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19h
answered Does the => truth table break mathematical induction?
23h
comment Is every axiom in the definition of a vector space necessary?
Ah, right, I didn't think of that.
1d
comment Is ordering of (possibly infinite) sets by cardinality a total ordering?
Ordering sets by cardinality isn't a total ordering. Take $A=\{1\}$ and $B=\{2\}$. Then neither $\left|A\right|<\left|B\right|$ nor $\left|B\right|<\left|A\right|$. What you probably meant is that the cardinalities themselves are totally ordered.
1d
answered Showing a function $f$ cannot be surjective
1d
revised Showing a function $f$ cannot be surjective
TeX formatting, fixed typos
1d
comment Is every axiom in the definition of a vector space necessary?
See also this question
1d
comment Is every axiom in the definition of a vector space necessary?
@DanielV: $1\ne 0$ is a field axiom; the field axioms are stated to hold by saying "a field $K$".
1d
comment Is every axiom in the definition of a vector space necessary?
Actually you can formulate Axiom 3* without naming $0'$: (Axiom 3*) $0_K\cdot x=0_K\cdot y$ for all $x,y\in V$.
Aug
27
reviewed Approve Equilateral triangle
Aug
27
comment Equilateral triangle
Hint: Use mirror symmetry to find the $y$ coordinate, and then Pythagoras for the $x$ coordinate.
Aug
26
comment solve for $\int_{0}^{{\alpha}{b}}(a^x-1)dx=\int_{{\alpha}{b}}^{b}(a^x-1)dx$
For $\alpha=1/2$, also $a=0$ is an obvious solution.
Aug
26
comment Is it true that $A \in A$?
No, with my definition, the sequence you give does not converge at all. It cannot converge to $\emptyset$ because $0\notin\emptyset$ but there is no $N$ so that for all $n>N$ we have $0\notin A_n$. Note that there's an "iff", not merely an "if".
Aug
26
comment Is it true that $A \in A$?
Note that for the union definition, both formulas fail, too (indeed, here the precondition is no longer fulfilled for $\{A_i\}$ resp. $X-A_i$). Anyway, why is (for the definition I gave) $X-\lim A_i \ne \lim(X-A_i)$?
Aug
26
comment Is it true that $A \in A$?
@Ariel: Sets without regularity would of course not be ZFC sets.
Aug
26
comment Is it true that $A \in A$?
I think there is an obvious natural definition of set limit: $\lim_{n\to\infty}A_n = A$ if for every $x\in\bigcup_{n\in\mathbb N}A_n$ there is a natural number $N$ so that for any $n>N$ we have $x\in A_n$ iff $x\in A$. However with that definition, for the sequence in question the limit is the empty set.
Aug
26
comment If $b \equiv 0 \pmod a$ and $c \equiv 0 \pmod b$, then $c \equiv 0 \pmod a$
Bad form $\ne$ incorrect proof. I'm all for discouraging bad form by calling it bad form. I'm all against discouraging bad form by incorrectly claiming it to be incorrect.
Aug
26
comment Is 4 the second or third digit of pi
Another fun fact: If you express the largest distance we can observe (46 million light years) in terms of the planck length (the smallest length that actually makes sense physically), you get that it is about $2.7\cdot 10^{58}$ planck lengths. Thus with $\pi$ to about 60 digits, you should be able to calculate the circumference of every circle in the universe to the accuracy permitted by physics (of course the objection by @MarcvanLeeuwen about GR still applies; even more so, since at that precision, every local object will change the geometry).
Aug
26
comment If $b \equiv 0 \pmod a$ and $c \equiv 0 \pmod b$, then $c \equiv 0 \pmod a$
He writes "$c=a\cdot(r\cdot s)$ so $c=ak$. That makes it absolutely clear that $k=r\cdot s$.
Aug
26
comment If $b \equiv 0 \pmod a$ and $c \equiv 0 \pmod b$, then $c \equiv 0 \pmod a$
He rewrites the statement "$a\mid c$" he wants to prove into "$c=ak$ for some $k$". That's nothing but another way to state the same. The statement "$c=ak$ for some $k$" is the statement he's going to proof. Note that he writes "Now I start my proof" below that line, so that line is not even part of the proof, so how can it make the proof wrong?
Aug
26
comment If $b \equiv 0 \pmod a$ and $c \equiv 0 \pmod b$, then $c \equiv 0 \pmod a$
He did define what $k$ is. It is the integer for which $c=ak$. That comes directly from the definition of $a\mid c$.