Fun proofs

In this video, I showed the futility of applying mathematical induction to non mathematical claims. I also showed that 1 less than the square of any odd number is divisible by 8.

[@AzharLatif-d4z]

This guy must be an ethical man, a man of pure vision in the societal, and personal spheres of his life .His love for all learners gleams in his style of teaching Mathematics, reaching the edges of chosen topics in real time. Stay Blessed.

[@PrimeNewtons]

Thank you. You sound very wise.

[@gkotsetube]

You should also hear him sing with the Bob Marley tribute band, the Junior Whalers.

[@georgesbv1]

(2k+1)^2 -1 = 2k * 2(k+1) = 4k*(k+1) It doesn't matter if k is even or odd, one of k or k+1 is divisible by 2 (and so is their product)

[@PrimeNewtons]

I knew I should that but somehow, I went the way of the wasteful 😂

[@wafflaaar1067]

I arrived at that logic too.

[@Grecks75]

That's what I came up with in less than a minute.

[@davidbrisbane7206]

​@@PrimeNewtons It's never a waste of effort to solve a problem a different way.

[@JustJulezJulezing]

I don’t know how to describe it but most teachers I had taught math in a very cold and sterile way. You explain things in such a warm compassionate manner. Even if I don’t grasp the concept right away, I don’t feel like an imbecile after watching. Really glad I found your channel.

[@crsmtl76]

Using mathematics to light up smiles is a such a blessing. Thank you Coach! 🙏

[@CasiMediocre]

I found it funny that he said "let's get into the video ☺️" almost halfway through the video

[@AbouTaim-Lille]

The proof is easy without induction.if a is odd then a-1, and a+1 are even numbers and since they are consecutive even numbers then one of them is divisible by 4. So that their multiplication is divosible by 8. And their multiplication is (a-1)(a+1) =a²-1

[@PrimeNewtons]

Smart!

[@robertlunderwood]

That was my proof.

[@darcash1738]

That's an awesome way of doing it!

[@schwobelabibalbol]

This is my preposition of a induction proof. I will apologize for any language mistake since I'm German. I'm 17 and am doing this kind of math for fun and to learn. If anyone spots any mistakes or the like, critique is greatly appreciated. Claim: a²-1 is divisible by 8 for every odd number a. P(1): (1)²-1=0 -> Divisible by 8 P(3): (3)²-1=24 -> Divisible by 8 ... P(2k+1): (2k+1)²-1=4k²+4k+1-1=4k²+4k. Now, we claim that this is divisible by 8. Thus, it can be expressed as such: 4k²+4k=8m Now for the next term k+1 P(2k+3): (2k+3)²-1=4k²+12k+9-1=4k²+12k+8 We can express this another way, by splitting 12k -> 4k²+4k+8k+8. As per the inductive hypothesis =>8m+8k+8=8(m+k+1). This is surely divisible by 8 since it's a multiple of 8. Thus, (2k+3)²-1 is divisible by 8 and the proof is complete.

[@WahranRai]

10:33 remark that 4(k^2 + k) = 4*k(k+1) ---> k and k+1 are 2 consecutive number ----> one of them is even ---> a^2 - 1 is divisible by 8. My demo is : Let a = 2k+ 1 an odd number ---> a^2-1 = (a+1)(a-1) = (2k+1 +1)(2k+1 -1) = (2k+2)(2k) = 4*(k+1)*k --->divisible by 8

[@bobbobby7654]

Oh, fascinating dive into the intricacies of odd numbers and their mystical relationship with the number 8! Reminds me of the profound lyrics from "We Need Cash" - "I bought an electric guitar, I rented a shiny car." Just like odd numbers, life's equations seem to dance in mysterious patterns. By the way, did anyone notice that elusive secret cat in the background? A subtle nod to the feline chorus singing "nya ;3" in the song. Life's oddities and hidden references, what a peculiar symphony! Keep on calculating the enigma of existence!

[@themathhatter5290]

I never expected a Tally Hall reference in a math teacher's comments.

[@jensberling2341]

Loveable and lovely. Enjoyable every second of the presentation

[@debjanimukherjee502]

Presentation is awesome

[@ChimezieFredAnaekwe]

With mathematical induction I might have the chance to live forever.

[@vaibhavsrivastva1253]

A quick way to prove that a^2 - 1 is divisible by 8, for every odd number a, is :- Let a = 2k + 1, where k is an integer. Let x = a^2 - 1 => x = (a+1)(a-1) = (2k + 1 + 1)(2k + 1 - 1) = 2(k+1) * 2k = 4 * k(k+1) => x = 8 * k(k+1)/2 Now, we know that n(n+1)/2 is just the sum of all natural numbers from 1 to n (the sum being another natural number), and since all natural numbers are also integers, we can say that the sum is also an integer (note : it is of no importance whether this integer is positive, negative, or even zero). So, x = 8 * [integer] => x / 8 = [integer] => (a^2 - 1)/8 = [integer] => (a^2 - 1) mod 8 = 0 Hence, a^2 - 1 is divisible by 8, for any and all odd numbers (here, 'a'). P.S. :- Food for thought : How would we prove that n(n+1)/2 is necessarily an integer for an integral n, given that n(n+1)/2 is a natural number for a natural n? Or, is this assumption wrong in itself?

[@gamesofsteffen9159]

You can do the second proof with induction, which seems useful in the context of the first proof, showing how it is done properly.

[@carltompkins3282]

That was a fun problem thanks mate 😊 I like these simpler ones I can understand and can get engaged in. ✅

[@darkkupo5162]

clarification, the error lies in the assumption that P(K+1) = P(K) + P(1). You can assume that you survive day K and if you can show you survive day K+1 then it would have to be the case that you can live forever. The reason your joke proof doesn't work is because the function cannot be broken the way you claim, not because you can't assume K.

[@JMcMillen]

I noticed a sequence I'm familiar within this proof. If A=2K+1 then (A^2-1)/8 = K*(K+1)/2 It's the triangle numbers. It even holds up if K is a negative number.

[@diogochadudmilagres4533]

Newton, I choose a² -1 = (a + 1)(a - 1). So, a + 1 = 2k + 2, a - 1 = 2k, and a² -1 = (a + 1)(a - 1) = (2k + 2)*2k = 4k² + 4k. After, putting 4k in evidence, we have 4k(k + 1). So, if k is odd, k + 1 is even, and if k is even, k + 1 is odd. Any "Even times Odd" resullts in a Even. Hence, we have 4k(k + 1) = 4.Even = 4.2t = 8t, divisible by 8

[@chrisrybak4961]

Or just note that (a^2-1) = (a-1)(a+1), so if a=2n+1 (ie an odd number), then (a^2-1)=2n.2.(n+1) which always contains 2x2x2 whether n is odd or even => divisible by 8

[@Arkapravo]

Can you consider doing some basic Topology - I never grasped that subject!

[@PrimeNewtons]

Ha, that's a big ask. I'll see if I can get some friends to fill in for me

[@alposmanagaoglu7352]

…we conclude that life is non-linear 😊

[@dububo]

a = 2k + 1. a² - 1 = (a + 1)(a - 1) = (2k + 1 + 1)(2k + 1 - 1) = (2k + 2)(2k) = 4k(k + 1). If k is even, then 4k(k + 1) = 4(2q)(k + 1) = 8q(k + 1), which is divisible by 8. If k is odd, then k + 1 is even, so 4k(k + 1) = 4k(2q) = 8kq, which is divisible by 8.

[@rcnayak_58]

Once we have noted that a² - 1 = 4k² + 4 k, we can, taking 4k as common, write a² - 1 = 4k (k+1) = 4 . k(k+1). Notice that k(k+1) is the product of two consecutive numbers, the product of which is always an even number, say 2t , where t is any even or odd number. Then our a² - 1 becomes 8t, that is divisible by 8.

[@idkman640]

If a²-1 is divisible by 8, Then, (a²-1)/8=n =>a²-1=8n -(i) (n belongs to set of integers) {(a+2)²-1}/8=k =>(a+2)²-1=8k -(ii) (k belongs to set of integers) From (i) & (ii), It is clear that (a²-1) and {(a+2)²-1} are even numbers as they both equate to a certain integer being multiplied by 8. So, =>(a+2)²-1-(a²-1)=2x =>a²+4a+4-1-a²+1=2x =>4a+4=2x =>a=(x/2)+1 Here,For every value of x for which a is an integer is where a is odd. As,a is odd when a²-1 is divisible by 8 then conversely a²-1 is divisible by 8 when a is odd.

[@chaosredefined3834]

If a = 1, then a^2 - 1 = 0, which is divisible by 8. Assume that k is an odd number such that, if a = k, then a^2 - 1 is divisible by 8. Consider a = k+2. So, a^2 - 1 = (k + 2)^2 - 1 = k^2 + 4k + 4 - 1 = k^2 - 1 + 4(k+1). Since k^2 - 1 is divisible by 8, we only need to show that 4(k+1) is divisible by 8. Note that k is odd, so k+1 is even. 4 times an even number is a multiple of 8, so 4(k+1) is a multiple of 8. Since it's true for a = 1, then by what we just argued, it's true for a = 3. Therefore, it's true for a = 5. etc...

[@flight7218]

It is sufficient to study the remainders of a modulo 8 which are 0,1,2,3,4,5,6,7

[@Curufin1984]

Yes, that was also my own though. And you even only have to check 1,3,5,7 which I would call trivial. The "problem" is you have to give an argument that modulo arithmetic works (i.e. a*b (mod n) = a (mod n) * b(mod n) and a+b (mod n) = a (mod n) + b (mod n)). Not hard, but not self-evident.

[@Uriboica]

U got me addicted to maths

[@Alessandro-1977]

Grouping 4k(k+1) in 4(k^2+k), it easily turns out that k(k+1) is always even, either k is odd or even (if k is even, k+1 is odd and vice versa, thus k*(k+1) is always even)

[@mosespeters5546]

Hi good day Prime Newtons. I love your videos and i have a question. How does this proof implies that a^2-1 is divisible by 8, for values of a which are odd only? Is it that an odd number squared is always odd?

[@nanamacapagal8342]

if a is odd, then a^2 is odd. but the -1 makes it even. the proof in the video is to write a = 2k + 1 where k is an integer. a = 7 can be written as 2*3+1. a = 13 can be written as 2*6+1. a = -5 can be written as 2*(-3)+1. hope that helps clarify a few things

[@mosespeters5546]

@@nanamacapagal8342 Hey there , thank you

[@Jianlong-xp5li]

Can you please make a video about log and natural log

[@PrimeNewtons]

That's a wide topic. Is this algebra or calculus?

[@Jianlong-xp5li]

Both

[@Jianlong-xp5li]

Maybe calculus

[@darcash1738]

and it seems like if you were to do it by induction would it be (2(k+1)+1)^2 - 1 that you need to check. so, (2k+3)^2 - 1. It's pretty nice since it follows basically the same process in factoring out an 8 as this one. (2k+3)^2 - 1 = 4k^2+12k+8 = 4(k^2+3k+2) if k is even, m is some other interger: = 4(even + even + 2); (even numbers squared are even; 2k1(2k2+1) = 4k1k2+2k1 = 2(2k1k2+k1), so even times odd is even) = 4(2m1+2m2+2) = 8(m1+m2+1) if odd, q for some other interger: = 4(odd+odd+2); (odd nums squared are odd, odd*odd = odd) = 4(2q1+1+2q2+1+2) = 8(q1+q2+2)

[@schwobelabibalbol]

This is not really induction, rather you simply analyze like the creator did in the video. This is my preposition of a induction proof. I will apologize for any language mistake since I'm German. Claim: a²-1 is divisible by 8 for every odd number a. P(1): (1)²-1=0 -> Divisible by 8 P(3): (3)²-1=24 -> Divisible by 8 ... P(2k+1): (2k+1)²-1=4k²+4k+1-1=4k²+4k. Now, we claim that this is divisible by 8. Thus, it can be expressed as such: 4k²+4k=8m Now for the next term k+1 P(2k+3): (2k+3)²-1=4k²+12k+9-1=4k²+12k+8 We can express this another way, by splitting 12k -> 4k²+4k+8k+8. As per the inductive hypothesis =>8m+8k+8=8(m+k+1). This is surely divisible by 8 since it's a multiple of 8. Thus, (2k+3)²-1 is divisible by 8 and the proof is complete.

[@sirak_s_nt]

It obvious that for any number n, n²= 1,4,0 mod 8, 1 if n is odd and 4,0 when n is evenm Since a is odd a²= 1 mod 8 Thus a²-1 is divisible by 8.

[@SpennyBoi]

Start off with an integer a Double it Add one (this is k) Square it Minus 1 And divide it by 8 Following this you get a 2a 2a + 1 4a^2 + 4a + 1 4a(a + 1) and now it should be clear that when we divide it by 8 to get a(a + 1)/2 either a or a + 1 is even because a is an integer showing that we can divide one of those by 2 and multiply the remaining numbers to form a new integer.

[@DrEeveeYT]

The second proof you did in the video is similar to a proof I did in a university project. I was deriving the formula for something in Knot Theory. Essentially, the formula needed to output an integer, and I actually had two parts of it. One of the parts hypothetically produced double of what I was needing (so I multiplied that by 1/2) but I didn't have a guarantee that it was even. The only guarantee I found was if I could show (x^2 - x) was always even, so I did something similar to what you did. I later realised that you could factor (x^2 - x) into x(x-1). If the input to that function is an integer, then x(x-1), or in your case x(x+1) is the product of two consecutive integers. This would prove its even, as I think there's a theorem out there where for any n consecutive integers, you are guaranteed that one of them will be divisible by n. When n = 2, one of the 2 consecutive integers is divisible by 2, meaning whatever product of the consecutive integers would have 2 as a factor, making it even. The way we both did it is fine, but that was another way it could be proven.

[@Blade.5786]

My solution was simply: a² - 1 = (a + 1)(a - 1) Since a is odd, (a + 1) and (a - 1) are even. In any two consecutive even numbers, one of them will be a multiple of 4. Therefore, their product is divisible by 2 x 4 = 8.

[@nasrullahhusnan2289]

a is odd --> a=2k±1 a²-1=(2k±1)²-1 =4k²±4k =4k(k±1) Note that 2 | (k±1) Hence 8 | a²-1

[@goldeneagleyt8652]

Beautiful

[@xyz9250]

You could also just factor k^2+k into k(k+1) one of them will be even number.

[@WhiteGandalfs]

Well: (5:20) Just by adjusting from an undisputed "survive every day" to a probability of surviving a day (maybe even in dependence from age and other factors), we make a valid scientific base for survivability analysis. Nonetheless, IF you INDEED survive every daily unit, you INDEED live forever. That is absolutely true - even if that "if" is not at all undisputed in real life. But IF it would hold, the "life forever" holds as well. You are allowed to make assumptions in your models. You just have to track what you do for what purpose. I don't count that as an "illogical" construct. It just is an example for the responsibility of a scientist for judgement of his data source and the judgement of the conclusions he draws from his data. You just showed in your model that infinite life is possible by infinitely surviving every day - completely analogue to counting to infinity or to dividing into infinitely small differentials - which you may instantiate as a model for certain examinations, DESPITE never being able to factually do so in real life.

[@klementhajrullaj1222]

Can you prove that: n^3 - n is divisible by 6, where n is a natural number?

[@ALIEAKAMARA-gy2vo]

Sir I need your help please

[@novasleepy]

Can you do the Cbrt of i

[@nanamacapagal8342]

we're gonna need some euler's formula e^(iT) = cos(T) + isin(T) get ready

[@AbulfazMehdizada-q4t]

a^2-1=(a-1)(a+1)=2k * (4+2(k-1))=8k+4k*(k-1). we know that k(k-1) is divisible by 2, so we proved

[@dankaxon4230]

what if you were a stillborn

[@AzharLatif-d4z]

But Souls are eternal as fire-fly.

[@Bowl_of_Moths]

Simply, P(1) would not be true.

[@bart2019]

Then you wouldn't have been watching this video.

[@adamhanna9940]

2k+1 , k>1. As the first + odd is 3

[@davidbrisbane7206]

Let a = 8k + r, where k and r are integers and 0 ≤ r ≤ 7. But since a is odd, then r = 1, 3, 5, or 7. So, a² = (8k + r)² = 8(8k² + 2kr) + r² So a² - 1 is divisible by 8 if r² - 1 is divisble by 8. We note that the only possible values of r² - 1 are 0, 8, 24, 48 which are all divisible by 8. So a² - 1 is divisible by 8 if a is odd.

[@embracinglogic1744]

Assume my head is frozen on day k.

[@xenumi]

I think I found a counter-example

[@arthurkassis]

As a is suposed to be odd, a is equal to 2n+1, (n belongs to naturals) aˆ2=4nˆ2+4n+1, aˆ2-1=4nˆ2+4n so, 4nˆ2+4n/8=nˆ2+n/2 if n is even, nˆ2 is even and even + even is even if n is odd, nˆ2 is odd and odd + odd is even therefore, nˆ2+n=2k, (k belongs to naturals) 2k/2 is k

[@rob876]

since a is odd, a = 2n + 1 where n is an integer so a^2 - 1 = (a - 1)(a + 1) = 2n(2n + 2) = 4n(n + 1) but n(n + 1) is divisible by 2 because either n or n + 1 is even. therefore 4n(n + 1) is divisible by 8

[@adamhanna9940]

I meant k> 0

[@reamartin6458]

The answer is: [Life = Yeshua]

[@Jianlong-xp5li]

Hello

[@ivanhuertas5307]

it is guaranteed of course it is that you can live forever John 11:25 Jesus said unto her, I AM the resurrection and the life; he that believes in me, though he is dead, yet shall he live; John 11:26 and whosoever lives and believes in me shall never die. Believest thou this?

[@glorrin]

I have a better way to (maybe not) live forever. I. You have the ability to live for 1 day. II. Let's assume you have the ability to live for k days. What about k + 1 days ? well + is commutative, so k+1 = 1 +k We know you have the ability to live for 1 day because of I. it remains k days, and thanks to II. we knoe youhave the ability to live for k days. therefor you can live for 1+k days Conclusion you will live forever.