Unusual Induction Inequality Proof (3 of 3: By exhaustion)

[vinitasrivastava]

I am in 9th currently and at the same time an IITian aspirant.
Your videos have helped me to complete my whole 10th syllabus.
Thanks a lot for your hard work!!!

[xristossamiotis]

I love how easy it makes it and the energy he has. He looks even happy that makes the viewer feel good!!
Great work good job!!

[mikegalway]

i have very basic maths experience from college that included algebra and calculus. seeing you work through those problems in the 3episodes was so interesting and well described as far as my basic understanding is aware!!!..would love to see more videos like these. i love its loical process, just a bit slow on the uptake!! Great video mate

[matarli]

I have read your book translated to mandarin! As a university student study in science department, I really admire your contribution ! What a interesting book about math ~~~~

[sibusisondimande]

I already did my post grad qualification, just wanted to revise stuff for the hell of it, I can't help it. Why didn't I have you back in my day !!

[wheatley]

Hey Eddie, really good video and was well explained! Thanks for the help!

[justinpark]

I tried something else myself, wanted to see if its valid.

Manipulating the inequality, you can show that (hopefully) a^(k+1)+a^-(k+1)>=a^k+a^-k is equivalent to (a-1)(a^2k-1)>=0

we will assume that a>0

we will also establish that a-1>a^2k-1, when a<1 (by showing that it leads to a>a*a^2k -> 1>a^2k which is true if a<1). Similar working leads to the findings that a-1=a^2k-1 when a=1 and a-1<a^2k-1 when a>1

for the first case, we will use the case of 0<a<1. This case implies that a-1<0, which further implies that a^2k-1<0, thus lhs>0
for the second case, we will use a=1. Obviously, from the first line, 2=2 so therefore the inequality is true, but also since a-1=0, then lhs=0 thus valid.

for the third case, we will use a>1. This case implies that a-1>0, which further implies that a^2k-1>0 thus lhs>0

For all cases, a>0, the k+1 case holds, meaning that a^n+a^-n>=a^(n-1)+a^-(n-1), by induction, is true (proceeds to draw nicely shaded rectangle :) )

[jowprower]

Hi Mr Eddie...love your teaching skills...and BTW I'm in the 12th grade from eastern country called Lebanon...I know my question may be unprecise but...when we see an enauality with a polynomial from level 2 is greater than 0...um...the domaines that you mentioned which are a>o or 0<a<1 or a=1...when we use this manner exactly which we but cases and write these certain domains...uh..I dunno

[sanelezuma]

Hey Eddie, love your teaching strategies. I have included them in my teaching and gotten some positive results from my learners. I truly thank the Lord Christ for you.

Side question: what program are you writing on? Recently brought a writing tablet but haven't found a good program to write on.

[adeebaaqil]

Can you make a video on your present setup of video recording, the tablet you using for writing and the app you are writing on. ??

[JohnSmith-rftx]

During the examination of cases in the secondary proof, I think it was a bit handwave-y to just state as obvious (i.e. without showing or proving) that (a-1/a) was negative for 0<a<1. IMO, the right way to do this would have been to multiply the expression by a/a and rewrite it as (a^2 - 1)/a. Factorizing the top via difference of squares: [(a-1)*(a+1)]/a. Since a>0, the denominator is always positive and therefore irrelevant for determining the sign of entire expression. And now the numerator very clearly shows it is going to be negative when a<1 as negative*positive=negative.

[duckymomo]

Is this high school math in AUS?

In the US we can’t do this because it’s too difficult

[forthrightgambitia]

You should have used 'lemma' instead of secondary proof to sound more distinguished. ;)

[fareedabifarraj]

@Eddie Woo
What kind of drawing pad are you using?
I would love to buy a similar one for my online teaching
And can you please tell me what whiteboard application are you using?

[Shadabacademy]

Sir please tell the whiteboard application you are using.