(x+1)^2 = x^2 + 1 - "you were right in Z2 but wrong in R" at 19:30 didn't seem appreciated by the students, but I appreciate it. I have math teaching experience so that was hilarious! Should've brought the house down.
Hello Sir ,At 12:45 you are saying this is because of this theorem that we have to take degree of zero polynomial as not defined.But ,this theorem,talks about only those g(x) which are non zero .So ,we will be simply neglecting the degree of zero polynomial while applying this theorem. How does the degree of zero polynomial affects this theorem in any way? thanks
@@billkinneymath Yes sir ,a few doubts from the solution of last question. 1)Why x cannot be in A ( at 23:54 ) 2) In equation at 25:00 ,why have we selected the term ax+ b ,means the mathematical logic behind it ? 3) At 25:44 x^2 +1= 0 ,I am not sure why it is so ? Sorry ,if these questions are trivial ,but I couldn't understand these things . Thanks for your lectures .
@@sherryj3046 The most important thing to realize is that f(x) + A = g(x) + A if and only if f(x) - g(x) is an element of A. Since A is the principal ideal generated by x^2 + 1, this is equivalent to saying that f(x) - g(x) is a polynomial multiple of x^2 + 1 (when you divide by x^2 +1 you'll get a remainder of zero). This answers your third question because x^2 + 1 is a multiple of itself. Next, note that elements of Z3[x]/A can be written in the form ax + b + A. The reason is that if you have an element, for example, like x^3 + x^2 + x + 2 + A, you can divide it by x^2 + 1 (mod 3) and get a remainder of the form ax + b so that x^3 + x^2 + x + 2 + A = ax + b + A (in this case, x^3 + x^2 + x + 2 + A = 1 + A because the remainder is 1 (try the long division...for any example like this, the remainder will always have degree less than 2)). This answers your second question. This can also answer your first question. For x to be in A, then x would have to be (x^2 + 1)*(ax + b) mod 3, for some a and b. But (x^2 + 1)*(ax + b) = ax^3 + bx^2 + ax + b. There is no choice of a and b that will make this equal x.
@@billkinneymath Thanks a lot sir .I am clear with my doubts now .I have no words to thank you for your efforts .It's selfless people like you who make this earth a better place .May God bless you.
(x+1)^2 = x^2 + 1 - "you were right in Z2 but wrong in R" at 19:30 didn't seem appreciated by the students, but I appreciate it. I have math teaching experience so that was hilarious! Should've brought the house down.
I think they smiled...it was a very early morning class (7:40 am - 8:50 am). :-)
Hello Sir ,At 12:45 you are saying this is because of this theorem that we have to take degree of zero polynomial as not defined.But ,this theorem,talks about only those g(x) which are non zero .So ,we will be simply neglecting the degree of zero polynomial while applying this theorem. How does the degree of zero polynomial affects this theorem in any way?
thanks
You are right. I think when I said this I didn't think about the fact that it already says g(x) is nonzero in the theorem statement.
@@billkinneymath thanks for your reply sir
Sorry to say sir ,but last few minutes of lecture were not so good. Apologize,but this is what i felt.
Was there something confusing that I can try to clear up?
@@billkinneymath Yes sir ,a few doubts from the solution of last question.
1)Why x cannot be in A ( at 23:54 )
2) In equation at 25:00 ,why have we selected the term ax+ b ,means the mathematical logic behind it ?
3) At 25:44 x^2 +1= 0 ,I am not sure why it is so ?
Sorry ,if these questions are trivial ,but I couldn't understand these things .
Thanks for your lectures .
@@sherryj3046 The most important thing to realize is that f(x) + A = g(x) + A if and only if f(x) - g(x) is an element of A. Since A is the principal ideal generated by x^2 + 1, this is equivalent to saying that f(x) - g(x) is a polynomial multiple of x^2 + 1 (when you divide by x^2 +1 you'll get a remainder of zero). This answers your third question because x^2 + 1 is a multiple of itself. Next, note that elements of Z3[x]/A can be written in the form ax + b + A. The reason is that if you have an element, for example, like x^3 + x^2 + x + 2 + A, you can divide it by x^2 + 1 (mod 3) and get a remainder of the form ax + b so that x^3 + x^2 + x + 2 + A = ax + b + A (in this case, x^3 + x^2 + x + 2 + A = 1 + A because the remainder is 1 (try the long division...for any example like this, the remainder will always have degree less than 2)). This answers your second question. This can also answer your first question. For x to be in A, then x would have to be (x^2 + 1)*(ax + b) mod 3, for some a and b. But (x^2 + 1)*(ax + b) = ax^3 + bx^2 + ax + b. There is no choice of a and b that will make this equal x.
@@billkinneymath Thanks a lot sir .I am clear with my doubts now .I have no words to thank you for your efforts .It's selfless people like you who make this earth a better place .May God bless you.