Nice presentation! Note, however, that it’s not necessary to use math induction. Claim: If n>3, then n! >(n(n+1))/2. Multiplying both sides by 2 and dividing by n, we get the equivalent inequality 2(n-1)!>n+1. As 2(n-1)!>2(n-1), to prove the claim it will suffice to show that 2(n-1)>n+1, for n>3. But this is immediate: 2(n-1)>n+1 iff 2n-2>n+1 iff 2n-n>2+1 iff n>3.
This is a good example to demonstrate, how proof by induction works. Although there is a quicker and relatively simple way to proof this directly. If n>3 => 2n>n+3 => 2n-2>n+1 => (n-1)>(n+1)/2 Obviously (n-1)! is also greater than n-1. Therefor we can safely conclude that (n-1)!>(n+1)/2. We now multiply both sides of the inequality by n>0: n(n-1)!>n(n+1)/2 => n!>n(n+1)/2. (q. e. d.)
Great video! It is a pretty interesting way to show how matematicians use demostrations techniches such as proof by induction, and in a curious problem! I love your videos
Excellent lecture ! Induction works smartly, with ± heavy writings… Let’s try directly : in order to prove that n ! > n (n + 1) / 2 for any n ≥ 4, equivalent to 2 (n-1) ! > n + 1, we notice that (for n ≥ 4 ) : 2 (n-1) ! > 2^(n-1) because 2 (n-1) ! = 2x2x3x…x(n-1), and 2^(n-1) > n + 1 by studying f(x) = 2^x - x - 2 for x ≥ 3 (easy stuff…). Then we’ve got it… Thank you for your interesting videos ! 🙂
An other way is : let U_n = 2 (n-1) ! and V_n = n +1. Then U_n+1 / U_n = 2 n and V_n+1 / V_n = (n+2) / (n+1). As U_3 = V_3 = 4, and for n ≥ 1, U_n+1 / U_n > V_n+1 / V_n, Then V_n < U_n for n ≥ 4.
Because n! ≥ n*(n-1)*(n-2) and n*(n+1)/2 < n*(n+1) for all n≥4, all we need to show is that (n-1)*(n-2) > n+1 n² - 3n + 2 > n + 1 n² + 1 > 4*n n + 1/n > 4, which is true for n ≥ 4 since 1/n > 0
I did it similarly with one intuitive observation: n >= (n + 1) / 2: Because (1 + 2 + ... + n) = n * (n + 1) / 2 = n! n >= (n - 1)! Knowing that factorials increase quickly in comparison to a linear function, only n = 1, 2, 3 is possible, check to find n = 1, 3.
I totally yield to your formality about this, but I know I'm not alone in intuitively knowing that 3, and after a second's thought, also 1, were the solution set.
I like to think that geometrically the LHS is a quadratic and the RHS is a Gamma function on the first quadrant so intuitively there should be 2 intersections that has two solutions
This can be done without using the formula for Σi=(n)(n+1)/2 since If n=1, then 1=1, If n=2, then LHS=3, RHS=2 if n=3, RHS=LHS=6 And if n ≥4, 1+2+3+…+n n-1+1=n, n! ≥n(n-1)*2>n*n=n^2> 1+2+…+n.
@Salko_ So to get (n+l)! > (n+1)(n+2)/2, he took n! > n(n+1)/2 and replaced n with n+ 1? And then to get (n+1)n! > (n+1)n(n+1)/2, he took n! > n(n+1)/2 and multiplied both sides by n+1?
@Myhair0_0 That's why we have to check whether the inequality holds for the base case (n=4 in this example) Then, the proof from the video, which @Salko_ summarized, shows that if the inequality holds for any integer >=4, then it holds for the next integer. Thus, after manually verifying that the inequality holds for n=4 and completing the short proof, we know that the inequality holds for n=4, n=4+1=5, n=5+1=6, and so on. Hope this helps
@@Salko_ I don't know why the freak my response got deleted last night, I was asking this: Are you saying (n+1)! > (n+1)(n+2)/2 come from replacing n with n+1 in the original n! > n(n+1)/2? And does (n+1)n! > (n+1)n(n+1)/2 cone from the original but after multiplying both sides by n+1?
Hello sir, Very interesting approach to the problem. But if we are looking for the condition of the equality happening is it not easier? What needs to happen so n*(n+1)/2=n! ? Since n!=0 we can divide by n so (n+1)/2=(n-1)! n+1=2(n-1)! n=2(n-1)!-1 If we replace n by k+1 to have a nicer number inside the factorial k=2k!-2 so k=2(k!-1) which means that k>k!-1 or k+1>k! and we can easily verify that this proposition is only true for very small values
thats a nice idear wait net me simplify the first one: (n+n²)/2 and the second one as well: n! ore use the stirling formular... 1:00 i stll like your intro! 2:53 ok you write it in an other way but still the same 6:00 no it becomes interesting show it! 7:59 so any Natural nubers higher or equal than/to 4 8:47 but now it looks easy 10:41 because we just deal with a n greater than 4 12:02 nice i think i repeat it on my own board
Very rough, not rigorous idea but... I figured that because the factorial grows much faster than the quadratic, after some point there won't be any more solutions, so there isn't any need to check every case. I remembered the 1 + 2 + 3 = 1 x 2 x 3 meme, so n = 3 n also = 1. That was basically my reasoning, because after n = 3 the factorial grows much faster than the quadratic so they will never ever intersect again, the only need is to check for answers within that range. I think the proof of the factorial outlasting the triangular numbers for n >= 4 was rigorous to help cement that idea that they can never be equal and hence no solutions for that region.
From the thumbnail: the solutions are n=1 and n=3 . For any n>3, the cumulative product is greater than and also will be increasing faster than the cumulative sum. I'm watching the video to see if you consider n=0 a solution or not (and why).
I immediately know of 1, and 3. I think those are the only 2 but proving that is a whole different issue. I might be able to do that but not sure exactly what I'd do maybe induction or contradiction but would take awhile to prove it.
All you have to do is prove that n! grows faster than n(n+1)/2 for all n>=4. You can prove that and thus since 24>10. That is 4!>10 and that n! grows faster than n(n+1)/2 then I can show that n(n+1)/2 can never catch up to n! thus only 1 and 3 are true
Sorry for my bad english, I'm french. In my induction, I decided to make a proof by contradiction. Bascially, we wanna show that since n >= 4, the sum of all the integers from 1 to n is not equal to the product of all the integers from 1 to n. So we pick a n superior or equal to 4, let's assume that our property is true for that n (induction hypothesis) And now let's suppose that the sum from 1 to n+1 is equal to the product of all the integers from 1 to n+1. (Hypothesis of the proof by contradiction, it's a part of the induction and we will prove this is false) Ok so we have the sum from 1 to n+1 = the product from 1 to n+1 Then we can say that the sum from 1 to n PLUS n+1 equals the product from 1 to n TIMES n+1. It's an equality so we put all the terms at the left and we have : (Sum from 1 to n) + (n+1) - (product from 1 to n) × (n+1) = 0 Since the sum can be wrote n(n+1)/2 we factorise : (n+1)(1 + n/2 - n!) = 0 Or n >= 4 so n+1 =/= 0 So 1 + n/2 - n! = 0 So 1 = n! - n/2 So 1 = n((n-1)! - 1/2) So 1/n = (n-1)! - 1/2 (since n not equal to 0) So (n-1)! - 1/2 < 1 So (n-1)! < 3/2 So 2(n-1)! < 3 Or n >= 4 So n-1 >= 3 So (n-1)! >= 6 So 2(n-1)! >= 12 And there is the contradiction So we have the sum from 1 to n+1 who is not equal to the product from 1 to n+1 and then we have proved it for n+1 Thanks for all the people who will tell me my blunders : in maths but also in english. I tried my best to be understandable, so please be nice, and don't hesitate to tell me my mistakes, at least it shows that you read my work and it is very nice of you.
There are exactly 77000 ordered quadruples (a,b, c,d) such that gcd (a,b, c,d) =77 and lcm (a,b, c,d) =n, What is the smallest possible value of n? Hello teacher. Could we look at this question? There were many solutions that i didn't understand well. I would like to see your approach
@@abulfazmehdizada In that case, there is no possible solution for n ; because the number of quadruples has to be a multiple of 16 ; and 77000 is not a multiple of 16 . A quadruple (a,b,c,d) cannot contain any 0, because in that case, n=0 and in that case there are an infinite number of quadruples, for example of the form (0, 77, 77, 77k) where k is any non-zero integer. Therefore, |a| , |b| , |c| and |d| must be positive (i.e. nonzero). Now, suppose (a, b, c, d) = (t, u, v, w) satisfies gcd(a,b,c,d) = 77 and lcm(a,b,c,d) = n , and t, u, v, w are positive integers. Then (t, u, v, w) represents a set of 16 _distinct_ quadruples that each satisfy the gcd and lcm conditions, namely (a, b, c, d) = (t, u, v, w), (t, u, v, -w), (t, u, -v, w), (t, u, -v, -w), (t, -u, v, w), (t, -u, v, -w), (t, -u, -v, w), (t, -u, -v, -w), (-t, u, v, w), (-t, u, v, -w), (-t, u, -v, w), (-t, u, -v, -w), (-t, -u, v, w), (-t, -u, v, -w), (-t, -u, -v, w), or (-t, -u, -v, -w). This is true for any quadruple (t, u, v, w) of positive integers that satisfies the conditions, hence the total number of quadruples must equal {16 times the number of distinct quadruples of only positive integers}.
@@abulfazmehdizada In that case, there is no possible solution for n ; because the number of quadruples has to be a multiple of 16 ; and 77000 is not a multiple of 16 . A quadruple (a,b,c,d) cannot contain any 0, because in that case, n=0 and in that case there are an infinite number of distinct quadruples, for example of the form (0, 77, 77, 77k) where k is any integer. Therefore, |a| , |b| , |c| and |d| must be positive (i.e. nonzero). Now, suppose (|a|, |b|, |c|, |d|) = (t, u, v, w) satisfies gcd(a,b,c,d) = 77 and lcm(a,b,c,d) = n , and t, u, v, w are positive integers. Then (t, u, v, w) represents a set of 16 _distinct_ quadruples that each satisfy the gcd and lcm conditions, namely (a, b, c, d) = (t, u, v, w), (t, u, v, -w), (t, u, -v, w), (t, u, -v, -w), (t, -u, v, w), (t, -u, v, -w), (t, -u, -v, w), (t, -u, -v, -w), (-t, u, v, w), (-t, u, v, -w), (-t, u, -v, w), (-t, u, -v, -w), (-t, -u, v, w), (-t, -u, v, -w), (-t, -u, -v, w), or (-t, -u, -v, -w). This is true for any quadruple (t, u, v, w) of positive integers that satisfies the conditions, hence the total number of quadruples must equal {16 times the number of distinct quadruples of only positive integers}.
Thank you for your video. It is possible not to use induction: as (n-1)! = (n+1)/2 is in Z, n must be odd and as (n-1)! = (n-1).(n-2)..2.1 > (n-1)(n-2).4 for any n >=7, we get (n-1)(8n-17) < 2 which is impossible as n>6.
@@ishanpurkait9124 It's a VERY difficult book. I'll recommend that you start with basic books about calculus, elementary number theory, classical plane geometry, linear algebra, and abstract algebra.
I thin, the o natural number satisfying this equation are n=1 and n=3. The left side can be substituted b n*(n+1)/2 (according to gauss forula for sumof the first n natural numbers) while the rigtside is equall to n! (accordingto the definition of factorial).So we seach values for n with n*(n+1)/2=n!. Since n! is per definition n*(n-1)!, we can transform this equation to (n+1)/2=(n-1)!. n=1 and n=3 are possiblesolutionsforthhis equation, because (1+1)/2=(1-1)! and (3+1)/2=(3-1)!. Equal numbers can not fullfill the equation, because the right side is alwas a natural number, while the left side is for even values of n neer an integer. For all odd numbers n greater than 3, (n-1)! is greater than (n+1)/2, so n=1 and n=3 are the only solutions. For me, it was obvious,that (n-1)! is greater than (n+1)/2 for an n>3, but nice, that you gave a proof ...
Something bugged me, in the induction, since we only needed n>=2 why couldn't we start at 2 ? well, 15 years after formerly learning about induction, I finaly understand how important the initial step is (base case). if we start at 2, initial step would be 2! = 2 2+1 = 3 2 is not greater than 3 So we can't start with 2, And we can't start with 3 either since we have shown it is equal. This is a marvelous Induction.
ATTEMPT: By inspection, N = 1, 3 1 = 1 1 + 2 + 3 = 1 * 2 * 3 = 6 N = 2 doesn't work. 1 + 2 = 3, 1 * 2 = 2 For N >= 4: N! > N(N-1) = N^2 - N = (N^2)/2 + N/2 + (N^2)/2 - 3N/2 >= (N^2)/2 + N/2 + 8 - 6 > (N^2)/2 + N/2 = N(N+1)/2 Which is the sum of all natural numbers up to N. Therefore for all natural N >= 4, 1 + 2 + 3 + ... + N < 1 * 2 * 3 * ... * N, and so the two sides cannot be equal. The only solutions are N = 1 and N = 3.
The one slight issue is the substitution of n = 4 in the (n²-3n)/2. We would just need to show that (n²-3n)/2 is bigger than 0 for n ≥ 4. But that's easy.
Me: "Obviously its n = 1 or n = 3" *Trying to prove that these are the ONLY values* Me: You got me there LMAO I thought about using induction to prove it since I saw that n = 4 didn't work and I knew for sure n > 4 didn't work either but I was a bit apprehensive knowing that it would've required some work.
@@robertveith6383 you are right, yes n=1 xor n=3 As a fact I can write you that in my language there is a word "albo" which suits the best here and it is equivalent to exclusive or which you don't use
n!=n(n+1)/2 n!=(n+1)!/2(n-1)! (n-1)!=(n+1)/2 Ř(n)=(n+1)/2 d/dn((n+1)/2)=1/2 Ř'(1)=-ř≈-.57 Ř'(2)=1-ř≈.43 Ř'(3)=3-2ř≈1.86>1/2 Ř(3)=2 4/2=2 the only solutions are n=1 and n=3
You are the best mathematics educator I've ever seen on TH-cam. You neither overexplain nor underexplain. Every step is 100% clear.
Nice presentation! Note, however, that it’s not necessary to use math induction.
Claim: If n>3, then n! >(n(n+1))/2. Multiplying both sides by 2 and dividing by n, we get the equivalent inequality 2(n-1)!>n+1. As 2(n-1)!>2(n-1), to prove the claim it will suffice to show that 2(n-1)>n+1, for n>3. But this is immediate: 2(n-1)>n+1 iff 2n-2>n+1 iff 2n-n>2+1 iff n>3.
I realized that while watching the video. Thanks.
This is a good example to demonstrate, how proof by induction works.
Although there is a quicker and relatively simple way to proof this directly.
If n>3 => 2n>n+3 => 2n-2>n+1 => (n-1)>(n+1)/2
Obviously (n-1)! is also greater than n-1. Therefor we can safely conclude that
(n-1)!>(n+1)/2. We now multiply both sides of the inequality by n>0:
n(n-1)!>n(n+1)/2 => n!>n(n+1)/2. (q. e. d.)
Great application of the Principal of Mathematical Induction. We almost forget how powerful it can be at times 😅
Never stop learning, nice lecture, you are genious.
Great video! It is a pretty interesting way to show how matematicians use demostrations techniches such as proof by induction, and in a curious problem! I love your videos
Excellent lecture !
Induction works smartly, with ± heavy writings…
Let’s try directly : in order to prove that n ! > n (n + 1) / 2 for any n ≥ 4, equivalent to 2 (n-1) ! > n + 1, we notice that (for n ≥ 4 ) :
2 (n-1) ! > 2^(n-1) because 2 (n-1) ! = 2x2x3x…x(n-1), and
2^(n-1) > n + 1 by studying f(x) = 2^x - x - 2 for x ≥ 3 (easy stuff…).
Then we’ve got it… Thank you for your interesting videos ! 🙂
An other way is : let U_n = 2 (n-1) ! and V_n = n +1.
Then U_n+1 / U_n = 2 n and V_n+1 / V_n = (n+2) / (n+1).
As U_3 = V_3 = 4, and for n ≥ 1, U_n+1 / U_n > V_n+1 / V_n,
Then V_n < U_n for n ≥ 4.
Because n! ≥ n*(n-1)*(n-2) and n*(n+1)/2 < n*(n+1) for all n≥4, all we need to show is that (n-1)*(n-2) > n+1
n² - 3n + 2 > n + 1
n² + 1 > 4*n
n + 1/n > 4, which is true for n ≥ 4 since 1/n > 0
1/n is never < 0 as long as n is positiv and as n is greater than 3 and a natural number i think that is proof enought
I am thankful for your efforts, your videos are always nice and filled with benefits, it's our pleasure to watch your videos.
I am grateful for you
You have n(n+1)/2 = n! iff (n + 1)/2 = (n-1)!. But one has (n - 1)! >= n - 1 > (n +1)/2 for all n > 3. So you don't have to check beyond n = 3.
Very methodical explanation profesor! I'm very content to You! Bravo!
I like the i which appears and disappears in the central panel...
you noticed too 😂
Great video. While comparing n^2 + n with n + 2, you could also see that since n >= 4, n^2 >= 16 so n^2 + n >= n + 16 > n + 2
Nice point
It's really interesting sir make more videos like this sir 💯💯💯❣️💫✨🌟
Excellent job
I did it similarly with one intuitive observation: n >= (n + 1) / 2:
Because (1 + 2 + ... + n) = n * (n + 1) / 2 = n! n >= (n - 1)!
Knowing that factorials increase quickly in comparison to a linear function, only n = 1, 2, 3 is possible, check to find n = 1, 3.
is there a way to solve this using the tangent property from trignometry
I totally yield to your formality about this, but I know I'm not alone in intuitively knowing that 3, and after a second's thought, also 1, were the solution set.
I like to think that geometrically the LHS is a quadratic and the RHS is a Gamma function on the first quadrant so intuitively there should be 2 intersections that has two solutions
This can be done without using the formula for Σi=(n)(n+1)/2 since
If n=1, then 1=1,
If n=2, then LHS=3, RHS=2
if n=3, RHS=LHS=6
And if n ≥4, 1+2+3+…+n n-1+1=n, n! ≥n(n-1)*2>n*n=n^2> 1+2+…+n.
You are the best sir
9:42 where did that new n+1 on the RHS come from?
@Salko_ So to get (n+l)! > (n+1)(n+2)/2, he took n! > n(n+1)/2 and replaced n with n+ 1? And then to get (n+1)n! > (n+1)n(n+1)/2, he took n! > n(n+1)/2 and multiplied both sides by n+1?
@Salko_ but are we not proving that n! > n(n+1)/2 so using that fact in our proof is circular logic?
we multiply both sides of eq at 7:11 by n+1
@Myhair0_0 That's why we have to check whether the inequality holds for the base case (n=4 in this example) Then, the proof from the video, which @Salko_ summarized, shows that if the inequality holds for any integer >=4, then it holds for the next integer. Thus, after manually verifying that the inequality holds for n=4 and completing the short proof, we know that the inequality holds for n=4, n=4+1=5, n=5+1=6, and so on. Hope this helps
@@Salko_ I don't know why the freak my response got deleted last night, I was asking this:
Are you saying (n+1)! > (n+1)(n+2)/2 come from replacing n with n+1 in the original n! > n(n+1)/2? And does (n+1)n! > (n+1)n(n+1)/2 cone from the original but after multiplying both sides by n+1?
Hello sir,
Very interesting approach to the problem. But if we are looking for the condition of the equality happening is it not easier?
What needs to happen so n*(n+1)/2=n! ?
Since n!=0 we can divide by n so (n+1)/2=(n-1)!
n+1=2(n-1)!
n=2(n-1)!-1 If we replace n by k+1 to have a nicer number inside the factorial
k=2k!-2 so k=2(k!-1) which means that k>k!-1 or k+1>k! and we can easily verify that this proposition is only true for very small values
I did the same 😂
Thank you my best professor but I didn't understand the end part of the solution especially on how you got that of n+n to be equal to (n+1)!
Nice proof by induction.
thats a nice idear wait net me simplify the first one: (n+n²)/2 and the second one as well: n! ore use the stirling formular...
1:00 i stll like your intro!
2:53 ok you write it in an other way but still the same
6:00 no it becomes interesting show it!
7:59 so any Natural nubers higher or equal than/to 4
8:47 but now it looks easy
10:41 because we just deal with a n greater than 4
12:02 nice i think i repeat it on my own board
Very rough, not rigorous idea but...
I figured that because the factorial grows much faster than the quadratic, after some point there won't be any more solutions, so there isn't any need to check every case.
I remembered the 1 + 2 + 3 = 1 x 2 x 3 meme, so n = 3
n also = 1.
That was basically my reasoning, because after n = 3 the factorial grows much faster than the quadratic so they will never ever intersect again, the only need is to check for answers within that range.
I think the proof of the factorial outlasting the triangular numbers for n >= 4 was rigorous to help cement that idea that they can never be equal and hence no solutions for that region.
That’s obvious and as you said not rigorous
From the thumbnail: the solutions are n=1 and n=3 . For any n>3, the cumulative product is greater than and also will be increasing faster than the cumulative sum.
I'm watching the video to see if you consider n=0 a solution or not (and why).
I immediately know of 1, and 3. I think those are the only 2 but proving that is a whole different issue. I might be able to do that but not sure exactly what I'd do maybe induction or contradiction but would take awhile to prove it.
All you have to do is prove that n! grows faster than n(n+1)/2 for all n>=4. You can prove that and thus since 24>10. That is 4!>10 and that n! grows faster than n(n+1)/2 then I can show that n(n+1)/2 can never catch up to n! thus only 1 and 3 are true
Thanks!:)
Ready for it
can you use gama function to figure out the equation ∑n=n!？
Best wishes for you and your family Professor 🎉🎉😊😊❤❤
In year 2025
For all n>4, n!>n(n-1)(n-2). So n(n+1)/2>n(n-1)(n-2) which is same as 2n^2-7n+3
Fun fact, graph f(n) = (n(n+1))/2 and g(n) = g!
Sorry for my bad english, I'm french.
In my induction, I decided to make a proof by contradiction. Bascially, we wanna show that since n >= 4, the sum of all the integers from 1 to n is not equal to the product of all the integers from 1 to n.
So we pick a n superior or equal to 4, let's assume that our property is true for that n (induction hypothesis)
And now let's suppose that the sum from 1 to n+1 is equal to the product of all the integers from 1 to n+1. (Hypothesis of the proof by contradiction, it's a part of the induction and we will prove this is false)
Ok so we have the sum from 1 to n+1 = the product from 1 to n+1
Then we can say that the sum from 1 to n PLUS n+1 equals the product from 1 to n TIMES n+1.
It's an equality so we put all the terms at the left and we have :
(Sum from 1 to n) + (n+1) - (product from 1 to n) × (n+1) = 0
Since the sum can be wrote n(n+1)/2 we factorise :
(n+1)(1 + n/2 - n!) = 0
Or n >= 4 so n+1 =/= 0
So 1 + n/2 - n! = 0
So 1 = n! - n/2
So 1 = n((n-1)! - 1/2)
So 1/n = (n-1)! - 1/2 (since n not equal to 0)
So (n-1)! - 1/2 < 1
So (n-1)! < 3/2
So 2(n-1)! < 3
Or n >= 4
So n-1 >= 3
So (n-1)! >= 6
So 2(n-1)! >= 12
And there is the contradiction
So we have the sum from 1 to n+1 who is not equal to the product from 1 to n+1 and then we have proved it for n+1
Thanks for all the people who will tell me my blunders : in maths but also in english. I tried my best to be understandable, so please be nice, and don't hesitate to tell me my mistakes, at least it shows that you read my work and it is very nice of you.
n²+n isn't always bigger than n+n as it is smaller at n=(0,1)
but n is greater then 3
I know but i said that it isn't always, but it is in that case
@@pizza8725 💀
10:25 only if n>1, which is understood here...
Great! Thank you
There are exactly 77000 ordered quadruples (a,b, c,d) such that gcd (a,b, c,d) =77 and lcm (a,b, c,d) =n, What is the smallest possible value of n?
Hello teacher. Could we look at this question? There were many solutions that i didn't understand well. I would like to see your approach
What restrictions are placed on a, b, c, d ? For example, can a, b, c, or d be a negative integer?
@yurenchu there's no restrictions I think
@@abulfazmehdizada In that case, there is no possible solution for n ; because the number of quadruples has to be a multiple of 16 ; and 77000 is not a multiple of 16 .
A quadruple (a,b,c,d) cannot contain any 0, because in that case, n=0 and in that case there are an infinite number of quadruples, for example of the form (0, 77, 77, 77k) where k is any non-zero integer. Therefore, |a| , |b| , |c| and |d| must be positive (i.e. nonzero).
Now, suppose (a, b, c, d) = (t, u, v, w) satisfies gcd(a,b,c,d) = 77 and lcm(a,b,c,d) = n , and t, u, v, w are positive integers. Then (t, u, v, w) represents a set of 16 _distinct_ quadruples that each satisfy the gcd and lcm conditions, namely
(a, b, c, d) = (t, u, v, w), (t, u, v, -w), (t, u, -v, w), (t, u, -v, -w), (t, -u, v, w), (t, -u, v, -w), (t, -u, -v, w), (t, -u, -v, -w), (-t, u, v, w), (-t, u, v, -w), (-t, u, -v, w), (-t, u, -v, -w), (-t, -u, v, w), (-t, -u, v, -w), (-t, -u, -v, w), or (-t, -u, -v, -w). This is true for any quadruple (t, u, v, w) of positive integers that satisfies the conditions, hence the total number of quadruples must equal {16 times the number of distinct quadruples of only positive integers}.
@@abulfazmehdizada In that case, there is no possible solution for n ; because the number of quadruples has to be a multiple of 16 ; and 77000 is not a multiple of 16 .
A quadruple (a,b,c,d) cannot contain any 0, because in that case, n=0 and in that case there are an infinite number of distinct quadruples, for example of the form (0, 77, 77, 77k) where k is any integer. Therefore, |a| , |b| , |c| and |d| must be positive (i.e. nonzero).
Now, suppose (|a|, |b|, |c|, |d|) = (t, u, v, w) satisfies gcd(a,b,c,d) = 77 and lcm(a,b,c,d) = n , and t, u, v, w are positive integers. Then (t, u, v, w) represents a set of 16 _distinct_ quadruples that each satisfy the gcd and lcm conditions, namely
(a, b, c, d) = (t, u, v, w), (t, u, v, -w), (t, u, -v, w), (t, u, -v, -w), (t, -u, v, w), (t, -u, v, -w), (t, -u, -v, w), (t, -u, -v, -w), (-t, u, v, w), (-t, u, v, -w), (-t, u, -v, w), (-t, u, -v, -w), (-t, -u, v, w), (-t, -u, v, -w), (-t, -u, -v, w), or (-t, -u, -v, -w). This is true for any quadruple (t, u, v, w) of positive integers that satisfies the conditions, hence the total number of quadruples must equal {16 times the number of distinct quadruples of only positive integers}.
@@yurenchu answer is 27720
Can I send you a question
Thank you for your video. It is possible not to use induction: as (n-1)! = (n+1)/2 is in Z, n must be odd and as (n-1)! = (n-1).(n-2)..2.1 > (n-1)(n-2).4 for any n >=7, we get
(n-1)(8n-17) < 2 which is impossible as n>6.
thank you teach us some strong induction
Let's get into the video
sir , can you record me some books to learn advanced mathematics
Begin with "Algebraic Geometry" by Hartshorne.
@Tommy_007 thank you
@@ishanpurkait9124 It's a VERY difficult book. I'll recommend that you start with basic books about calculus, elementary number theory, classical plane geometry, linear algebra, and abstract algebra.
An older series of books that are recommended for interested high school students: New Mathematical Library.
thank you ,can you help me choose between calculus by stewart ( republished by clegg and watson ) and thomas , both the early transcendental version
@ Prime Newtons n = 1 *OR* 3, not 1 "and" 3.
Hence _tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3) = 180°_
from every single letter he choosed i (imaginary number)
Also used as the "index" variable.
I thin, the o natural number satisfying this equation are n=1 and n=3. The left side can be substituted b n*(n+1)/2 (according to gauss forula for sumof the first n natural numbers) while the rigtside is equall to n! (accordingto the definition of factorial).So we seach values for n with n*(n+1)/2=n!. Since n! is per definition n*(n-1)!, we can transform this equation to
(n+1)/2=(n-1)!. n=1 and n=3 are possiblesolutionsforthhis equation, because (1+1)/2=(1-1)! and (3+1)/2=(3-1)!. Equal numbers can not fullfill the equation, because the right side is alwas a natural number, while the left side is for even values of n neer an integer. For all odd numbers n greater than 3, (n-1)! is greater than (n+1)/2, so n=1 and n=3 are the only solutions.
For me, it was obvious,that (n-1)! is greater than (n+1)/2 for an n>3, but nice, that you gave a proof ...
Hmm, let's see. I'm gonna guess 1 and 3.
*Suppose **_∃n ≥ 4 : _Σ⁽ⁿ⁾ᵢ₌₁{i} = Π⁽ⁿ⁾ᵢ₌₁{i}_*
⇒ _½n(n + 1) = n!_
⇒ _½(n + 1)!/(n - 1)! = n!_
⇒ _2n!(n - 1)! = (n + 1)!_
⇒ _2n[(n - 1)!]² = (n + 1)! = (n + 1)n(n - 1)!_
⇒ _2(n - 1)! = n + 1 = (n - 1) + 2_
⇒ _2(n - 2)! = 1 + 2/(n - 1) < 2_ since _n ≥ 4_
⇒ *_(n - 2)! < 1_** which is impossible for any factorial.*
∴ *_Σ⁽ⁿ⁾ᵢ₌₁{i} ≠ Π⁽ⁿ⁾ᵢ₌₁{i} ∀n ≥ 4_*
By inspection equality holds when *_n = 1 or 3 but not 2._*
Something bugged me,
in the induction, since we only needed n>=2
why couldn't we start at 2 ?
well, 15 years after formerly learning about induction, I finaly understand how important the initial step is (base case).
if we start at 2, initial step would be
2! = 2
2+1 = 3
2 is not greater than 3
So we can't start with 2,
And we can't start with 3 either since we have shown it is equal.
This is a marvelous Induction.
By inspection, there is only one answer, n=3
n=1
What about n = 1?
Wrong notation. Why i? That is n.
It’s can be any variable as long as you define it that way.
n=1 and n=3.
Sir pls make video on fermat's last theorem ❤ lost of love from India ❤❤
If i=1, then n(Σ) and n(Π)= {1- ♾️ \♾️ }ℕ
n = (1;3)
ATTEMPT:
By inspection, N = 1, 3
1 = 1
1 + 2 + 3 = 1 * 2 * 3 = 6
N = 2 doesn't work.
1 + 2 = 3, 1 * 2 = 2
For N >= 4:
N! > N(N-1)
= N^2 - N
= (N^2)/2 + N/2 + (N^2)/2 - 3N/2
>= (N^2)/2 + N/2 + 8 - 6
> (N^2)/2 + N/2
= N(N+1)/2
Which is the sum of all natural numbers up to N.
Therefore for all natural N >= 4,
1 + 2 + 3 + ... + N < 1 * 2 * 3 * ... * N, and so the two sides cannot be equal.
The only solutions are N = 1 and N = 3.
The one slight issue is the substitution of n = 4 in the (n²-3n)/2. We would just need to show that (n²-3n)/2 is bigger than 0 for n ≥ 4. But that's easy.
n = 1 *or* 3.
Yes but you have to prove it (i mean the fact that the sum from 1 to n is inferior to the product from 1 to n)
Me: "Obviously its n = 1 or n = 3"
*Trying to prove that these are the ONLY values*
Me: You got me there LMAO
I thought about using induction to prove it since I saw that n = 4 didn't work and I knew for sure n > 4 didn't work either but I was a bit apprehensive knowing that it would've required some work.
No, it's n = 1 *OR* 3.
By observation: n can be 1 and 3. But not 2, 4....
No, n can be 1 *or* 3.
log( 1 + 2 + 3 ) = log(1) + log(2) + log(3)
that's actually true
Those are not natural numbers
log( j1 x j2 x j3 x ... jn) = log(j1) + log(j2) = log(j3) + ...+ log(jn), from j1 + j2 + j3 + ... + jn = j1 x j2 x j3 x ... jn, log( j1 + j2 + j3 + ... + jn) = log(j1) + log(j2) = log(j3) + ...+ log(jn)
n=1 and n=3 and in my opinion that's all possibilities
n = 1 *or* 3
@@robertveith6383 you are right, yes n=1 xor n=3
As a fact I can write you that in my language there is a word "albo" which suits the best here and it is equivalent to exclusive or which you don't use
This is amazing!!! 🔥🫶😜
Naive
n!=n(n+1)/2
n!=(n+1)!/2(n-1)!
(n-1)!=(n+1)/2
Ř(n)=(n+1)/2
d/dn((n+1)/2)=1/2
Ř'(1)=-ř≈-.57
Ř'(2)=1-ř≈.43
Ř'(3)=3-2ř≈1.86>1/2
Ř(3)=2
4/2=2
the only solutions are n=1 and n=3
n! | T(n)
1!=1 | T(1)=1
2!=2 | T(2)=3
3!=6 | T(3)=6
4!=24 | T(4)=10
5!=120 | T(5)=15
6!=720 | T(6)=21
7!=5040 | T(7)=28
8!=40320 | T(8)=36
n! is already 3 orders of magnitude larger than T(n)
my friend, please answer! ! Are you from South Africa ? I LOVE MANDELA AND BLACK PEOPLE !!!!!
Nigeria 🇳🇬
Original poster, stop yelling in all caps. A color of a person is not to be loved. That is not logical.
THE INTEGRAL.
1/(1+x⁴).
THE THIRD WAY.
WHERE IS ITTT??
Stop yelling your post in all caps.
@robertveith6383 I just wonder what the 3rd way is😞😞