Discrete Math II - 5.1.1 Proof by Mathematical Induction

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  • เผยแพร่เมื่อ 25 ม.ค. 2025

ความคิดเห็น • 13

  • @JosephCote-y3j
    @JosephCote-y3j 9 หลายเดือนก่อน +5

    You should take the sum of I.H (k^2) and substitute that in for 1 + 3 + 5 + … (2k - 1) since we are assuming they equal each other. So you can now do, k^2 + 2k + 1 = (k + 1)^2.

  • @artemkrazhan8603
    @artemkrazhan8603 3 หลายเดือนก่อน

    ur the goat prof kimberly thank you so much for this video

  • @Userpng77y
    @Userpng77y ปีที่แล้ว +4

    Thanks for the course, but this should be seen according to the roden of the youtube playlist or according to the number of the videos "5.1.1, 5.1.2......"?

  • @XxNGameCubexX
    @XxNGameCubexX 2 หลายเดือนก่อน

    Slight mistake on the well-ordering principle slide, I think the professor meant to say S is a subset of Z+, Q+, or R+, respectively, not element of.
    Thanks for your lectures!

  • @karqoa3968
    @karqoa3968 2 ปีที่แล้ว +6

    thank you very very much for these videos

  • @AdrianMeyer8
    @AdrianMeyer8 หลายเดือนก่อน

    Very appreciative for making the making these courses available, Kimerbly! But - however I am a bit confused, how does 1 + 3 + 5 + ... + (2k - 1) + (2k +1) = (k + 1)^2? I understand if you would swap "1 + 3 + 5 + ... + (2k - 1)" with the "I.H" witch states "k^2". But teacher in the video doesn't explain this. It would be a lot more time saving if she would adress that when writing 1 + 3 + 5 + ... + (2k - 1) is just the I.H and the I.H is k^2 and next step is to add (2k+1).

    • @SawFinMath
      @SawFinMath  หลายเดือนก่อน +1

      I'll take another look at that video. Thanks!

  • @sedibe
    @sedibe ปีที่แล้ว +3

    you r the best

  • @caitbenn365
    @caitbenn365 ปีที่แล้ว +1

    Great video. Why did you add 2k+1 to both sides?

    • @SawFinMath
      @SawFinMath  ปีที่แล้ว +4

      That is the definition of an odd integer

  • @feltonkaonga
    @feltonkaonga 2 ปีที่แล้ว +1

    ❤👍

  • @kario360
    @kario360 ปีที่แล้ว

    very complex video