Discrete Math - 5.1.2 Proof Using Mathematical Induction

Discrete Math - 5.1.2 Proof Using Mathematical Induction - Inequalities

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@tailypo7929 Жыл бұрын

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.

@SawFinMath Жыл бұрын

Haha. One of my children!

@justinliebenberg2321 Жыл бұрын

"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

@TheAngryMaskSalesman64 Ай бұрын

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.

@bryncurry3562 2 жыл бұрын

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!

@Sam-by3uk 2 жыл бұрын

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

@ptree1694 4 жыл бұрын

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!

@AsatsuyaH 3 жыл бұрын

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

@williamlunsford5966 3 жыл бұрын

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.

@espon360 Жыл бұрын

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 7 ай бұрын

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

@danlupu5943 Жыл бұрын

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!

@joseblua6566 3 жыл бұрын

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

@fsxaviator 3 жыл бұрын

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

@TheBanana821 7 ай бұрын

Exactly

@mmuopeter7214 2 жыл бұрын

You are a really great teacher

@Shahroz_A1 4 ай бұрын

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

@ashrafulislam7825 3 жыл бұрын

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?

@argav625 Жыл бұрын

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?