Proof by Induction | Explanation + 3 Examples

In this video, I explain the proof by induction method and show 3 examples of induction proofs! :D
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[@financialservicesprofessio7377]

I have watched hundreds of KZbin videos on this topic and I have never seen anyone made it so easy, until now. You are D'Man!!!

[@mahfoozali1162]

Trust me guys, he has the best way of presenting the topic!. I've been searching for discrete math vids, and finally found the best one. keep doing what u do✌

[@dylanrobichaud3676]

i’m a junior in college and you covered this material significantly better and more efficiently than my professor. you have a great way of explaining. keep it up man💪

[@saturnrobinson5057]

Thank you!!! Taking a course on linear algebra and one on number theory without taking an intro proof class so we're just kinda vibing over here!

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[@cohenharley1462]

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[@kasenwade5386]

@Cohen Harley happy to help :)

[@MichiDavis09]

Where did the additional 2 come from in the numerator?

[@SomenoobIoI]

Great video, the third example helped me quite a bit!

[@zanekovac726]

Thank you, this video made things so clear, the explanation is nice and detailed.

[@arjay_2002]

is the basis step the base case?

[@rupanbaro5849]

Nice presentation for beginners in

[@randydollente9651]

Thank you; this helped me so much.

[@financialservicesprofessio7377]

Can you do one or a few on Inequalities?

[@evank0385]

12:40 why do you ignore the far left side of the equation? you skip the 1+5+9 part and just say that 4n-3=2n^2-1 for n=1 but don’t include that extra 15?

[@deepstudios3832]

at 13:56, we we are writing what we want to show, why is the 4k - 3 present in addition to plugging k + 1 into k. if we supposed that n = 1, i understand putting 4k + 3, but if we Want to show P(k + 1), wouldnt we just plug in k + 1 wherever n is present (like we did for the right side)?

[@BrainGainzOfficial]

Good question. Yes, you can just replace k with k+1 and simplify (I also did this). Writing 1+5+...+(4k+1) is the same as writing 1+5+...+(4k-3)+(4k+1). I just made the term before 4k+1 more explicit because I think it helps us see how we can show what we want to show (by adding (4k+1) to both sides). If you can do without it then it's still perfectly fine.

[@k.8597]

I used (7^k = 3 + 4^k) so you end up with 7(7^k)-4(4^k) = 7(3+4^k) - 4(4^k) = 21 + 7(4^k) - 4(4^k) = 21 + 3(4^k) = 3(7+4^k) and since k is a positive integer and the integers are closed under addition and multiplication, 3(7+4^k) = 3z where z is an integer. Bit different but is it still valid?

[@gerardsagliocca6292]

I think you should organize your videos by similar topics

[@airylharridzuan6377]

At 24:12 how did you came up with 3 . 7^k+ 4 . 7^k + 4 . 4^k ? Btw thank you so much for your videos! Helped me a lot

[@glansingColt]

Can you make a vid on Strong induction?

[@quelloroberts6310]

im mirin the brain gainz brah

[@kamogeloxavier8581]

you're a GOAT

[@AkhonaAlbertIsaacMali]

Thanks a lot for this. I just want to ask how you got to 23:56. I'm a bit lost at that step. But I understand how you got from there onward.

[@AkhonaAlbertIsaacMali]

Oops. Never mind. I understood just as I typed this comment out. You the man! Awesome videos

[@chrisrodriguez6357]

Can you be my professor?

[@gerardsagliocca6292]

Should have gotten more board space. It looks crammed.

[@AmineDahmani-v6c]

Mcil bba❤

[@glansingColt]

am paying million dollars for these in college. shame on me

[@hamzarehan1838]

At 18:15 how do you get 2k^2 + 4k + 2 - k -1 from 2k^2-k+(4k+1)? This doesn't make any sense for me.

[@handsanitizeroos5235]

Maybe someone smarter than me can figure it out. But, to me, it looks like he made a mistake that he mentioned earlier in the video. He manipulated the conclusion and accidentally used the second expanded form from what he wanted and put it back into the proof to get the expression you mentioned. I'm not even sure the end result of the proof even factors nicely, I got 2k^2+3k+1 for the induction step and the expanded form of the right hand side completely and stopped there without factoring back into 2(k+1)^2-(k+1).