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This is a masterpiece.
Why did you use the division algorithm and assume that r = 0?
In the statement of the division algorithm, the condition for the existence of r is that 0
Beautiful lecture series, thank you!
Super helpful🙏
Very helpful
Seems like in your proof at 10:00 you need a strictly positive number. If you take 0, you get a trivial ideal instead of something useful.
I was wondering about that, thanks for clarifying that (as all ideals must contain the zero element).
if you keep on watching, Michael says exactly that around 14:00
Thank u
this was helpful thank you
I want to ask you, does x-c, c in C belongs to ideal generated by x^2+1 in C[x] that is I=? I think it doesn't belong to, because degree of x-c is 1
the ideal is made out of polynomials of the form (x^2+1)*p(x) for a generic polynomial p(x), so x-c is not in the ideal
Thank youHave you an email?
In the first proof you didn't show 0∈(a) and x+y∈(a) for all x,y∈(a). An ideal is a subgroup of R with respect to +!
This is a masterpiece.
Why did you use the division algorithm and assume that r = 0?
In the statement of the division algorithm, the condition for the existence of r is that 0
Beautiful lecture series, thank you!
Super helpful🙏
Very helpful
Seems like in your proof at 10:00 you need a strictly positive number. If you take 0, you get a trivial ideal instead of something useful.
I was wondering about that, thanks for clarifying that (as all ideals must contain the zero element).
if you keep on watching, Michael says exactly that around 14:00
Thank u
this was helpful thank you
I want to ask you, does x-c, c in C belongs to ideal generated by x^2+1 in C[x] that is I=? I think it doesn't belong to, because degree of x-c is 1
the ideal is made out of polynomials of the form (x^2+1)*p(x) for a generic polynomial p(x), so x-c is not in the ideal
Thank you
Have you an email?
In the first proof you didn't show 0∈(a) and x+y∈(a) for all x,y∈(a). An ideal is a subgroup of R with respect to +!