Discrete Math - 5.1.2 Proof Using Mathematical Induction - Inequalities

More practice on proof using mathematical induction. These proofs all prove inequalities, which are a special type of proof where substitution rules are different than those in equations.

Video Chapters:
Introduction 0:00
Proving our First Inequality 0:09
Proving our Second Inequality 4:47
Up Next 9:44

Textbook: Rosen, Discrete Mathematics and Its Applications, 7e

[tailypo]

I love that at 3:15, once you start explaining "The part you are going to hate with a fiery passion, " a small child starts aimlessly howling in the background. You and I are on the same page, little buddy.

[TheAngryMaskSalesman]

You are such a great teacher! for so many reasons but one of them is because you humanize this math. You make this math seem less cold and less sterile, into something more natural.

[justinliebenberg]

"You're either going to love or hate with a fiery passion"..."Most people hate it with a fiery passion".... The few that do love it are psychopaths

[bryncurry]

I had my teacher, TA, and textbook all try to explain this, and your video made it click in just a few minutes. Thank you so much!

[AsatsuyaH]

Thank you!!!! This is the only video that actual helped me understand how 2k+1 can become 2k + 2k and so on.

[Sam-byuk]

Thank you for explaining this properly. The other inequality proofs I found on YouTube were bugging me, because it felt like they were skipping steps or cheating. Now it makes sense thanks to you!

[ptree]

Thank you so much for the in depth explanation on inequalities and how you converted 1 to 2^k to 2^(k+1). Your videos are helping a lot!

[espon]

Thank you so much for this video, this class is the only one I've been struggling with and I feel like I'm finally understanding it through your videos

[refathasanabir]

thank you very much for sharing your knowledge.May almighty bless you.

[joseblua]

Great video, thank you very much for uploading videos like these! You're a really skilled teacher

[williamlunsford]

3:34-4:46 is her employing logic from the solution established at the basis step: 1 < 2^1. This solution is then modified to 1 < 2^k on the inductive step, where the inequality is then used to say that if 1 is less than 2^k, then 1 can be replaced with 2^k in an inequality where 1 is greater than some value: it implies that 2^k is also greater than that value, and so the inequality holds.

[melvinkwakuyawlui]

I didn't see that as cheating for real. That's math genius 😂🔥

[danlupu]

For the second proof, could you not just substitute k! or 2^k like so

2^(k+1) < (k+1)!
2*2^k < k! * (k+1) (using consecutive neighbours)

sub in k! for 2^k or 2^k for k!, both will work, I choose 2^k in this case (remember we assume 2^k < k!

2*(k!) < k! * (k+1)

cancel out k!

2 < (k+1)

Therefore since we stated the predicate function was true for all k part of the set of positive integers and k equal to or more than 4, k+1 will always be more than 2 as the minimum k+1 can be is 5

Would this be wrong and if so could anyone point out why, maybe I am using circular reasoning or something, but I don't think I am since I am not assuming 2^(k+1) < (k+1)! is true, i am proving it is.

If you could please explain whether this would be right that would be amazing, thank you for the video by the way!

[fsxaviator]

So essentially we're showing that P(k) = P(k+1) right?

[Shahroz_A]

Mam in my book they havent gave any of the value for the second example so what should i suppose to do?

[argav]

8:20 I wonder why would you come up with the "k+1>2" ? I mean, yes I understand you want to get to the "(k+1)!", but why it must be "k+1" to be greater than '2'? Is "(k+1)>2" just the assumption or something?

[ashrafulislam]

Thank you so much for this!
However, I had a question. At 8:50, if we write 2^k+1 < 2k! < (k+1)k! then 2^k+1 < (k +1)k! this is also okay, right?

[maximocarrillo]

I never met God until I found this video

[briandorant]

you lost me at 3:34, would help if you be clearer instead of saying 'that' or 'this'