Video

Where is warm up questions ?

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Thank you!

‏‪20:27‬‏ i tried with myself and found that the first number that has two different partitions with parts 3(mod4) is 14 bc 14=11+3, 14=7+7

Hello, thank you very much for your course follows Book of Proof. This full video also contains an introductory part from 00:00 to 1:12. It would be great to add this introductory part to your playlist "Proof Writing" (czcams.com/play/PLVMgvCDIRy1x00m7Oo9XzEkDDACeEK_m-.html&si=XY1SA5MzKS58Zd3Q) based on this full video =)

Okay lectures

Just wanted to say thanks for making this playlist ❤😅

Thanks I try to learn by myself

learning for midsem exam from nit ap. love from india

Prof., what text are you making use of?

Call me a geek but I was smitten by the proposition that L/I is abelian iff <=>L' is in I. Lie algebras are fascinating.

‏‪36:40‬‏ why and how it possible to divide both sides by 4 ?

Where is warm up Questions ?

‏‪4:39‬‏ professor penn didn't proof the rule that we can use p/q instead of using q/p with a minus sign or not

The derivative function that maps differentiable functions to continuous functions is surjective because every continuous function has an antiderivative.

At 5:00 I think it should be x^{-1}

Very thorough lecture!

Thanks alot, this course has been super helpful, especially all the worked examples.

18:20 minor sign error, it should be -pi+epsilon and pi - epsilon

mom

suck

naga

Can you recommond a textbook for this course? Thank you.

Much easier than I thought. Random question, is gluing computational? Thanks.

At 8:28 we divide by r^b but why and how? When we can divide both sides of the modular and the statement still true?

At 3:30 we know r^phi(n) is congruent with 1(mod n) then r^phi(n) *y is congruent with y(mod n) not 1 (mod n) ?

Thank you so much good sir

‏‪19:18 phi(2p^k)=phi(p^k) is true for all prime numbers but it isn't true for p=2 because the multiplicative property that phi function has meaning: phi(mn)=phi(m)*phi(n) but the gcd (m,n)=1 if p=2 then gcd(2,2^k) doesn't equal to 1

‏‪14:19‬‏ why L+1 must be equal to K not K-1 as written in the equation d|p^K-1(p-1( ?

The introduction of IDa and IDb at about 0:58 is somewhat skimpy, and confused me further on. I had to rewatch that part multiple times and dredge up old memories to realise IDa and IDb are the identity functions! Given the formalism of like videos, I think concepts like IDa and IDb should be treated more formally.

answers to the warmup questions????

f(x)€ Z[x] means f is a polynomial function with integer coefficients and integer roots or with integer coefficients only !! see ‏‪8:18 and ‏‪ ‏‪0:31

Book of Proof by Richard Hammack.....and.....How to Prove it by Daniel Velleman. Fantasic!

Any idea about what to do with the third example in 1 and 2 at the end?

ty

At the end of the video there is a warm up question saying that: show that 2*(p-3)! Is congruent with -1 mod p? my solution is the following: We know: (p-1)! Is congruent with -1 mod p but (p-1)!=(p-1)(p-2)(p-3)! Is congruent with -1 mod p . Put p-1 is congruent with -1 mod p , p-2 is congruent with -2 mod p so (p-1)*(p-2) is congruent with 2 mod p therefore (p-1)(p-2)(p-3)! is congruent with 2*(p-3)! which is congruent with -1 mod p

Thank you Sir

in the definition itself why didn't you mention bi linear property of the lie bracket?

Gold

At 5:18, another way to find all generators of <r^15> <= D_21 is to use the applicaable theorem in the lecture. Since D_21 has order of 42, we know from the theorem that <r^15> = <r^gcd(15,42)> = <r^3>.

The transpose is most certainly *not* the inverse in SL2(Z), the adjugate matrix is the inverse (since the determinant is 1). I’m sure all can be proved without this mistake. Micheal man, you need to fix this. The course is incomplete

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Thank you!

Nice session sir

yemin ederim ki sabah akşam sadece ve sadece bunlara çalışacağım

Warm up problems sollotions ?

wow thanks you so much, it's definitely a very useful video