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 Sep 10 awarded Yearling Jul 19 answered Symbol for set of strictly positive real numbers? May 7 answered Is an $R$-module $A$ a module over the image of a homomorphism $f:R\rightarrow{f(R)}? May 7 awarded Informed May 7 comment Prove a specific basis exists satisfying certain conditions with an endomorphism For intuition's sake, have you tried writing$T$as a matrix with respect to$\{e_1, \dots, e_n\}$? Can you compute the eigenvalues of$T$? May 7 comment What is the Dedekind cut for 2? You're right,$\beta$does have the property that there is no largest element. May 2 comment Euclid's proof for the existence of infinitely many prime$p_i$divides both$q$(by definition of$p_i$) and$P$(because we have assumed that$p_i$is one of the factors making up$P$), so it divides$q-P$, which is equal to$1$. But there isn't a prime number that divides$1$. This contradicts our assumption (that$p_i$is a factor of$P$), so the assumption must be false. So$p_i$must be a new prime number we didn't already have in our list. May 2 comment Split extension for semi-direct and direct products. Can a split extension be exact? I don't know what you mean by either "equivalent" or "isomorphic" here, because elements can't be those things. You are right that they have to be equal. We have a composite map$H\to N\times H \to H$which sends$h\mapsto (1,h) \mapsto h$, and$h$is definitely very much equal to$h$. May 2 comment Split extension for semi-direct and direct products. Can a split extension be exact? No, that is correct. The map$N\times H\to H$is the obvious one,$(n,h)\mapsto h$, and as a 'splitting' map, we may take$H\to N\times H$to be$h\mapsto (1_N,h)$. May 2 comment Split extension for semi-direct and direct products. Can a split extension be exact? The extension that we are testing for split-ness is$1\to K\to G\to H\to 1$, not$H\to G\to H$(which is not an extension anyway). May 2 answered Split extension for semi-direct and direct products. Can a split extension be exact? Apr 14 answered Prove that$\mathbb{Q}(r+s\sqrt{t})=\mathbb{Q}(\sqrt{t})$. Apr 14 comment Prove that$\mathbb{Q}(r+s\sqrt{t})=\mathbb{Q}(\sqrt{t})$. I see. That object that user222031 is defining isn't a priori a field (note the use of square brackets rather than round brackets - in general,$F(u)$and$F[u]$are different things), but it happens to be in this case. I'll write up an answer. Apr 14 comment Defining an operation on a quotient set Oh, strange. I don't know the book. In any case, if f isn't a homomorphism, there's no guarantee that Q (or Gf) is a group, so that must be what he means. Apr 14 comment Prove that$\mathbb{Q}(r+s\sqrt{t})=\mathbb{Q}(\sqrt{t})$. @Skull-Face You're going to have to tell us what definitions your class is working from. As far as I'm concerned, what user222031 gave is the definition of$\mathbb{Q}[x]$- but your class might be doing things in a different order. What do you understand$\mathbb{Q}(x)$to mean? Apr 14 comment Defining an operation on a quotient set No, it means that f is a group homomorphism. (That is, if * is the operation on G and # is the operation on G', we have f(g * h) = f(g)#f(h) for all g and h in G.) But group homomorphisms do have the property that images are still groups. Apr 14 comment Defining an operation on a quotient set 1. G' also has a binary operation. 2. Isn't Gf the image of G under the group map f? In which case, it's a subgroup of G'. Apr 14 comment Proving$(a_1,b_1)\times (a_2,b_2)\times\cdots\times(a_n,b_n)$is open in$\mathbb{R}^n$. Using the distance formula makes it harder than it needs to be. In particular,$y\in B(x;r)$implies that$|x_i-y_i| < r$for each$i$separately. This reduces it to the one-dimensional problem: does an open interval of radius$r$about$x_i$lie inside the interval$[a_i, b_i]$? (Answer: yes, by your choice of$r\$.) Apr 13 awarded Nice Answer Apr 11 answered At what points of r in real number is f continuous?