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.

"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

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!

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

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!

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

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

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

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

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.

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

You are a really great teacher

Ty for this video

Thank you... Very helpful

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?

you are the best

谢谢，我明白了

Thank tou Ma'am 😍

WITCH CRAFT! I love this so much

Thanks!

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

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!

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

thank u so much ☺️

i love you

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?

I never met God until I found this video

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