Billy
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 May2 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. May2 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 $h$. May2 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 $h\mapsto (1_N,h)$. May2 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). May2 answered Split extension for semi-direct and direct products. Can a split extension be exact? Apr14 answered Prove that $\mathbb{Q}(r+s\sqrt{t})=\mathbb{Q}(\sqrt{t})$. Apr14 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. Apr14 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. Apr14 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? Apr14 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. Apr14 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'. Apr14 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 $r$.) Apr13 awarded Nice Answer Apr11 answered At what points of r in real number is f continuous? Apr11 reviewed Approve Right and left hand limits. Apr8 comment What does this notation mean? $x \mapsto f(x)$ @YoTengoUnLCD: if you consider it tautologous, fair enough. (But there are people who would write their functions as $x\mapsto xf$ or $x\mapsto x^f$ or similar. Not stating this can lead to genuine confusion - does $fg$ mean "do $f$ then $g$" or "do $g$ then $f$"? Writing explicitly that you will denote the image of $x$ under $f$ by $f(x)$ avoids this potential confusion. Also, for instance, with more arguments, you might write $g(x,y)$, but would prefer to write $x*y$ rather than $*(x,y)$; writing this out explicitly avoids confusing your reader.) Apr8 comment Find all Gaussian integers $α, β, γ$ such that $αβγ = α + β + γ = 1$ "If any of α,β,γ are equal from that list they cannot add to 1" - apart from the case (1,1,-1) which is easily checked to fail. Mar27 reviewed Approve Let f(x),g(x) be complex polynomials, if f(x) | g(x) and g(x) | f(x) then.. Mar27 comment Does a terminating recurrence relation diverge? As you've noticed, the expression $u_{k+1} = \frac{4}{u_k + 2}$ doesn't make sense when $k = 6$ (so there should really be a caveat in the definition). What is your definition of 'sequence', and does $\{u_k\}$ fit it? If this isn't even a sequence (in this case, because it fails to be infinite), it doesn't make sense to ask whether or not it converges. Mar27 comment Can I make an assumption about arbitrary numbers in a proof? No, you may not assume anything that you can't prove (or your lecturer / teacher hasn't proved). Can you prove that $Q$ exists? (Hint: try to find $P_1$ with $P_1(x_1) = y_1$ and $P_2$ with $P_2(x_2) = y_2$, and combine them somehow to find $P$ with $P(x_1) = y_1$ and $P(x_2) = y_2$.)