Hi may i ask when you are checking for whether z%2 is a field, what happens when you add two expressions that each come to 1. eg 3mod2+4mod2=1+1 = 2 or 10?
In abstract algebra ops could be anything so say if we take natural numbers n subtraction op, it won't be closed under 2-5=-3 so it won't even be a group?
you proved integers mod 2({0,1}) to be a finite field but it does not satisfy additive inverse property of field? ..the additive inverse of 1 is -1 which is not present
Very nice overview professor, thank you. Some remarks: (1) in the field example (ua-cam.com/video/TPW4_Z5kiRw/v-deo.html) you mention that it is commutative. How is that relevant, provided that on ua-cam.com/video/TPW4_Z5kiRw/v-deo.html you say a ring does not need to be commutative, and a later that a field is an extension of a ring and you do not mention commutativity at ua-cam.com/video/TPW4_Z5kiRw/v-deo.html? (2) minor but at ua-cam.com/video/TPW4_Z5kiRw/v-deo.html a typo: 'P (uppercase) ... where p (lowercase) is'.
Thank you it was very clear
This video should appear first in youtube , when i search "Group , Ring and Field".... Thank You sir...
Very clear ,crisp in a flow explanation for groups, fields and rings. To good!
such a great explanation 💚😊
thank you , it was very clear
Thanks for sharing, it helpful
Happy to know it helped. Thanks
Hi may i ask when you are checking for whether z%2 is a field, what happens when you add two expressions that each come to 1. eg 3mod2+4mod2=1+1 = 2 or 10?
Thank you for the explanation
5 mod 2 = 1, mentioned wrong while adding :)
Thanks for pointing the error and for posting the comment 👍
wonderful explanation .. May I have your PowerPoint that you explain on it?
thank you sir
5 mod 2 + 3 mod 2 = 2 . Since 2 is not part of { 0 , 1 } ......How can we take it as a field?
And basic law additive inverse of 1 not exist so not even a group how can it be field or ring...
In abstract algebra ops could be anything so say if we take natural numbers n subtraction op, it won't be closed under 2-5=-3 so it won't even be a group?
Yes, even with addition it is not a group, because every element needs an inverse
you proved integers mod 2({0,1}) to be a finite field but it does not satisfy additive inverse property of field? ..the additive inverse of 1 is -1 which is not present
Very nice overview professor, thank you. Some remarks: (1) in the field example (ua-cam.com/video/TPW4_Z5kiRw/v-deo.html) you mention that it is commutative. How is that relevant, provided that on ua-cam.com/video/TPW4_Z5kiRw/v-deo.html you say a ring does not need to be commutative, and a later that a field is an extension of a ring and you do not mention commutativity at ua-cam.com/video/TPW4_Z5kiRw/v-deo.html? (2) minor but at ua-cam.com/video/TPW4_Z5kiRw/v-deo.html a typo: 'P (uppercase) ... where p (lowercase) is'.
Sir Please Share notes of DES and AES algorithm.
at 24:19 you made a misake by making 5 mod 2=0