# Lec 2 | MIT 6.042J Mathematics for Computer Science, Fall 2010

### 100 thoughts on “Lec 2 | MIT 6.042J Mathematics for Computer Science, Fall 2010”

• February 20, 2016 at 4:52 am

"All horses are the same color." Zebra(h)s complaining about racism …

P(1) is trivially true, let's say FOR NOW that P(2) is also TRUE. Proving P(3) should not be difficult:
P(2) is true =>
=> color{h1, h2} = color{h1} = color{h2} & color{h2, h3} = color{h2} = color{h3} (1)
=> color{h1} = color{h2}= color{h3} = c (2)
=> color{h1, h2, h3} = c
=> P(3) is true
"equal to" is transitive so (1) implies (2). We could go on for a couple of more proofs (P(3) => P(4), P(4) => P(5), …) same principle.
Remember that P(2) wasn't proven to be true (I assumed…). The main problem is that we cant prove P(1) => P(2), let's see why:
P(1) is true =>
=> color{h1} = c1 & color{h2} = c2
c1 could be equal to c2 but certainly does not have to be. Just think about it, "P(n): In any set of n horses, the horses are all the same color.", the main thing is to interpret this predicate correctly.

I had the need to add an explanation because I assumed that the professor lost some students @ aprox 52:00. Nevertheless great Professor.

My sqrt(2) cents.

• March 6, 2016 at 9:16 am

Awesome lecture.

• April 4, 2016 at 6:44 pm

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

• April 8, 2016 at 12:09 pm

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.

• April 12, 2016 at 3:54 am

Fuuuuuuuuuuuuuuck this.

• April 24, 2016 at 12:28 pm

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

• May 12, 2016 at 1:06 pm

thx,mit,this course really helps

• May 16, 2016 at 3:01 pm

7:34
(2/a) how that ?!

• May 18, 2016 at 2:38 pm

they are indepenndet sets yellow and red

• May 21, 2016 at 7:09 am

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?

• May 29, 2016 at 7:07 pm

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?

• May 31, 2016 at 1:51 pm

• June 10, 2016 at 10:56 pm

What's with this guy and the God talk?

• June 30, 2016 at 10:38 pm

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 ?

• July 11, 2016 at 7:22 pm

killing that pythogarian was quite irrational… 😛

• July 11, 2016 at 7:51 pm

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)

• July 11, 2016 at 8:06 pm

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

• July 14, 2016 at 6:55 pm

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

• July 19, 2016 at 3:43 am

absolutely brilliant class

• July 26, 2016 at 12:21 pm

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.

• July 26, 2016 at 9:02 pm

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

• August 7, 2016 at 8:13 pm

professor explained things simply

• August 21, 2016 at 7:44 am

• September 4, 2016 at 6:41 am

deepthroat

• September 17, 2016 at 9:18 pm

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.

• September 21, 2016 at 11:41 pm

Man, that horse proof was a mindfuck…

• September 29, 2016 at 3:34 pm

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?

• October 3, 2016 at 1:02 am

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

• December 17, 2016 at 10:04 am

Thank you MIT, for your education.

• December 24, 2016 at 12:09 pm

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

• January 20, 2017 at 6:58 am

• January 27, 2017 at 4:13 pm

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.

• February 4, 2017 at 9:02 am

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?

• February 4, 2017 at 9:04 am

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

• March 1, 2017 at 9:11 pm

This is awesome 🙂

• April 4, 2017 at 12:31 am

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!

• April 29, 2017 at 8:50 am

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

• May 15, 2017 at 9:06 am

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

• May 18, 2017 at 8:22 am

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?

• May 30, 2017 at 10:54 pm

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.

• June 29, 2017 at 3:52 pm

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

• July 2, 2017 at 2:54 am

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

• August 11, 2017 at 12:49 am

What a great teacher

• August 20, 2017 at 12:56 pm

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

• August 20, 2017 at 7:44 pm

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.

• September 7, 2017 at 7:40 am

Nobody puts Billy in a corner!

• September 14, 2017 at 9:39 am

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

• September 19, 2017 at 11:23 pm

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?

• September 30, 2017 at 11:52 pm

15:30
why this dotted line bigger than two?
2/10 = x/9
=> x < 2 that's the truth

• October 15, 2017 at 8:48 pm

prove that proof by induction works.

• October 28, 2017 at 1:30 pm

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 🙂

• February 7, 2018 at 12:12 am

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

• February 16, 2018 at 6:11 am

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.

• April 30, 2018 at 12:25 pm

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

• May 21, 2018 at 5:18 pm

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?

• May 27, 2018 at 3:50 am

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

• June 4, 2018 at 4:50 pm

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

• June 24, 2018 at 5:35 am

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

• June 24, 2018 at 3:38 pm

do u guys know that this prof is a multimillinoire?

• June 27, 2018 at 1:51 pm

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?

• June 30, 2018 at 11:36 am

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

• July 12, 2018 at 6:08 am

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

• July 16, 2018 at 8:41 am

Are there any lecture notes available?

• August 2, 2018 at 5:00 pm

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

• August 27, 2018 at 11:28 am

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

• August 28, 2018 at 5:55 pm

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

• September 21, 2018 at 5:06 am

Why is this in my suggested

• September 27, 2018 at 3:53 pm

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

• September 29, 2018 at 2:48 pm

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

• October 3, 2018 at 12:29 pm

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

• October 7, 2018 at 7:55 pm

The number is one and two in line.

• October 8, 2018 at 7:45 pm

This guy knows nothing
Don't believe it

• October 13, 2018 at 9:20 pm

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.

• November 2, 2018 at 12:02 am

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

• November 3, 2018 at 7:45 pm

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.

• November 24, 2018 at 7:22 am

feeling blessed after joining MIT open courses 😉

• November 29, 2018 at 12:39 am

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

• December 18, 2018 at 3:05 pm

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…

• December 26, 2018 at 12:41 am

Do we have recitation playlist as well?

• January 6, 2019 at 7:00 am

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

• January 16, 2019 at 8:50 pm

What a great lecture. Thanks!

• January 31, 2019 at 8:17 am

We students in China learn induction when we are 15…

• February 24, 2019 at 8:43 pm

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

• March 5, 2019 at 8:40 pm

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.

• March 19, 2019 at 5:51 am

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

• March 22, 2019 at 8:25 pm

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

• March 30, 2019 at 6:49 am

oh math is a real art

• April 3, 2019 at 3:56 pm

So, i'm learning from MIT…Thank you ❤

• May 10, 2019 at 10:32 am

• May 20, 2019 at 7:50 pm

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

• June 28, 2019 at 5:17 pm

L tile problem gone above my head …..!

• August 10, 2019 at 2:11 pm

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

• September 14, 2019 at 6:17 pm

I wish, i were studying in this institution.

• September 25, 2019 at 8:29 am

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

• October 4, 2019 at 8:22 pm

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

• October 5, 2019 at 3:26 pm

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)

• October 14, 2019 at 8:13 am

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.

• October 30, 2019 at 2:47 pm

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??????????