I used desmos to graph the three functions anx actually the equation holds for x>=0.6782 whitch is the lower bound value that has been found. Great video as always.
I know that this isn't totally relevant because we are accepting all positive real values for n, but I thought it was interesting to note that 9n+8 can never be a aquare number for any integer valuenof n. 0,1,4,7 are the only square numbers mod 9
@@TheArtOfBeingANerdLet 9n+8=s² for some natural number s. Now look at the equation modulo 3: 0+2=s² so s²=2 (mod 3) which is not true because all perfect squares are 1 modulo 3.
@@TheArtOfBeingANerdAnother way to prove is to use @graf_paper method. Look at all the possible remainders of s² when divided by 9 and you'll notice that 8 is not in that list. s can leave a remainder of (0,1,2...8) when divided by 9 so just use modular arithmetic to find the remainders of s², here are some examples: If s=7 (mod 9) then s²=49=4(mod 9) If s=8 (mod 9) then s²=64=1(mod 9)
Using the well-known inequalities between geometric, arithmetic and squared means for not equal a, b, c: cbrt(abc) < (a+b+c)/3 < sqrt((a^2+b^2+c^2)/3) /cbrt stands for cube root/ for sqrt(n-1), sqrt(n), sqrt(n+1), we get 3*(cbrt(sqrt((n^3-n)) < sqrt(n-1)+sqrt(n)+sqrt(n+1) < 3*sqrt((n-1+n+n+1)/3) = sqrt(9n). For n>=4, it's easy to see that sqrt(9n-1) < 3*(cbrt(sqrt((n^3-n)), or equivalently (9n-1)^3 < 3^6(n^3-n): it leads to the quadratic 0 < (9n)^2 - 28*9n + 1/3 = (9n-14)^2 - 14^2 + 1/3, clearly true for n>=4. So for n>=4, sqrt(9n-1) < sqrt(n-1)+sqrt(n)+sqrt(n+1), and the same is easy to check "manually" for n=2, 3. Replacing n by n+1, we get the two desired inequalities: sqrt(9n+8) < sqrt(n)+sqrt(n+1)+sqrt(n+2) < sqrt(9n+9) implying the identity in question. Q.E.D.
How was this problem come up with? The '9' coefficient makes since (1+1+1)^2 = 9, but the constant at the end seems tricky to pinpoint, and I assume there are other non-integer constants which still satisfy the inequality. Is it posssible to generalise?
I found this posed as a problem to solve - there's a link in the description. Not sure how it was discovered though. It should be possible to generalise to include more terms. For example, some trial and error with replacing the "+8" term leads to sqrt{16n + 23} < sqrt{n} + sqrt{n+1} + sqrt{n+2} + sqrt{n+3} < sqrt{16n + 24}, so a similar result holds for floor{sqrt{16n + 23}}.
Great problem! Heer's a question: 3 sqrt(n) < sqrt(n) + sqrt(n+1) + sqrt(n+2) is true for all natural numbers; and sqrt(9n+8) < 3 sqrt(n+1) for all natural numbers. So why can't you just compare floor(3 sqrt(n)) < = floor(3 sqrt(n+1)), which is true for all natural numbers; hence, the stated problem is true for all natural numbers.
I used desmos to graph the three functions anx actually the equation holds for x>=0.6782 whitch is the lower bound value that has been found. Great video as always.
always love seeing your video, especially ones about floor function.
Thank you!
I know that this isn't totally relevant because we are accepting all positive real values for n, but I thought it was interesting to note that 9n+8 can never be a aquare number for any integer valuenof n.
0,1,4,7 are the only square numbers mod 9
I remember seeing this problem in a floor function list but n was a natural number so that observation is actually crucial in the proof.
How do you show this result?
@@TheArtOfBeingANerdLet 9n+8=s² for some natural number s.
Now look at the equation modulo 3:
0+2=s² so s²=2 (mod 3) which is not true because all perfect squares are 1 modulo 3.
@@TheArtOfBeingANerdAnother way to prove is to use @graf_paper method.
Look at all the possible remainders of s² when divided by 9 and you'll notice that 8 is not in that list.
s can leave a remainder of (0,1,2...8) when divided by 9 so just use modular arithmetic to find the remainders of s², here are some examples:
If s=7 (mod 9) then s²=49=4(mod 9)
If s=8 (mod 9) then s²=64=1(mod 9)
Using the well-known inequalities between geometric, arithmetic and squared means for not equal a, b, c: cbrt(abc) < (a+b+c)/3 < sqrt((a^2+b^2+c^2)/3)
/cbrt stands for cube root/ for sqrt(n-1), sqrt(n), sqrt(n+1), we get
3*(cbrt(sqrt((n^3-n)) < sqrt(n-1)+sqrt(n)+sqrt(n+1) < 3*sqrt((n-1+n+n+1)/3) = sqrt(9n).
For n>=4, it's easy to see that sqrt(9n-1) < 3*(cbrt(sqrt((n^3-n)), or equivalently (9n-1)^3 < 3^6(n^3-n):
it leads to the quadratic 0 < (9n)^2 - 28*9n + 1/3 = (9n-14)^2 - 14^2 + 1/3, clearly true for n>=4.
So for n>=4, sqrt(9n-1) < sqrt(n-1)+sqrt(n)+sqrt(n+1), and the same is easy to check "manually" for n=2, 3.
Replacing n by n+1, we get the two desired inequalities: sqrt(9n+8) < sqrt(n)+sqrt(n+1)+sqrt(n+2) < sqrt(9n+9)
implying the identity in question. Q.E.D.
You're a very good presenter; I'm enjoying your channel. Thank you for sharing your talents.
Thank you so much!
Your number theory videos are my favourites! Cool techniques used here!
Thank you!
Using the techniques used to bound a, and c
Beautiful solution, I always love inequality work.
Your videos are wonderful
How was this problem come up with? The '9' coefficient makes since (1+1+1)^2 = 9, but the constant at the end seems tricky to pinpoint, and I assume there are other non-integer constants which still satisfy the inequality. Is it posssible to generalise?
I found this posed as a problem to solve - there's a link in the description. Not sure how it was discovered though. It should be possible to generalise to include more terms. For example, some trial and error with replacing the "+8" term leads to sqrt{16n + 23} < sqrt{n} + sqrt{n+1} + sqrt{n+2} + sqrt{n+3} < sqrt{16n + 24}, so a similar result holds for floor{sqrt{16n + 23}}.
@@DrBarker Problem E3010 appeared on Vol. 93, issue No. 7, page 483. It's been published in 1983. Still loved your solution, though.
Beautiful problem and solution!
Great problem! Heer's a question:
3 sqrt(n) < sqrt(n) + sqrt(n+1) + sqrt(n+2) is true for all natural numbers; and sqrt(9n+8) < 3 sqrt(n+1) for all natural numbers. So why can't you just compare floor(3 sqrt(n)) < = floor(3 sqrt(n+1)), which is true for all natural numbers; hence, the stated problem is true for all natural numbers.
You would need floor(3 sqrt(n+1)) < = floor(3 sqrt(n)) that is not true.
You can't conclude that sqrt(9n+8)
Is this similar to Part (3) of Problem 723 of Ramanujan’s 3rd notebook:
floor(sqrt(n)+sqrt(n+1))=floor(sqrt(4n+2))?
It’s Friday already ???