Zev Chonoles
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 1h revised Can someone help me why this equation equals zero? deleted 51 characters in body; edited tags 1h answered Partition on a Closed Set A= [2,3] 1h answered Subgroups and subsets 1h comment A Sequence That has No Upper Bound But Does Not Tend To Infinity "Converges" is not the same as "it is false that $a_n\to\infty$". A divergent sequence could also diverge to (hint, hint...) 3h answered Beginner level : What is the intuitive meaning and what are the steps to prove for an injective function 1d comment How can the Hausdorff measure be nonzero? For some sets $X\subset\mathbb{R}$, it is the case that $\inf(X)\notin X$... 1d comment How can we tell from looking at a problem that multiplication principle fails to solve it? And why does MP fail(?) in the first place? If you choose the secretary first, you have 3 options: Ann, Bob, Dan, so I already have no idea where you're getting a 2 from. Moreover, note that if you choose Bob for secretary then you have 3 options for treasurer whereas if you choose Ann or Dan for secretary then you have 2 options for treasurer (since it cannot be Bob) - that's the non-interchangeability in action. 1d comment How can we tell from looking at a problem that multiplication principle fails to solve it? And why does MP fail(?) in the first place? In fact, the possibilities are 1[A] 2[B] (only one option remains if I put A in 1) or 1[B] 2[A] (only one option remains if I put B in 1) or 1[C] 2[A] or 1[C] 2[B] (two options remain if I put C in 1). 1d comment How can we tell from looking at a problem that multiplication principle fails to solve it? And why does MP fail(?) in the first place? The point of my answer is that you can't "solve by multiplication". Here's a simpler case: I have two boxes labeled 1 and 2. I have three letters, A, B, and C, and I want to put one letter in each box. If I make no restrictions on what can go where, then my options are (A, B, or C for the first box) $\times$ (whichever two are left for the second box) = $3\times 2$ = $6$. However, if I declare that C cannot go in box 2, then it is incorrect to say "whichever two are left" because sometimes there aren't two left that can go in box two. 1d comment How can we tell from looking at a problem that multiplication principle fails to solve it? And why does MP fail(?) in the first place? The manner in which you phrase the non-interchangeability (e.g., "alien overlords have decreed that Bob cannot be a secretary") is irrelevant to the mathematics. If not all objects have the same options available to them, you then have to keep track of which objects goes in what slot. 1d answered How can we tell from looking at a problem that multiplication principle fails to solve it? And why does MP fail(?) in the first place? 1d comment integral domains and field of fractions You could also take a look at Wikipedia, where it explicitly says The field of fractions of $R$ is characterised by the following universal property: if $h:R\rightarrow F$ is an injective ring homomorphism from $R$ into a field $F$, then there exists a unique ring homomorphism $g:\mathrm{Quot}(R)\rightarrow F$ which extends $h$. 1d comment integral domains and field of fractions It's not clear to me what you're confused about: what you just wrote down is a universal property. Apr23 comment What is vectors straddle a plane mean? Also, my initial guess would be that "straddle" means "linearly span". Apr23 comment What is vectors straddle a plane mean? More context would help - how about copying for us the entire paragraph where this phrase occurs? Apr23 comment Polynomial: Is there a theorem that can save my proof when $K$ doesn't include $\mathbb C$ It isn't quite correct to say that any number field is a literal subset of $\mathbb{C}$. The correct statement is that any number field can be embedded in $\mathbb{C}$. Apr23 comment Polynomial: Is there a theorem that can save my proof when $K$ doesn't include $\mathbb C$ If you're uncomfortable with your argument so far, just recall that any number field of degree $n$ has $n$ different embeddings into $\mathbb{C}$, choose one of them to "move" the problem over to $\mathbb{C}$ and then use your (correct) argument. Apr23 awarded Nice Answer Apr23 answered Trig and derivatives: If condition holds for derivative, does it hold for the original equation? Apr23 answered Are the Eisenstein integers the ring of integers of some algebraic number field? Can this be generalised?