I'm a total math-illiterate trying to learn discrete mathematics. And these videos of yours makes so much more sense to me than a lot of the learning material that i try to dig into. Thank you for taking the time!
Great stuff! Absolutely enjoyed watching you proving it . When I first was taught PMI , I didn't like it and had a really hard time with it . But now , I have a much better understanding of it . Thank you very much the Math Sorcerer .
Genial, locamente, con los videos explicativos en español sentí que no entendía nada :( Solamente habían ejemplos muy difíciles. Fuiste muy amable para explicar
Yooo this exact question was in my review session today and I'm 100% sure its going to be on my exam tomorrow. Thank you dear teacher, for you have shown me the way!
It’s a nice proof and a great use of proof by induction, but I don’t think induction is necessary to prove the result. A much simpler proof is that since x^2-1=(x-1)(x+1), 2^(2n)-1=(2^n+1)(2^n-1) Now upon division by 3, 2^n cannot have a remainder of 0 (for it has no prime factor 3) it must either have a remainder of 1 or 2. Suppose it has a remainder of 1. Then 2^n-1 is divisible by 3. Likewise, if it has a remainder of 2, then 2^n+1 is divisible by 3. Hence 2^(2n)-1 must be divisible by 3.
Nice proof! I also proved it but instead using modular arithmetic: 1) Proving 3|2^(2n)-1 is equivalent to prove that 2^(2n)≡1 (mod 3) 2) Observe that 2^(2n)=(2^2)^n=4^n 3) a≡b (mod m) implies a^n≡b^n (mod m), therefore it follows from 4≡1 (mod 3) ---> 4^n≡1^n (mod 3) ---> 4^n≡1 (mod 3). By the way, I had an exam this semester where I needed to prove that 2^(2^n) always ends in 6 for natural n greater than 1. Is there a way to prove it without using strong induction?
Proof w/o Induction Let q=2^n. Then 2^(2n)-1=(2^n-1)(2^n+1)=(q-1)(q+1) Now consider the numbers q-1, q, and q+1. These are 3 consecutive integers, so one of them must be a multiple of 3. But we know that number cant be q, since q=2^n. Therefore 3 divides either q-1 or q+1 which means 3 definitely divides (q-1)*(q+1) and were done
![ลองสั่งของจากแอพ temu ถูกจริงหรอ?? ( part : 1/2 ) [โกงไหมครับ ep.89 ] | DOM](http://i.ytimg.com/vi/RTcZSQTiDZw/mqdefault.jpg)
I'm a total math-illiterate trying to learn discrete mathematics. And these videos of yours makes so much more sense to me than a lot of the learning material that i try to dig into. Thank you for taking the time!
Watching your videos keeps the blade of proofs sharp. This channel isn't nearly as watched as it should be.
thanks man!!!!!!!
Great stuff! Absolutely enjoyed watching you proving it . When I first was taught PMI , I didn't like it and had a really hard time with it . But now , I have a much better understanding of it . Thank you very much the Math Sorcerer .
Few weeks ago, I had been doing exercises of arithmetic & geometric series. Now mathematical induction today. Great.
awesome! Yeah induction is great. I will be posting more induction stuff, mainly stuff I don't already have on my channel.
Thank you! I was struggling with PMI, and you made it click.
Genial, locamente, con los videos explicativos en español sentí que no entendía nada :(
Solamente habían ejemplos muy difíciles.
Fuiste muy amable para explicar
que buene que este video te ayudo! me alegro mucho:) gracias por dejar un mensaje!!
:)
This helped me understand induction with divides thank you so much
Yooo this exact question was in my review session today and I'm 100% sure its going to be on my exam tomorrow. Thank you dear teacher, for you have shown me the way!
God bless you teacher, your question came out in my online exam.. Thank you very much
It’s a nice proof and a great use of proof by induction, but I don’t think induction is necessary to prove the result. A much simpler proof is that since x^2-1=(x-1)(x+1), 2^(2n)-1=(2^n+1)(2^n-1) Now upon division by 3, 2^n cannot have a remainder of 0 (for it has no prime factor 3) it must either have a remainder of 1 or 2. Suppose it has a remainder of 1. Then 2^n-1 is divisible by 3. Likewise, if it has a remainder of 2, then 2^n+1 is divisible by 3. Hence 2^(2n)-1 must be divisible by 3.
Nice proof!
I also proved it but instead using modular arithmetic:
1) Proving 3|2^(2n)-1 is equivalent to prove that 2^(2n)≡1 (mod 3)
2) Observe that 2^(2n)=(2^2)^n=4^n
3) a≡b (mod m) implies a^n≡b^n (mod m), therefore it follows from 4≡1 (mod 3) ---> 4^n≡1^n (mod 3) ---> 4^n≡1 (mod 3).
By the way, I had an exam this semester where I needed to prove that 2^(2^n) always ends in 6 for natural n greater than 1. Is there a way to prove it without using strong induction?
way simpler : 2^(2n) - 1 = (2^2)^n - 1 = 4^n - 1 = (4-1)Q(n) = 3Q(n) .
idk how much I can thank you
thank you so much
You are welcome!
Proof w/o Induction
Let q=2^n. Then 2^(2n)-1=(2^n-1)(2^n+1)=(q-1)(q+1)
Now consider the numbers q-1, q, and q+1. These are 3 consecutive integers, so one of them must be a multiple of 3. But we know that number cant be q, since q=2^n. Therefore 3 divides either q-1 or q+1 which means 3 definitely divides (q-1)*(q+1) and were done
Very cool induction proof that I understood for once! Great hair!
Thank you so much man.You are really awesome.
Just a reminder that I ONLY like math because of my luck in having you as a teacher.
aww thanks man!!!!
We can also do just modulo 3 and get: 4^k-1==1^k-1==0 (mod 3)
Thanks ✨
A multiple but it look so real I am a bit late i can understand so much from this sense
Well, now I know why it is like x+3x. I feel so dumb 😂. Thank you
Hehe
Plz prove 4^n=4mod6
6:39 My Ex?! Oh, dear...