If sqrt(a), sqrt(b), and sqrt(c) are part of a geometric sequence with ratio r, it's easy to show that so are a, b, and c (with ratio r²) and vice versa. So the square roots in the problem statement aren't really relevant and just make it slightly more tedious.
Prove that Sqrt[2],Sqrt[5],Sqrt[7] cannot be in the same geometric progression. You can’t Square root any prime number and expect a rational number back.
I think we can simply consider the possibility of a geometric progression containing theses values so Let a=√2 , √5=a*r^m and √7=a*r^(m+n) where m and n are positive integers Then it results by division that: r^m=√(5/2) and √(7/5)=r^n So r=(5/2)^(1/2*1/m)=(7/5)^(1/2*1/n) raising to the power 2mn : (5/2)^n=(7/5)^m 5^(m+n)=2^n*7^m which is impossible for m and n integers because 2,7 and 5 are relatively prime so the hypothesis was wrong So by reducing to absurd the original statement is false
but is it correct to compare intervals between different terms of the same geometric sequence ? Isn't it better to compare sqrt(5)/sqrt(2) with sqrt(7)/sqrt(2) ?
I believe there's another way to prove the contradiction other than checking parity. 2, 5, and 7 are prime numbers. The form that was given is similar to prime factorization Since a single number cannot have two different sets of prime factors, it shows the contradiction
That was actually what I did. I did the same thing but I went a bit overkill and said that the prime factorisation of a number is unique, so all the equations are false because they imply that there exist such a number that has 2 completely unique ways of factoring it in primes. which is not true. Hence, no solutions exist, no natural triplets of the numbers exist, and hence there is no way the three surds can exist in the same geometric progression.
Geometric progressions are always growing exponentially as long as r and a_n are both greater than 1 but sqrt(2), sqrt(5), sqrt(7) are increasing in that order but the amount they increase decreased from sqrt(2) to sqrt(5) to sqrt(5) to sqrt(7).
@budderzmonahan6215 There could be numbers between them but it doesn't matter, the fact that there is 3 points and ant smooth path connecting them would be increasing at a lesser rate means that it is not growing exponentially when going from sqrt(2) to sqrt(5) and then to sqrt(5) to sqrt(7). They don't need to be consecutive.
@@maxhagenauer24 That only holds as long as the common ratio is at least 2. For a simple counterexample, consider a geometric progression starting from 1 with a ratio of 1.5. The first few values are 1, 1.5, 2.25 and 3.75. The difference between the first and the third is entries 1.25 and between the third and the fourth 1.125. It's still growing exponentially but the increase between successive entries after the second one is now always _less_ than all the earlier increases added together. With a ratio of 1.5, it tends asymptotically towards half the earlier contributions.
@jounik First, why does the common ratio have to be strictly greater than 2? It just needs to be greater than 1 otherwise it's not growing exponentially. Second, the 4th term in a geometric progression with starting value 1 and common ratio 1.5 is 3.375, not 3.75 but your difference of 1.125 between the 3rd and 4th terms is still correct. Third, your example is not doing the sane as what I was doing, you skipped from the 1st value to the 3rd value and got that different but the numbered values are not consecutive when you did that. When you look at consecutive values like from 1st to 2nd or 3rd to 4th or 7th to 8th, their distances increase as you go, they would have to for it to be growing exponentially and that works of r < 2 as long as it's greater than 2.
@@maxhagenauer24 Copying error aside, the point was that the ratio would have to be >=2 in order for any _non-consecutive_ terms always having a smaller difference than the next consecutive one. You did say that "there could be numbers between them but it doesn't matter" after all.
Sir please do a limit question which was came in JEE Advanced 2014 shift-1 question number 57 it's a question of a limit lim as x-->1 [{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4 You have to find the greatest value of a It has 2 possible answers 0 and 2 But I want the reason that why should I reject 2 and accept 0 Because final answer is 0 Please help 😢
Surely a_0 is the first term? Because r^0 = 1, thus a_n when n=0 is a_0 which is the first term. But if a_1 is the first term then for any r =/= 1, a_1 = a_1 x r^1 =/= a_1 Am I missing something? (Also, sorry about the formatting)
I think you can do the same, replace the numbers with p1, p2, p3, and show that the prime factorisation of a number is unique, so there can never be a number with two ways of factorising it in primes, and hence there is a contradiction, and hence no geometric progression can contain 3 primes.
Worst case 1: First term is prime. Ratio of r suggests that the subsequent terms will always be divisible by r if r is an integer. So the next term cannot be prime, nor the next term cannot be prime. If r is a rational or irrational number then there is no point talking about primes. Worst case 2: First terms are p1 and p2. p1, p2, p3 common ratio of p1 and p2 is p2/p1. so p3 = p2 ^2 / p1 p1p3 = p2^2. Once again, illogical conclusion because of uniqueness of prime factorisation.
Can't it prove like that: If √2, √5 and √7 are in geometry progression, then its geometric mean should be √5. After some calculations, you will see √5 wont be equal to geometry mean, thus the above sequence wont be in geometric
A question- why did you use ‘i’ when you substituted? If it was me I’d steer clear of ‘i’ When you deal with square roots imaginary numbers can come into play. In similar proofs, although not this one, it could be confusing. Most mathematicians do it!
Bro it is a number theory question. Since imaginary numbers are not generally used in these type of questions, noone will misunderstand i as the imaginary unit.
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
It's great to see you do hungarian math problems, we really apprechiate it
It’s important to note that n and m are not zero since the roots on the middle board are not 1. Thus, we DO have odd equals even and not 1=1.
He notes that i != j != k and m = j - i , n = k - j so m and n cant be 0
Also even if the only possibility was n = 0 for all numbers it would cause a practically degenerate case where r = 1/0 and hence division by 0 occurs
@@jokou8223 Oh. Yeah, you’re right.😅
an interesting expansion. If you are given 3 numbers x,y,z with x
The logic of math is always satisfying.
You’re an awesome teacher!
твёрдое доказательство, простое для понимания и усвоения!
мой лайк каналу
متألق كالعادة
شكرا أستاذ
where is the 3rd way of the integral?
If sqrt(a), sqrt(b), and sqrt(c) are part of a geometric sequence with ratio r, it's easy to show that so are a, b, and c (with ratio r²) and vice versa.
So the square roots in the problem statement aren't really relevant and just make it slightly more tedious.
Prove that Sqrt[2],Sqrt[5],Sqrt[7] cannot be in the same geometric progression. You can’t Square root any prime number and expect a rational number back.
I don't think geometric progressions require a rational number.
1, sqrt(2), 2, 2sqrt(2), 4...
I think we can simply consider the possibility of a geometric progression containing theses values so
Let a=√2 , √5=a*r^m and √7=a*r^(m+n) where m and n are positive integers
Then it results by division that:
r^m=√(5/2) and √(7/5)=r^n
So r=(5/2)^(1/2*1/m)=(7/5)^(1/2*1/n) raising to the power 2mn :
(5/2)^n=(7/5)^m
5^(m+n)=2^n*7^m which is impossible for m and n integers because 2,7 and 5 are relatively prime so the hypothesis was wrong
So by reducing to absurd the original statement is false
but is it correct to compare intervals between different terms of the same geometric sequence ? Isn't it better to compare sqrt(5)/sqrt(2) with sqrt(7)/sqrt(2) ?
Yo, I need help in finding the primes p such that p^3 - p + 49 is a perfect square.
I believe there's another way to prove the contradiction other than checking parity.
2, 5, and 7 are prime numbers.
The form that was given is similar to prime factorization
Since a single number cannot have two different sets of prime factors, it shows the contradiction
That was actually what I did. I did the same thing but I went a bit overkill and said that the prime factorisation of a number is unique, so all the equations are false because they imply that there exist such a number that has 2 completely unique ways of factoring it in primes. which is not true.
Hence, no solutions exist, no natural triplets of the numbers exist, and hence there is no way the three surds can exist in the same geometric progression.
Geometric progressions are always growing exponentially as long as r and a_n are both greater than 1 but sqrt(2), sqrt(5), sqrt(7) are increasing in that order but the amount they increase decreased from sqrt(2) to sqrt(5) to sqrt(5) to sqrt(7).
That is true but they aren’t necessarily consecutive numbers, for example there could be multiple numbers in between sqrt (2) and sqrt (5)
@budderzmonahan6215 There could be numbers between them but it doesn't matter, the fact that there is 3 points and ant smooth path connecting them would be increasing at a lesser rate means that it is not growing exponentially when going from sqrt(2) to sqrt(5) and then to sqrt(5) to sqrt(7). They don't need to be consecutive.
@@maxhagenauer24 That only holds as long as the common ratio is at least 2. For a simple counterexample, consider a geometric progression starting from 1 with a ratio of 1.5. The first few values are 1, 1.5, 2.25 and 3.75. The difference between the first and the third is entries 1.25 and between the third and the fourth 1.125. It's still growing exponentially but the increase between successive entries after the second one is now always _less_ than all the earlier increases added together. With a ratio of 1.5, it tends asymptotically towards half the earlier contributions.
@jounik First, why does the common ratio have to be strictly greater than 2? It just needs to be greater than 1 otherwise it's not growing exponentially. Second, the 4th term in a geometric progression with starting value 1 and common ratio 1.5 is 3.375, not 3.75 but your difference of 1.125 between the 3rd and 4th terms is still correct. Third, your example is not doing the sane as what I was doing, you skipped from the 1st value to the 3rd value and got that different but the numbered values are not consecutive when you did that. When you look at consecutive values like from 1st to 2nd or 3rd to 4th or 7th to 8th, their distances increase as you go, they would have to for it to be growing exponentially and that works of r < 2 as long as it's greater than 2.
@@maxhagenauer24 Copying error aside, the point was that the ratio would have to be >=2 in order for any _non-consecutive_ terms always having a smaller difference than the next consecutive one. You did say that "there could be numbers between them but it doesn't matter" after all.
Sir please do a limit question which was came in
JEE Advanced 2014 shift-1 question number 57
it's a question of a limit
lim as x-->1
[{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4
You have to find the greatest value of a
It has 2 possible answers 0 and 2
But I want the reason that why should I reject 2 and accept 0
Because final answer is 0
Please help 😢
Surely a_0 is the first term?
Because r^0 = 1, thus a_n when n=0 is a_0 which is the first term.
But if a_1 is the first term then for any r =/= 1, a_1 = a_1 x r^1 =/= a_1
Am I missing something? (Also, sorry about the formatting)
Sqrt[5/2]=r^m, Sqrt[7/5]=r^n
Since 5 is prime, 5 to any natural power will not have 2 or 7 as a multiple to begin with, even before comparing parity.
Great work ❤❤❤❤ keep up
Great Video, I probably would not use i since it could lead to confusion since i also indicates complex numbers ❤😊
Next problem would be: A geometric progression never contains three primes.
I think you can do the same, replace the numbers with p1, p2, p3, and show that the prime factorisation of a number is unique, so there can never be a number with two ways of factorising it in primes, and hence there is a contradiction, and hence no geometric progression can contain 3 primes.
Worst case 1:
First term is prime. Ratio of r suggests that the subsequent terms will always be divisible by r if r is an integer. So the next term cannot be prime, nor the next term cannot be prime.
If r is a rational or irrational number then there is no point talking about primes.
Worst case 2:
First terms are p1 and p2.
p1, p2, p3
common ratio of p1 and p2 is p2/p1. so p3 = p2 ^2 / p1
p1p3 = p2^2.
Once again, illogical conclusion because of uniqueness of prime factorisation.
Thought provoking.
Aula incrivel
Very good proof ! 🙂
a sub n =a sub1*r^ (n-1) no?
That’s what I’m thinking, I’m sure it is.
True, because as it is now, it says a1=a1*r
Wow! Don't often get a wow out of a maths proof.
Can't it prove like that:
If √2, √5 and √7 are in geometry progression, then its geometric mean should be √5.
After some calculations, you will see √5 wont be equal to geometry mean, thus the above sequence wont be in geometric
00:48 - 13:43
Sqrt[5/2]=0.5Sqrt[10]
r^j/r^i=r^(j-i)
now do the problem
3^(1/2) 5^(1/2) 7^(1/2) cannot b in the same geometric progression and cannot b in the same arithmetic progression.
A question- why did you use ‘i’ when you substituted? If it was me I’d steer clear of ‘i’ When you deal with square roots imaginary numbers can come into play. In similar proofs, although not this one, it could be confusing. Most mathematicians do it!
i Is often used as index
@ I also find that confusing.
Bro it is a number theory question. Since imaginary numbers are not generally used in these type of questions, noone will misunderstand i as the imaginary unit.
@@a_man80-- "No one" is two words.
r^k/r^j=r^(k-j)
That was great
First drop from IIT bombay student 😊
Can't m and n be negative. You have proved that rt2 can't come before rt5 and rt7 in a geometric series.
You may have skipped the the part where I addressed that.
Sqrt[7/5]=0.2Sqrt[35]