Proof: a³ - a is always divisible by 6 (2 of 2: Proof by exhaustion)

[@beccabattalio]

I watched this when I first woke up this morning, and I must say it was super satisfying. Love a nice, tidy little proof.

[@aashsyed1277]

#EDDIEWOOLOVELYMATHS

[@williamgolove561]

I watched it at 7:30pm

[@freddyfozzyfilms2688]

the product of n consectutive integers will always be divisible by n factorial

[@matej_grega]

6×7×8×9×10 ≠ 5!

[@siddkumar8032]

@@matej_grega Freddy didnt say equal to 5!, but divisible. So if you do 6x7x8x9x10 and divide by 5! you get no remainder

[@muhammadhamad5191]

Pp

[@muhammadhamad5191]

P

[@muhammadhamad5191]

P

[@wyboo2019]

a^3-a is the product of three consecutive integers: multiples of 2 (even numbers) occur every other number, so either one or two of the products is even, so a^3-a is divisible by 2 multiples of three occur every third number, so exactly one of the three is divisible by 3 and thus a^3-a is divisible by 3 so a^3-a is divisible by both 2 and 3, and therefore also 2*3=6

[@KevinAPamwar]

& If 'a' is divisible by 6 by both first prime numbers 2 & 3 =6K then at least one of the numbers 6K-1 or 6K+1 is a prime number also e.g 5,6,7 11,12,13 17,18,19 23,24,25 29,30,31 35,36,37 41,42,43 47,48,49 53,54,55

[@jacoblepley9966]

Wow... as a senior in high school finishing up calc AB... this proof thoroughly amazed me... AWESOME!!!!!!

[@soba4523]

Me: You can change the odd expression with 2k -1 yeah? My math teacher: *nO.*

[@jack_papel]

For this proof it happens to not matter. Also your math teacher is unfortunately stubbornly wrong

[@lychenus]

a lot of math teacher sucks, and its a fact. no need proof am math grad.

[@cquirklesed7578]

Any three consecutive integers will contain a multiple of 2 and also a multiple of 3. Hence, the product P of the 3 consecutive integers must be a multiple of 2 and 3 => P is divisible by 2×3=6. Same goes for any n consecutive integers, i.e. n consecutive integers will contain an integer that is a multiple of all of {1,2,...n} and hence their product must be a multiple of 1×2×...×n=n!

[@pranavanil5420]

This question was eating me alive for weeks, thank you very much sir

[@youssefwahba6120]

the pigeonhole principle can be applied to prove that any three consecutive integers will have a multiple of 3. Since all elements of Z have a remainder of 0,1,2 when dividing by 3, and the three consecutive integers cannot have the same remainder, at least one of the three leaves a remainder of 0, thus at least one is a multiple of 3.

[@KirilF]

I feel like I can hear him all day and not get tired XD best teacher ever ;)

[@oyeajugando7262]

Sos un genio!! Me salvaste de caer depresión por Álgebra ❤

[@mangeshpuranik31]

Not just cube. I have just realised that this might hold true for any odd power, starting from 3 i.e. a^5, a^7 and so on! Numbers are beautiful 🔥

[@zafnas5222]

It does work for 5 and 7, not because they are odd but because they are prime. For example, it will not work for 9.

[@mangeshpuranik31]

@@zafnas5222 it does work for 9, mate.

[@ГеоргиГеоргиев-с3г]

U R basically substituting a for a^n in line one: a^(1/n)(a^2 -1); therefore it is true for a^odd-a

[@MG-hi9sh]

@@mangeshpuranik31 It indeed does, I’ve checked.

[@MG-hi9sh]

@@mangeshpuranik31 I’ve just done a proof by induction, and found that (a^n)-a is divisible by 6 for all n that are odd positive integers.

[@Piglet0123456789]

a^3 - a = (a-1)a(a+1), I.e three consecutive integers Among any three consecutive integers, there must be at least one even number (hence divisible by 2) and at least one number divisible by three. Split into cases 1. WLOG if one of a-1, a or a+1 is divisible by 2 and 3 simultaneously, I.e. divisible by 6, then clearly the product (a-1)a(a+1) is divisible by 6 and we are done. 2. If none of a-1, a or a+1 is divisible by 6, then one must be a multiple of 3 and another must be a multiple of 2 . Therefore a factor of 3 and a factor 2 can be factored from the product (a-1)a(a+1), and hence overall divisible by 6

[@syuliya802]

Well, I want to do more proofs with the same type of materials you have, it looks sooo fantastically easy to orgonize and create 6 cases. :)

[@skilz8098]

n = 1 1^3-1 = 0 which also equals -1 * 0 * 1 = 0 0/6 = 0 2^3-2 = 6 which also equals 1 * 2 * 3 = 6 6/6 = 1 3^3-3 = 24 which also equals 2*3*4 = 24 24/6 = 4 4^3-4 = 60 which also equals 3*4*5 = 60 60/6 = 10 5^3-5 = 120 which also equals 4*5*6 = 120 120/6 = 20 n, n+1, n+2, n+3, n+4 each raised to the third power and subtracted by n is a multiple six since 0,1,4,10, and 20 are integers, therefore a^3-a is divisible by 6. We don't need any more examples due to the properties of modulo arithmetic. What more proof do you need?

[@skilz8098]

Now, geometrically speaking how and why does this modulo arithmetic work for the expression a^3-a is divisible by 6? If we take the polynomial and look at the term with the highest order of magnitude... a^3 and disregard the -a term... a^3 is a cubic function... This function within spatial terms gives you volume. Now, let's take any arbitrary cube regardless of the length of its side "a", and ask yourself this... How many faces or sides does it have? A Cube has 6 Sides! You are taking the Volume of the cube and you are subtracting the length or magnitude of one of its vectors from it... This resulting value will always be a multiple of 6 since a Cube has 6 faces or sides! And there is your geometrical proof!

[@IM-mc7iw]

At 5:50, RHS is product of 3 consecutive numbers: 2k,2k+1,2k+2. One of them must be a multiple of 3. Therefore LHS must be both divisible by 2, as well as 3, therefore 6. QED.

[@IM-mc7iw]

Actually the above is a completion of the induction proof at where you left off.

[@ragingmagikarp7190]

That’s exactly what he covers in Part 1

You can use part induction, part exhaustion for less work: ((a+1)³-(a+1))-(a³-a) =(a+1)³-a-1-a³+a =a³+3a²+3a+1-a-1-a³+a =3a²+3a if a is even, a=2b: 3(2b)²+3(2b)=12b²+6b if a is odd, a=2b+1: 3(2b+1)²+3(2b+1)=12b²+12b+3+6b+3=6×(2b²+3b+1) Or you could just do the double induction, it's not too bad: (3(a+1)²+3(a+1))-(3a²+3a) =3a²+6a+3+3a+3-3a²-3a =6a+6

[@balthazarbeutelwolf9097]

the divisibility by 3 part is much simpler by induction, as in the first part of the video, as you then do not need the 3-way case distinction. Yes, you still have to do the negative numbers, but that's just (-a)^3-(-a)= -(a^3) -(-a)= -(a^3-a). Thus if a^3-a is divisble by any number k then (-a)^3-(-a) is also divisible by k.

[@andyiswonderful]

Yay, I did this one in my head! No Alzheimer's yet.

[@cyprienchabin3540]

Another way is by using Fermat's little theorem plus the Chinese Remainder Theorem. With this method you can also prove that x^5-x is divisible by 30 for all x

[@fabioleonardo4680]

Hi!!!! Thanks for the teaching!!! Congratulations!!! What program are you using as a digital board?

[@wikiPika]

Would modular residue classes work here as well? 3 consecutive integers fulfills class mod 3 -> one value is divisible by 3 -> x | 3, 2+ consecutive does the same for x | 2, and x|3 with x|2 implies x|6

[@nyxfyi]

What app are u using? The writing looks very good c:

[@camilorafael5488]

P sure it’s notability

[@nyxfyi]

Thanks for ur responses ☺️ I am using GoodNotes since about a year but well I am still open to other apps too. Noteabilty doesn’t look bad. Maybe I am gonna try it out for some time. Kinda need a better handwriting somehow lol

[@AkshaySinghJamwal]

I'm probably missing something here, but why is the second half of the proof required at all? If we've already proved that a^3-a is the product of 3 consecutive integers, one of those integers has to be even.

[@johnnath4137]

a³ ≡ a (mod 3) by Fermat’s Little Theorem, ∵ 3 is prime. So a³ - a is divisible by 3. And a³ - a is divisible by 2 for the reasons you gave. So a³ - a is divisible by 3 x 2 = 6.

[@jordanweir7187]

Rly great explanation of this type of problem bro, here's a spinoff problem for others, prove a^5 -4a^3 + 3a is also divisible by 6

[@emmanueld90]

Great video! What program your writing on?

[@Mathskylive]

we have a ^ 3-a = a (a-1) (a + 1) is divisible by 2 and 3 but (2; 3) = 1 so a ^ 3-a is divisible by 6.

[@Mathskylive]

Do you have a lecture on inequality? give a, b> 0 proof. 1 / a + 1 / b is greater than or equal to 4 / (a ​​+ b)?

[@MegaWinner16]

Rewriting gives (a+b)/ab >= 4/(a+b) (a+b)^2 >= 4ab (a-b)^2 >= 0 Ineq often come down to rewriting/factorising and using well-known ineqs (sometimes in a clever way). E.g. AM-GM, Cauchy-Schwartz, Chevyshev, or even Jensen

[@elis2994]

Wow, love this! You are great at explaining :D

[@kqnrqdtqqtttel1778]

Or, by Fermat’s Little Theorem, a^3 == a (mod 3), so a^3 - a == a - a == 0 (mod 3). Furthermore, a^2 == a (mod 2), a^3 - a == a^2 - a == a - a == 0 (mod 2).

[@dineshmathsclasses2514]

Nice explanation , thank you for such a crystal clear proof.

[@mathew_pang]

any pro here know what app he is using?

[@pianoapp]

I think it's notability but not sure tho, anyways it's a great App for such purposes *Edit, 90% Sure it is Notability

[@CMBR]

Can we not prove divisibility by 3 in a more simple manner? We know a number is divisible by 3 of the sum of its digits is divisible by 3. So the sum of the digits of (a-1)(a)(a+1) = a-1+a+a+1= 3a which is divisible by 3 so (a-1)(a)(a+1) is divisible by 3.

[@aliaghaei5732]

Can you please let us know about the app or software you record your videos ?

[@gustavlindell2753]

Very easy way to prove this. Since a^3 - a = (a-1)(a)(a+1), and since one number of three consecutive ones will always be a multiple of three, (just like one number in four consécutives will be multiple of four). And since at least one of the numbers will be pair, you can multiply the factors 2 and 3, to make 6. Get it?

[@filipelqj]

Anyone knows which software/website he is using to write down the notes?

[@Efrain0465]

Que´ programa usas?

[@Azfa96]

proof Mr Eddie is a genius person!

[@mikaras]

So following that logic what about 12? How do you define the lower bounds for a in this case. 12 is divisible by 2,3 and 6 so there must be a generic way to prove the lower bounds of a=3 for divisibility of 12.

[@sagarpanchal_in5002]

Which note application is this???

[@chenpaul1132]

Does it goodnotes?

[@KevinAPamwar]

& If 'a' is divisible by 6 by both first prime numbers 2 & 3 =6K then at least one of the numbers 6K-1 or 6K+1 is a prime number also e.g 5,6,7 11,12,13 17,18,19 23,24,25 29,30,31 35,36,37 41,42,43 47,48,49 53,54,55

[@punitasingh8448]

or u could use induction ...that will require fewer steps.

[@chulodyman9138]

Why don’t we just let a=6k,6k+1,…,6k+5 and check divisibility by 6?

[@RJSRdg]

Not only that, but if a is odd (and >1), then a3-a is divisible by 24.

[@lithograhp]

Put a = 6k, 6k+1, 6k+2, 6k+3, 6k+4 or 6k+5 , (a-1)a(a+1) = 6 * M

[@toanhien494]

it's actually an easy proof but you did it nicely, I enjoyed the video.

[@tadejsivic534]

Cool video. I love these proof videos

[@avivwachman3207]

Can you do the same with 6k,6k+1,6k+2,6k+3,6k+4,6k+5?

[@MrRyanroberson1]

yep. 6k, 6k+2, 6k+4 are even, 6k, 6k+3 are threeven, and so the product of three consecutive terms = a^3-a = a multiple of two and of three

[@MyLifeEducation..]

I want to create videos like this one.How to enlarge my face window to such level?

[@littlefermat]

The problem that appears in all Number Theory books!

[@muhammadismail2300]

What’s this app that he is using?

[@schmeckoh]

I didnt need this, but I like it. Starting calc 1 next term

[@byronwatkins2565]

8 is divisible by 2 but not 6 and 9 is divisible by 3 and not 6. It is true that neither are a^3-a; however, this demonstrates that numbers can be divisible by 2 or 3 and yet not be divisible by 6. I cannot agree that you have proven that numbers divisible by 2 and 3 must be divisible by 6. Having N divisible by 2 means that N=2k for some integer k. Having N divisible by 3 means that 2k=N=3n for some integer n=N/3=2(k/3). Since n is an integer and 2 is not divisible by 3, k must be divisible by 3, and k=3m for some integer m. Then N=2k=2(3m)=6m for some integer m. Therefore, N is divisible by 6 if it is divisible by the prime factors of 6.

[@xoticfn3455]

Which grade is this?

[@alphabet601]

Anyone know what tablet/ipad he is using in this? Thanks

[@amaanali9525]

15 minutes to prove this? What!!!? I'll prove it now. This can also be written as a^3-a^1 is always divisible by 6. You take away the powers 3-1=2. Now you multiply 2 by the first power(3) 2×3=6. This shows that it will work.

[@3uryale964]

For your last rush, I'm not okay with your logic: you said it was an iff case but didn't prove it was ! You should have considered a number n being equal to 3k AND 2p at the same time and then you should prove it indeed means it's equal no 6m. (or you can just talk about relative primes but you didn't so...) Have a good day Mr. Woo ! (was a great video tho)

[@theaureliasys6362]

Well.. depends on whether 0 is divisible by every number.

[@maxime_weill]

it is. for any k, 0 = 0*k

[@arshadsiddiqui8485]

What software you are using

[@GooogleGoglee]

Really nice and neat!

[@aashsyed1277]

are you google>>>>>>>>me??????

[@GooogleGoglee]

@@aashsyed1277 I do

[@ei1864]

works for a^3 - a, but what about: Is a^p - a, where p is prime, always divisible by p? ;)

[@PedroCristian]

Indeed, it's called Fermat's little theorem. It comes from the fact that all residues are inversible but 0, i.e. there is p-1 elements in the modulo p multiplicative group, hence x^(p-1)=1 (mod p)...

[@clintonpham8743]

is this ext 2

[@particleonazock2246]

Woohoo, Woo is doing number theory, the queen of mathematics. Woo himself is the King of Mathematics, so Huzzah@!

[@mistersingh6470]

Maza aa gya🙌

[@sydneyriekert5726]

what if a=1?

[@robertveith6383]

0 is divisible by 6.

[@EngSeifHabashy]

You should say that a belongs to Z+ Not just Z because it may be negative and it would be meaningless to say that a negative no. Is odd or even

[@zafnas5222]

It is not meaningless for a negative integer to be even or odd. Even just means that a is divisible by 2. For example, -4 is even because -4=2(-2).

[@damianflett6360]

Negative integers are not excluded from the sets of evens and odds.

[@EngSeifHabashy]

@@zafnas5222 OK thanx

[@abisarwan20]

What branch of math is this?

[@abisarwan20]

@Bruno Mendes Silva thanks ^^

[@himanshusingh5056]

My only 2 brain cells died after seeing this video

[@rotemb9525]

Awesome

[@roshanoinam7103]

Euclid's Division Lemma

[@hassine3401]

But he is explaining so good 😊 to all

[@hassine3401]

I also solved this problem by this method last year in my school

[@michaelempeigne3519]

How to prove this : a / ( a + d ) < ( a + d ) / ( a + 2d )

[@michaelempeigne3519]

note : for all values of a > 0 and d > 0

[@divyapathak1665]

Firstly multiply (a+d) both side. And then multiply (a+2d) both side. After that you will get a(a+2d)0 Hence above equation is true. I hope it will help you.

[@OMGclueless]

I don't think your last step was a valid proof. You proved that if a number is divisible by 6 then it is divisible by 2 and by 3. But you didn't prove "if and only if". For example consider the following "proof": Any number divisible by 8 can be written as 8k. But 8k = 2[4k], 8 = 4[2k]. n divisible by 8 n divisible by 4 and 2. This is a false statement (for example, n=4 is a counter-example), but I have "proven" it using the exact same method you used (13:10).

[@pbenikovszky1]

I don't like the end of the proof: you never prooved that if a number is divisible by 2 and 3 then it is divisible by 6, you just stated it. True, if a= 6k then you can factor it like a = 2(3k) and 3(2k), but it means that if a number is multiple of 6 then it is a multiple of 2 and 3. Like if a number is divisible by 8 then it is a multiple of 2 and 4, but if a number is divisible by 2 and 4 then it does not mean it is divisible by 8. You should have proved that if a number is divisible by n and m, then it is also divisible by nm/gcd(n,m). As gcd(2,3)=1 the number is divisible by 2*3/1=6 ;)

[@slawomirdrapinski4538]

That's like saying he never proved 2x3=6.

[@deepakgoyal3250]

Can also use methamatical induction

[@Tejvir7]

Nice

[@aashsyed1277]

EDDIE PLEASE REPLY THIS COMMENT

[@TeamCosmicalEN]

I can't math X-2x=??? X2-2z=??? 2z-3n+3x-51x=??? How you very good at math

[@okanuroktavianda3887]

Of course, it will be divided by 6 yaa..

[@particleonazock2246]

Every time I watch one of your videos, I reward myself with a new Pokemon Trading Card. Last time, it was Magikarp. :[

[@reagancarbaugh9156]

ngl that's a very odd form of motivation to me

[@aryanpaul8435]

In India we literally do these type of question (and there was this exact question also) in 10th grade as our first chapter. And btw first chapter is considered super easy

[@4th_Dimension_I2b]

It's taught in high school mathematics in India.

[@Iamblindanddeaf23]

Wow

[@mdyunus3320]

By following this we can say that (a^2-1) is always divisible by 3.

[@Tchy]

That's not true. If a=3 => (3^2 -1) = (9 -1) = 8 8 is not divisible by 3.

[@mdyunus3320]

thanks

[@midoent]

eddie woo

[@anuragguptamr.i.i.t.2329]

Please make a video on this: instagram.com/p/CPF3aIiDlsA/?

[@bollyfan1330]

Pretty long winded proof, will be trivial with modulo arithmetic.

[@akvmaths]

Why 6 is divided by 3 and 2 because they are facter of 6 apply this basic concept 👍🏻

[@yumo]

Nice