Lec 2 | MIT 6.042J Mathematics for Computer Science, Fall 2010
Вставка
- Опубліковано 30 гру 2012
- Lecture 2: Induction
Instructor: Tom Leighton
View the complete course: ocw.mit.edu/6-042JF10
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu
IDK about you, but this is better than Netflix
Can you prove that
@@codegirl2069 h
Hgh
Hgh
Th h
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 !
Thanks to all the mathematic professors who helped me hate mathematics over the past years... I'm loving mathematics again :)
I thought I was the only one… apparently not.
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
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 feel like a wizard when successfully doing a proof by induction!
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.
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
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
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 😊
I think it is the consensus among all CS students around the world, that this is the class that's kills us all.
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
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.
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!!!
I like how when he says: I guarantee you that you'll make a mistake if you use/do X, and then actually makes all the student fall for the trick
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.
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
Great lecture! I am absolutely in love with MIT open course ware.
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.
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.
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
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?
Thank you MIT, for your education.
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.
killing that pythogarian was quite irrational... :P
get out
Dude 😬😬
I seeee what you did there
Pun intended
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.
proof by induction starts here :
28:31
Indeed, fabulous. Learned all the induction from here. Thank you so much!
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
Teaching is art ! Great lecture By Tom Leighton .
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 really admit that I am having an interesting experience with these lessons you offer
@1:01:05 "I'm not supposed to reveal his name so let's call him Bill" LOL
And its Bill Gates
@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.
"If you don't succeed at first, try something harder". Hahaha great quote.
This actually works believe it or not.
Great introduction of inductive proofs. The use of invalid inductive processes is very instructive.
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 ....
finally i hav no word to say its realy greate lecture thaks sir and thanks MIT
Amazing professor most people just learn math and know nothing of the history definitely added to my rewatch list
absolutely brilliant class
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.
What a great teacher
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?
well done, thanks for mit offer me such high quality lesson.
Finally induction is clear... Thanks a ton!
"beat me, it doesn't look like a multiple of 3"
Ali Abdul-Kareem loves his response to that as well haha
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
professor explained things simply
True. Made it easy to understand 🙂
thx,mit,this course really helps
that 90 > 92 thing really got me... be hold, the splash zone!
What a great lecture. Thanks!
22:16 is the best explanation of induction in my life.
Indeed!
feeling blessed after joining MIT open courses ;-)
Motivation : read the comment before watching lecture
This will engage you till end of the courses
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
Awesome lecture.
Best professor ever
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.
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.
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.
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
Thanks, great explanation.
holy shit! why I didnt discover this video when I was in school, it is so intersting and intriguing!
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
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
I like watch this it's so very good for my studies
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.
I like these staff...
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.
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.
Proof by induction is great.. I didn't fully grasp the tile problem, but I guess it will come to me by some time..
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.
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.
GREAT WORK
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.
very helpful lecture
Wow that is a great lecture :)))) I love Discrete Math now
This is awesome :)
This really helps me fall asleep thx
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.
Whenever i cant sleep, i watch this video so i can go asleep, thanks math video
Holy shit, these lectures make me mind blow I might have to try and do the readings first and take some time to eat some of these ideas but holy hell
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 .
Thank you.
Best forever
At the inductive step, shouldn't be a double arrow between p(n) and p(n+1)? because at the end when you prove that p(n+1) is true that makes the expression p(n) also true
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.
In the last proof of 2^n x 2^n tiles, can I make a proposition with "any corner" instead of "anywhere" ? This way I will have a more powerful P(n) without any need of lemma?
Hey guys, can't we just nest a square of size 2^n x 2^n on which the inductive hypothesis has been applied in the centre of a square of size 2^(n+1) x 2^(n+1)? In that case, p(n) => p(n+1) right? Meaning every subsequent block up to n+1 will have a bill in the centre.
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
oh math is a real art
Can someone help me here :
In the last problem (bill's), we wanted to prove p(n) but weren't able to do so, so we changed the p(n) and then we proved it . Right?
But then we didn't prove it for the original p(n) still?!! So didn't the very proposition we set out to prove still remained unproved?
If only I had a lecturer to teach at my university...
34:45 OMG!! Stunning! Magic!
6 divides [(2^2n)-4] for n>1 Is a more elegant proof for the last question.
@58:30 you can prove some great stuff if you don't check the base case.
I'm a bit lost with "Bill's" problem... Do I need to study matrices properties with powers of 2 ?
@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.
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?