I have horrible ADHD and I thought I would never be able to learn this class, but when I am able to rewind the vid when my mind fazes, this becomes such an easy class. Thank goodness for online vids and a good teacher! Thanks

The one thing that I find interesting is that we are assuming the antecedent in the conditional to be true for every inductive proof, but in reality, the logic if a then b (ie, a implies b; or, a => b) is true only when either a is F or b is T. Sure, both a and b can be T, but why necessarily assume that the antecedent is always true when proving by induction? Couldn't we just as easily assume the antecedent to be false and still prove that the conditional is true?

In the question about the horses,we get P(2) implies P(3),P(3) implies P(4) and so forth.Suppose,I frame a statement "Given that 2 horses are of the same colour,any set of horses greater than 2 must be of the same colour",wouldn't this become true?Which is wrong.So,what's wrong with my statement?

In the horses example, we chose the wrong base case. we should have started with 2. But couldn't we just say the base case is 3 or 4 ? and that would yield a correct proof? What governs our choice of the base case ?

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 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.

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?

Math is still a religion. It would have been so awesome to see the looks on their faces when irrational numbers were realized. Of course, these numbers don't really exist so we come full circle.

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?

it's time to restore order, and make the square root of 2 rational again, by killing everyone who thinks that it's irrational. Who's with me??? I've already proved that the Earth is flat, so a number thing will be much easier!

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?

How 3|[(0^3)-0] be true in base step at #39:09 ? It surely looks false to me since n|0 is not defined. Therefore there should be limit for n to be greater than 0.

The proof for Horses was wrong for a different reason. For his proof, n is 'any' set… which means any arbitrary set you can choose. But in his induction step he treated n as a variable which is using the proposition to prove itself. False by circular reasoning.

Question for the horse problem: Why didn't his base case prove/imply that H2 was true? From what I see, it looks the same as his other base cases, he proved that P(1) is true, and then went on to the the inductive step. What am I missing that makes it wrong?

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 🙂

i am understanding something with the horse thing wrong so please correct me if you think i am wrong and why?… ok so, why i think the horse was wrong is that when we assume that p(n) is true we also defined p(n) to be the set from h1,h2,…,hn but when we examined p(n+1) we also assumed that p(n) is {h2,h3,..,hn+1} is true but that means that p(n) has two different sets which is against the math logic that p(n) (a function) has only one answer….. so thats why i think it is wrong not that we didn't test it for p(1)=>p(2) …. so if you think i am wrong please tell me why ,thank you.

k j m E.g take b2 we will have k,j,j,m etc the same will have k,m,j,m then c will have k,k,j,m etc j m k take c2(b) we will have kjjm the same will have k,m,j,m then c will have kkjm etc j k m one letter k generates 3 options, the second the same when we get to row 2 the same m k j bearing in mind the letters the options generated by the column from 1 to 4 must differ j k m even if I what to go up to 100. which branch of maths is this? k m j from k alone in my example I will get three more columns, j the same getting these results, e.g m j k k again the same k to produce b2 k b2 k b2 k after k I will start with j,m ,j,m ,j,k,m b3 j c3 m d3 k the j k j m j k m. b4 j b4 j b4 j b5 m b5 m b5 m b6 j b6 j b6 j b7 k b7 k b7 k b8 m b8 m b8 m

Axioms as in a comparatively permanent and consistent position, reinforced through repetition and usage. Based on some set of previously deduced conclusions?

What about a field of interacting tenets and premises in which all elements are variable, even though recognizable ones do crop up again and again?

So long as over arching agenda is not one based on propose, dispose, oppose or depose? Where royal markers and gold standards are the targets?

Because then it just becomes a game of conkers! Nuts on a string bashing at one another. No matter how fancy the nuts you are using as wrecking balls sure, they are just nuts at the end of the day.

Maybe to prove negative value bias of anti-establishment mind sets, that academia is supposed to be famous for. That statesmen and regulators are willing to fund to prove maybe.

In determining psychological profiles of tomorrows republics. Once we've settled on the fact that inherited birth rights are virtually useless

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.

I teach maths+logic to computer science students. But I could not do these broken/repairable inductive proofs with my undergraduate students, I'd just get a wall of blank stares back. You need to have some super-bright students to have them spot that the Bill-in-the-corner proposition can be used as a lemma to prove the main theorem. That the broken horse-proof flummoxed an earlier cohort in their homework just goes to show that even at MIT you can push the boat out too far…

That is the most absurd explanation of why Pythagoras was killed. He was killed because of his philosophical convictions about the soul of man. Nothing to do with his math skills.

the decimal place value system developed in 5th century by aryabhatta but yeah greek knew about it altough we have no proof about it . i totally belive this story

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

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)

basis: 2×2 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.

wait so if we assume that any pair of horses are the same color or P(2) is true, then we can say that all n horses in a set are the same color given n>=2??????????

How does the horse proof hold for the case where n>=2? The text says "You should think about how to explain to such a student why this claim would get no credit on a Math for Computer Science exam."

Just to be sure, can someone articulate the explanation?

Awesome lecture.

I have horrible ADHD and I thought I would never be able to learn this class, but when I am able to rewind the vid when my mind fazes, this becomes such an easy class. Thank goodness for online vids and a good teacher! Thanks

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.

Fuuuuuuuuuuuuuuck this.

Can anyone tell me that how b is even when 2 | b^2 @ (8:08) Thanks in advance

thx,mit,this course really helps

7:34

(2/a) how that ?!

they are indepenndet sets yellow and red

The one thing that I find interesting is that we are assuming the antecedent in the conditional to be true for every inductive proof, but in reality, the logic if a then b (ie, a implies b; or, a => b) is true only when either a is F or b is T. Sure, both a and b can be T, but why necessarily assume that the antecedent is always true when proving by induction? Couldn't we just as easily assume the antecedent to be false and still prove that the conditional is true?

In the question about the horses,we get P(2) implies P(3),P(3) implies P(4) and so forth.Suppose,I frame a statement "Given that 2 horses are of the same colour,any set of horses greater than 2 must be of the same colour",wouldn't this become true?Which is wrong.So,what's wrong with my statement?

https://www.facebook.com/photo.php?fbid=10208289402307051

What's with this guy and the God talk?

In the horses example, we chose the wrong base case. we should have started with 2. But couldn't we just say the base case is 3 or 4 ? and that would yield a correct proof? What governs our choice of the base case ?

killing that pythogarian was quite irrational… 😛

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)

and hey don't Natural Numbers start from 1 and it is Whole Numbers that start from 0 (zero) ?? " 21:36

http://www.virtualnerd.com/algebra-2/equations-inequalities/real-numbers/number-types/natural-and-whole-numbers-definition

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

absolutely brilliant class

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.

In the induction step, why can we assume the induction hypothesis to be true?

professor explained things simply

very helpful lecture

deepthroat

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.

Man, that horse proof was a mindfuck…

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?

I wasn't aware there were mathematical whistleblowers in ancient Greece ! haha . Great lecture though.

Thank you MIT, for your education.

At 34:30, shouldn't that be "for all n greater than zero" as opposed to greater than or equal to zero?

I wanna to know the reading assignments. There is no information about this on the website of the course. Thx.^ ^

Math is still a religion. It would have been so awesome to see the looks on their faces when irrational numbers were realized. Of course, these numbers don't really exist so we come full circle.

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?

Also, I still can't figure out how come a harder case can be proved even more easily?

This is awesome 🙂

it's time to restore order, and make the square root of 2 rational again, by killing everyone who thinks that it's irrational. Who's with me??? I've already proved that the Earth is flat, so a number thing will be much easier!

"If you don't succeed at first, try something harder". Hahaha great quote.

If only I had a lecturer to teach at my university…

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?

How 3|[(0^3)-0] be true in base step at #39:09 ? It surely looks false to me since n|0 is not defined. Therefore there should be limit for n to be greater than 0.

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

@7:45 How is 'a', a multiple of 4? How did he arrive at that?

What a great teacher

@1:01:05 "I'm not supposed to reveal his name so let's call him Bill" LOL

The proof for Horses was wrong for a different reason. For his proof, n is 'any' set… which means any arbitrary set you can choose. But in his induction step he treated n as a variable which is using the proposition to prove itself. False by circular reasoning.

Nobody puts Billy in a corner!

If you want to save a lot of time, play this video 2x faster along with subtitles.

Question for the horse problem:

Why didn't his base case prove/imply that H2 was true?

From what I see, it looks the same as his other base cases, he proved that P(1) is true, and then went on to the the inductive step.

What am I missing that makes it wrong?

15:30

why this dotted line bigger than two?

2/10 = x/9

=> x < 2 that's the truth

prove that proof by induction works.

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 🙂

here a 2018 learner pleaseeee keep these videos alive!!! excellent job thank!!!! so

i am understanding something with the horse thing wrong so please correct me if you think i am wrong and why?… ok so, why i think the horse was wrong is that when we assume that p(n) is true we also defined p(n) to be the set from h1,h2,…,hn but when we examined p(n+1) we also assumed that p(n) is {h2,h3,..,hn+1} is true but that means that p(n) has two different sets which is against the math logic that p(n) (a function) has only one answer….. so thats why i think it is wrong not that we didn't test it for p(1)=>p(2) …. so if you think i am wrong please tell me why ,thank you.

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

lol, if you can proof bill can be in the corner, you just need to rotate the quarter that bill is in to put him in the middle. is it wrong?

Can anybody tell me how a is even from 2*(b^2) = a^2 ??

I'm a bit lost with "Bill's" problem… Do I need to study matrices properties with powers of 2 ?

At 45:01, why can't we take the predicate – P(n) : n horses are of same color?

do u guys know that this prof is a multimillinoire?

in the problem All horses are the same color . the prof first applied p(n) to 1 to n horses then applied p(n) to 2 to n+1 horses. isnt that wrong?

Proof by induction is great.. I didn't fully grasp the tile problem, but I guess it will come to me by some time..

k j m E.g take b2 we will have k,j,j,m etc the same will have k,m,j,m then c will have k,k,j,m etc

j m k take c2(b) we will have kjjm the same will have k,m,j,m then c will have kkjm etc

j k m one letter k generates 3 options, the second the same when we get to row 2 the same

m k j bearing in mind the letters the options generated by the column from 1 to 4 must differ

j k m even if I what to go up to 100. which branch of maths is this?

k m j from k alone in my example I will get three more columns, j the same getting these results, e.g

m j k k again the same k to produce

b2 k b2 k b2 k after k I will start with j,m ,j,m ,j,k,m

b3 j c3 m d3 k the j k j m j k m.

b4 j b4 j b4 j

b5 m b5 m b5 m

b6 j b6 j b6 j

b7 k b7 k b7 k

b8 m b8 m b8 m

Are there any lecture notes available?

I think it is the consensus among all CS students around the world, that this is the class that's kills us all.

Indeed, fabulous. Learned all the induction from here. Thank you so much!

10:53 The Greeks didn't have decimal notation.

Why is this in my suggested

Who understood this horse problem? I am dreaming horses now.

Wow that is a great lecture :)))) I love Discrete Math now

Axioms as in a comparatively permanent and consistent position, reinforced through repetition and usage. Based on some set of previously deduced conclusions?

What about a field of interacting tenets and premises in which all elements are variable, even though recognizable ones do crop up again and again?

So long as over arching agenda is not one based on propose, dispose, oppose or depose? Where royal markers and gold standards are the targets?

Because then it just becomes a game of conkers! Nuts on a string bashing at one another. No matter how fancy the nuts you are using as wrecking balls sure, they are just nuts at the end of the day.

Maybe to prove negative value bias of anti-establishment mind sets, that academia is supposed to be famous for. That statesmen and regulators are willing to fund to prove maybe.

In determining psychological profiles of tomorrows republics. Once we've settled on the fact that inherited birth rights are virtually useless

The number is one and two in line.

This guy knows nothing

Don't believe it

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.

22:16 is the best explanation of induction in my life.

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.

feeling blessed after joining MIT open courses 😉

Very very bad math level for university students, I started by set theory and topology in my first year at university.

I teach maths+logic to computer science students. But I could not do these broken/repairable inductive proofs with my undergraduate students, I'd just get a wall of blank stares back. You need to have some super-bright students to have them spot that the Bill-in-the-corner proposition can be used as a lemma to prove the main theorem. That the broken horse-proof flummoxed an earlier cohort in their homework just goes to show that even at MIT you can push the boat out too far…

Do we have recitation playlist as well?

can someone explain why he said"even when there's a square missing, we're in trouble. say this 64…….." at 01:07:45?

What a great lecture. Thanks!

We students in China learn induction when we are 15…

Is there anyone who has solutions of "Mathematics for computer Science"?

That is the most absurd explanation of why Pythagoras was killed. He was killed because of his philosophical convictions about the soul of man. Nothing to do with his math skills.

the decimal place value system developed in 5th century by aryabhatta but yeah greek knew about it altough we have no proof about it . i totally belive this story

Very informative. Teachers in my school never explained why we take p(n) => p(n+1) at inductive step.

oh math is a real art

So, i'm learning from MIT…Thank you ❤

Bill, take your money away!

This is why I find discrete math is more challenging to grasp than any other branch of math.

L tile problem gone above my head …..!

holy shit! why I didnt discover this video when I was in school, it is so intersting and intriguing!

I wish, i were studying in this institution.

"beat me, it doesn't look like a multiple of 3"

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

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)

basis: 2×2 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.

wait so if we assume that any pair of horses are the same color or P(2) is true, then we can say that all n horses in a set are the same color given n>=2??????????

tq mit ocw

For the proof of root 2 is irrational, if a and b occurs to be even, they could be divided by two and be reduced to lowest fraction right?😮

How does the horse proof hold for the case where n>=2? The text says "You should think about how to explain to such a student why this claim would get no credit on a Math for Computer Science exam."

Just to be sure, can someone articulate the explanation?