- #1
QuasarBoy543298
- 32
- 2
i saw a proof that said “in R/{0} , the set [-1,0) has an upper bound ,but no least upper bound. no such set exists in linearly ordered R” ,but i could not understand it.
The main reason for this is the presence of a zero element in R, which does not exist in R/{0}. Isomorphic structures must preserve the algebraic properties of their elements, but since R/{0} does not have a zero element, it cannot be isomorphic to R.
The zero element plays a crucial role in the algebraic structure of R, as it satisfies certain properties such as being an additive identity. The absence of a zero element in R/{0} means that the two structures cannot be mapped onto each other in a way that preserves their algebraic properties.
No, even if we remove the zero element from R, the structures would still not be isomorphic. This is because the presence or absence of a zero element fundamentally changes the algebraic properties of the structures, making them incompatible for isomorphism.
Apart from the absence of a zero element, there are other differences between R and R/{0} that make them non-isomorphic. For example, R is a field, while R/{0} is not, as it does not have multiplicative inverses for all elements (since the zero element is excluded).
Yes, understanding the reasons for the non-isomorphism between R and R/{0} helps us gain a deeper understanding of the algebraic structure of these two sets. It also highlights the importance of certain elements, such as the zero element, in determining the properties and relationships within a mathematical structure.