) to make the proof work. I understand that it does work but what rule allowed him to throw it in after the initial proof was false. i.e (n^3 + 3n^2 + 2n) did not divide three, so he changed it to (n^3 - n +3n^2 + 3n) which proved the theorem true.
He simply restated +2n as -n +3n. The rest didn't change. Effectively, nothing changed. It's like restating 5 as 3+2. You can do it if you like, the value doesn't change.
For the set which contains all sets of length n, where all sets of length n+1 has only horses of the same colour, it was not shown that the set of length 1 is a member.
By the way, the set of natural numbers does not include zero!
Did this in school when I was 16 or 17 years old. Did a ton of math in my engineering degree. Worked as a software engineer for many years and worked in IT for 30 years.
Never used this topic in math once in all that time.
Some math is helpful for computer scientists, but not this.
2/(triangleHeight -2) :: (suppposed2+Width/triangleWidth)
triangleHeight = 10;
traingleWidth = 9
2/8 == supposed2+Width/9
9/4 = suppposed2+Width== 2.25
(triangleHeight -2)x(triangleWidth + supposed2+Width) == actual area
8x(9+2.25) = 90
how does that person @ 18:00 say 1.8 something and also the professor.. can anyone explain??
(a -2)(b + c) == ab (conservation of area)
c/b == 2/(a - 2)
feeling blessed after joining MIT open courses ;-)
Around 34:00, he shows that (n+1)/2+(n+2)=n^2+2n+n+2/2. ... What algebra method is used to determine this? I can see how the third and final result of (n+1)(n+2)/2 is derived from the reverse of the FOIL method, but how did he get from step 1 to 2?
jackobthesnakob by expanding "n+1 " with 2 , in Order to get 2(n+1)/2
regarding the horse question (I havent seen the solution):
p(1) itself cant be used as the base case because p(1) recursively proves itself True. since there is only 1 horse, if we define these variables: p == "H1 is Black" and q == "all horses in any set is Black", we conclude that p => q == 1 (always), because if H1 is in fact Black then: p == 1, and if we check q we see that "all horses" consists of only H1 thus q should be 1 too. now if H1 is NOT Black, then p == 0 and q should be 0 too because, again, there is only H1 so "all horses" means that we're just checking H1
let's see if I got if right!
I didnt :(
At 45:01, why can't we take the predicate - P(n) : n horses are of same color?
you can.
Could someone please explain to me how he got the (-n) into the equation at 40:35? I was following the reasoning until that point.
He added n and subtracted n at the same time, so:
(n^3 + 3n^2+2n) = (n^3 + 3n^2+2n) - n + n = (n^3 -n + 3n^2+2n+n) = (n^3 -n + 3n^2+3n) =( n^3 -n) + (3n^2+3n)
@@darbrzoz Thank you!
For the statue proof, how would one define in mathematical terms the L shape figure of squares used? I proved this is possible using modulo of 3 but it doesn’t specify the shape. Matrice notations maybe?
I thought that the bug was in the predicate itself as color is not a function of numbers so how can we make a predicate that you enter a number and it conclude something about colors.
I do not know whether this is right or wrong but I guess it makes sense.
and hey don't Natural Numbers start from 1 and it is Whole Numbers that start from 0 (zero) ?? " 21:36
www.virtualnerd.com/algebra-2/equations-inequalities/real-numbers/number-types/natural-and-whole-numbers-definition
I'm a business informatics student so I am taking introduction to computer science and computer programming,
and so far I hate everything that has to do with this major aside from these 2 courses which I am really enjoying so far,
so I am considering changing to a computer science and engineering major,
but I want to make sure this is what I will enjoy before I make this change so I am researching some stuff
relating to computer science online, one of which being discrete mathematics.
Do you think discrete maths gives good insight on how CS is like?
How about 'Theory of computation' and 'Intro to Artificial Intelligence'?
Do you have any other recommendations for courses or subjects that DO give good insight on CS and
do not have any course prerequisites so I can look into them and be able to follow? Thanks in advance to those
who take the time to answer and help :) Apologies if this has spammed you on more than one video. I'm just trying to get as much information as possible.
Kalernor
Good question. As I'm just learning discrete math myself, I can't say if it gives good insight, but I've been involved in Comp Sci in one way or another for 4-5 years now.
I think a good place to start would be to learn to program, if you haven't already. Programming is just a part of Computer Science, but I think it's going to pop up a lot along your journey towards the mastery of this topic. I'd personally recommend learning to program in Java or Python to start. Then, let your curiosity guide you to other languages from there (maybe a LISP, since you mentioned Artificial Intelligence and Theory of Computation). If you do start getting into a LISP, I'd recommended reading 'Structure and Interpretation of Computer Programs'. You can find the text for free online.
thank you both of you. I read both your comments and took it into consideration and have benefited from you guys. I made the decision to switch majors and thankfully I got accepted and now I believe this is the right move for me. (btw I already am learning Java and am really enjoying it as well, but I think the real meat for me would be the theories underlying programming and stuff like the courses Benjamin mentioned :D if not then I already know I enjoy programming and writing code )
40:41 I don't get it. Can someone explain? why'd he subtracted n from n^3+3n^2+2n? and how'd it ended up n^3-n+3n^2+3n.
+“miltonius” miguel juan adjudicator He added +n and -n (which is zero and has no affect on result, so you are not changing anythink by doing it) to have the fact (n^3-n) . For the +n, its added to 2n resulted 3n. Mathematicaly shown; n^3 + 3n^2 + 2n + n - n = (n^3 - n) + 3n^2 + 2n + n = (n^3 - n) + 3n^2 + 3n
The last but one theorem for the induction hypothesis was proven false for p(1) with one of the reasons contributing is 'In the inductive step, there are no horses for n+1=2'.
How he has proven the base case of the last theorem as we are not tiling for p(0).
I mean how could both of the following phrases are true at the same time for p(0)?
1. 'a way to tile' and 2. 'any square missing'
Also any square missing meant a 2*2 square only(for Bill)?
If so how could he have a 2*2 square for p(0)?
And as clearly not, does it mean the length of the square be the length of the 'L' tile. (That's too assuming the base case taken at p(1)).
heard oxford was the best place for math, now I kind of agree.