sqrt2, sqrt5 and sqrt7 cannot be terms of the same geometric progression.

[@GreenMeansGOF]

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.

[@jokou8223]

He notes that i != j != k and m = j - i , n = k - j so m and n cant be 0

[@assiya3023]

متألق كالعادة شكرا أستاذ

[@BartBuzz]

The logic of math is always satisfying.

[@RyanLewis-Johnson-wq6xs]

You’re an awesome teacher!

[@ruud9767]

Next problem would be: A geometric progression never contains three primes.

[@padla6304]

твёрдое доказательство, простое для понимания и усвоения! мой лайк каналу

[@RyanLewis-Johnson-wq6xs]

Sqrt[5/2]=r^m, Sqrt[7/5]=r^n

[@RyanLewis-Johnson-wq6xs]

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.

[@FunkyTurtle]

I don't think geometric progressions require a rational number. 1, sqrt(2), 2, 2sqrt(2), 4...

[@dan-florinchereches4892]

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

[@AmilQarayev41]

where is the 3rd way of the integral?

[@jay_sensz]

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.

[@Modo942000]

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

[@sr6424]

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!

[@LovePullups]

i Is often used as index

[@sr6424]

@ I also find that confusing.

[@craig4320]

Thought provoking.

[@RyanLewis-Johnson-wq6xs]

Sqrt[5/2]=0.5Sqrt[10]

[@RyanLewis-Johnson-wq6xs]

r^j/r^i=r^(j-i)

[@RyanLewis-Johnson-wq6xs]

r^k/r^j=r^(k-j)

[@andrejflieger4182]

Great Video, I probably would not use i since it could lead to confusion since i also indicates complex numbers ❤😊

[@frreinov]

That was great

[@IITIAN_dost]

First drop from IIT bombay student 😊

[@peshepard412]

a sub n =a sub1*r^ (n-1) no?

[@RyanLewis-Johnson-wq6xs]

Sqrt[7/5]=0.2Sqrt[35]