but i don't see how, when seeing a problem like this for the first time, you'd know to replace n^(2) by 2n + 1. I don't see how you would know to do that.

You solved exactly same thing I got for homework, heh :D Now I understand it completely.

How does one know the correlary piece they need to prove before hand?

You are an excellent teacher, and that would be an understatement.

Thanks! Best explanation I've found. Finally got it!

hi! its very helpful, but I'm wondering how can you come up with the 2n+1 in the first place?

great video, it really helped me out! :) The only problem i have is this: why do you assume that 2^n is greater than or equal to 2n+1?

Wait a minute! This is the same guy in the video about parity bits that I watched! Man, this guy is all everywhere.

8:36 I don’t get how you can say 7 is greater than/equal to 0...

If anybody was wondering why 2^n > n^2 becomes 2^n * 2 > n^2 * 2 is because in order to get 2^n+1 you need to multiply 2^n by 2 (hence 2^n * 2). Since it's an inequality n^2 becomes n^2 * 2.

@ 8:39, it should be > 0 not > or = 0 correct? if something is > or = 7, it is > 0 not > or = 0 right?

does this solution work for 2^n > n^2 ?? > not >=

Dude thank you. I needed this explanation ❤️

The best explanation I watched so far

Ek wil graag weet hoekom is 2 tot die mag ' n' groter en gelykaan ( 2n +1)? En by die einde van die bewys hoekom is k groter en gelykaan 4??(induction inequality example 6).

How do you prove this using minimum counterexample?

Thank You soo much for this, but could you also prove n

Thank you for this wonderful video. I learnt a lot:D

is it possible to start at step 2 if you had never worked this problem? Not likely

2.2power k How convert 2 power k + 2 power z Tell me

I clicked the video without looking at the channel but immediately recognized Eddie Woo the moment he said the word "here."

Why not use a chain for the first one? For example: assume: 2n + 1

Hi Eddie, I find your videos very helpful, I kind of wonder the same thing like some other viewers here, can you explain a little more in detail when you swap n^2 by 2n+1, and how I can get to this step in general, for instance if I have to prove 2^(n-1)bigger and equal to n^2. Thanks very much!

But how is 7 greater or equal to 0? How can we make that assumption, when 7 would never equal 0?

Hi Eddie! Your video helped my alot! But I still have one question. Why did you change the letter n to k (n=k) and didn't continue using the letter n? Thank you !

Nice video! One question though. Isn't the whole first part a bit unnecessary if you could simply prove that (2^k-(2k+1)) >= 0 for k>=4, which it is. Or am I not allowed to just insert 4 like that?

I'm already getting the feeling that this video is gonna be directly responsible for a love affair I' going to have with mathematics for the next year or so.

Can you do a video for 4^n > n^4 ??

You are my savior.

why did you start with 2^(n)>and equal to 2n+1

Good work, buddy.

Can we do it by putting the equation on one side and zero on another side for every inequality question??????

actually you didnt have to demonstrate past 6:04 when you write consider as we multiplied by 2 which is a positive number which won't change the inequality right ?

2 question I don't understand after consider. How plus 2 power k

Hi eddie, i'm currently working on an assignment whereby, my assignment ask for to prove by induction that 2^n > n^2 for n>=5. is it possible for me to prove it ur way as ur question is quite similar?

Would you consider the first proof the lemma that you need for the current proof or proof in consideration? Good job:

please solve 2n less than n! for, n greater or equal to 4

Thank you, I understand now :D

thanks for this video! now i am even more confused haahah

You are saving me in Discrete Math. Thank you so much.

thank you very much .that was helpful

7:54 - Where did the 2k - 1 come from? I thought it was 2k + 1.

you are honestly the best math teacher

Thank you :D

Wow. That was great

Sir, why m√a^n = a^n/m please explain

In my opinion, the flowchart at the beginning is more important to those who want to teach the subject whereas for students it may add another layer of difficulty since these videos are primarily watched by people who need reinforcement on the title/topic.

Thank you!!

i got it......Thanks a lot Sir.......

You can go like a heat seaking missile right to the end. No need for ”expeditions” here and there. 1) Show that it is true for a smallest value of p. Bla bla 2) Assume the statement to be true for p. 3) Show that under the above assumption it is also true for p + 1. That is (p + 1)^2 LE 2^(p + 1) expand on both sides p^2 + 2p + 1 LE 2 * 2^p that is p^2 + 2p + 1 LE 2^p + 2^p Make use of our assumption under 2). That is, it now suffices to show that 2p + 1 LE 2^p. But 2p + 1 LE p^2 since after we subtract 2p and add 1 we get 2 LE p^2 - 2p +1 LE (p - 1)^2, which is true for p = 4, 5, 6,… And again by our assumption under 2) 2p + 1 LE p^2 LE 2^p Q. e. d. Comment: What we are as a whole to prove, and our assumption is like a balance leaning over to the right. We make use of our assumption and take away on both sides. The balance must not flip over to the other side! That is what suffices to prove. (And to do that we use our assumption once again.) Hope you see what I mean with a balance. The kind Mdm Justitia uses.

BTW, RTP means required to prove

Thank you this video was very helpful. ..but am a little bit confused. ...how come 2^k+1 =2^k+2^k. ...help

I don't get how you have two 2^k in 2.2^k. I'm confused

Thank you soo much for ur explanation,but I have a question why do assume that 2^n is greater or equal to 2n+1. Thank you

Your explanation was very helpful. I have had this question before but could not just figure out the induction step . I did understand what you did in both the first case and second but I have a small question. I understand that 2 to the power k+1 =2 to the power n times 2 but I don not seem understand how and why you had to write 2 to the power n+ 2 to the power n. May you please help me on that. Thank you..

thanks !!!

Shouldn't it be "n = { 4,5,6,... }," not "{ n = 4,5,6,... }"?

More videos plz it's interesting

why is it 2k+3

I do it this way, hope it's correct. 2.2^k>=2.k^2 2^(k+1)>=(k+1)^2 To prove 2.k^>=(k+1)^2 2.k^2>=k^2+2k+1 k^2-2k-1>=o.......(1) True if k>2.4...

Assume 2^k > k^2 for k > 4. 2^(k+1) = 2(2^k) = 2k^2 > 2k^2 = (k+1)^2 + k^2-2x-1 > (k+1)^2, since k^-2k-1 = (k-1)^2-2 > 0 for k>4

I would just use 2^(k+1)>=2k^2 from the assumption, then it's a simple matter of proving 2k^2>=(k+1)^2 for each k>=4

why is 2^k>2k+1

Thanks

I love mathematics!

The one thing I don't get about induction: why just assume something is true for n, then show it's true for n+1, if what you wanted to prove in the first place is that it's true for n?

Hi, I don't know if you still do videos or not. I found your explanations to be very simple and clear and wondered if you could show me how to do the following proofs: (1) let lcm(a,b) = l so l=pa and l=qb Prove that gcd(p,q) = 1 (2) if gcd(a,b) = 1 prove that gcd(a+b, a-b) is either 1 or 2. (3) Prove that if a | (bc) and gcd(a,b)=1 then a|c (4) If d=gcd(a,b) and f is any other common divisor of a and b, prove that f | d Any help with any of them is appreciated. If you don't do this anymore no problem. Thanks for all your videos.

thanks boss

Ok anyone watching this video: before you go and comment about not understanding something, take the time to carefully watch the ENTIRE thing. It will all makes sense :D

I finally understand why you have proved that 2^n≥2n+1 is true at first, thank you so much!! Now I have another question, which is familiar with this one, that is: to prove that 3^n>n^3,{n=4,5,6...} Thank you.

Thank you so much! But I don't understand why you can't you do it directly. ie. assume true for n=k so 2^k is greater than k^2 then rtp: n= k+1 so 2^(k+1) is greater than (k+1)^2 LHS= 2 x 2^k then sub in the assumed stuff and becomes 2x2k^2 RHS=k^2+(2k+1) LHS is greater than RHS because k^2 is greater than 2k+1 for k>4 why can't you do it this way?

thank you from palistine

Sir why did you replace 2^k by K^2 ?

What level of algebra is this?

I don't get it

I am from India my ? How can you return as n square is equal to 2n+1 .but why

n^2 = 2n+1 ????

2*2^(k)-(2k+1)>=k^2-2k+1-2=(k-1)^2-2 ,since k>=4,(k-1)^2>=9 -->k^2-2k-1>=9-2=7>=0 so i don't think we need to prove n^2>=2n+1

In 3:38, you said that LHS is greater than or equal to RHS when you have only shown that LHS is greater than RHS, for n = 4. You have not necessarily convinced me about equality of the two sides.

I love you

2021

微分して増減調べればいいのでは？

thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you reeaaaaaalllllyyyyyyyy muuuuuuch!!!!!!

for any positive integer n, 6n - 1 is divisible by 5.

Demostró que 2^n > 2n+1. Pero debía demostrar 2^n > n². 😬

2^3 is smaller than 3^2

Let me get another beer ...

o ga o.. oponu oshi

@ 8:46 7 is >= 0 why can you make 2^(k+1) - (2k+3) >=0 does it mean that it can also be =1? since it's >=0? but it should also be >=7 right? I can somehow understand the whole video but I want to fully grasp the concepts. I think there's something wrong somewhere in my understanding. Enlighten me please.

Is induction the only way to solve this kind of problem... This is genuinely not an easy thing to grasp.

Thwis syde wud be bigga

problem is ur first teacher

how 2.2^k = 2^k+ 2^k ?????

That problem is too easy Prove this one (2n)! > n^n

wait wha?

Majorly overcomplicated. When you had 2^(k+1)-(k+1)^2 you could simply rearrange to 2(2^k-k^2) + (k-1)^2 - 2 which is certainly positive for k>=4

How can n^2 be replaced by 2n+1? You didn't explain the first step and this whole video is senseless

Not helpful at all man. Even made more confused