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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
8
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.)
Apr
8
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.
Mar
27
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.
Mar
27
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 $P(x_2) = y_2$.)
Mar
27
comment Let f(x),g(x) be complex polynomials, if f(x) | g(x) and g(x) | f(x) then..
@MattSamuel Depends on your definition of "divides", surely? In a commutative ring (here $\mathbb{C}[x]$) the natural definition seems to me to be "$g|h$ iff there exists $f\neq 0$ with $fg = h$". If you want to make an exception for $0$ somewhere, go ahead, but definitions vary from author to author, so a discussion of which is correct doesn't seem all that useful. I just flagged it up as a potential issue.
Jan
16
comment Why are primes considered to be the “building blocks” of the integers?
@zod Edited, thanks!
Jan
11
comment $X^2$irreducible but not prime
(Are $X$ and $x$ the same thing, and do you mean $X^i$ instead of $X_i$?) $X^2$ certainly looks prime here to me.
Jan
11
comment Group Duality with respect to Generators and Relations
(2/2) On the other hand, there certainly are some notions of duality that arose from mathematical questions, which you may be interested in looking up. A simple one would be the concept of a dual vector space (vector spaces are, of course, continuous groups too), which arises by considering functions on that space (or matrix transposition). A harder one: Pontryagin duality comes from trying to do Fourier analysis on certain groups, and crops up very naturally in a bunch of places, including e.g. the character theory of finite abelian groups.
Jan
11
comment Group Duality with respect to Generators and Relations
(1/2) I think my real point is: in order for such a definition to be mathematically meaningful, its conception needs to be mathematically well-motivated. You can't just say "I want duals to exist; where are they?". Better questions: what do you want $G^D$ to do? Or how do you want it to look relative to $G$? I see no mathematical reason to swap the generators and relations of a group (and no guarantee that it could be made to work). My geometric proposal was just something slightly more well-motivated (but, if you work through the details, you'll see it's also not very interesting).
Jan
10
comment Group Duality with respect to Generators and Relations
Suppose $G = \langle x | x^2 \rangle$. What does your proposed object $G^D$ mean? (But, if you want your 'dual' group to be the group of symmetries of the respective 'dual' polyhedron - which seems like quite a reasonable request to me - then the right definition is probably simpler than you think.)
Jan
10
comment How to prove that a set spans a plane
@Pow: aha, okay - take an element (x,y,z) on your plane. Supposing $c \neq 0$, we can more helpfully write this as (x,y,$\frac{-ax-by}{c}$). Now it suffices for you to find r and s (in terms of a, b, c, x and y) such that r(b,-a,0) + s(0,c,-b) = (x,y,$\frac{-ax-by}{c}$) - shouldn't be too hard. (Then of course you have to deal with the case $c = 0$...) :)
Jan
25
comment What's the difference between $\mathbb{R}^2$ and the complex plane?
(You might also find it helpful to note that the "natural" multiplication on $\mathbb{R}^2$, namely $(u,v)\cdot (x,y) = (ux, vy)$, does not agree with the natural multiplication on $\mathbb{C}$ for any choice of $g\in \mathrm{GL}_2(\mathbb{R})$.)
Jan
25
comment What's the difference between $\mathbb{R}^2$ and the complex plane?
@laovultai: Let $f$ be the map $\mathbb{R}^2\to \mathbb{C}$, $(a,b) \mapsto a+ib$, and let $g$ be any element of $\mathrm{GL}_2(\mathbb{R})$. Then $f\circ g$ is an isomorphism $\mathbb{R}^2\to \mathbb{C}$ as $\mathbb{R}$-vector spaces. Now simply "pull back" the multiplication from $\mathbb{C}$ to $\mathbb{R}^2$ along the map $f\circ g$ (e.g. when $g$ is the identity map, the multiplication inherited is $(u,v)\cdot (x,y)=(ux−vy,uy+vx)$), and you get an isomorphism of rings (or, equivalently, $\mathbb{C}$-vector spaces). Does that answer your question?
Jan
17
comment How should I understand $R[x]/(f)$ for a ring $R$?
f should be irreducible, otherwise it is not true that R[alpha] = R[x]/f.