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bio website blog.plover.com
location Philadelphia, USA
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visits member for 2 years, 11 months
seen 7 hours ago

I am Mark Dominus, an amateur.

I have a blog that often carries articles about mathematics.

You can email me at mjd@plover.com.


9h
revised The Present Worth of $169 due in 2 years at 4% Per Annum Compound Interest
edited tags
12h
comment The drawer has M different colored socks. What is the least amount of socks I that I need to draw to guarntee N pairs
There are general formulas, and it is not too hard to find them. I suggest that instead of considering fifty pairs of socks, you consider two pairs. And instead of considering 50 colors, you consider 3 colors. You may get some ideas that will lead you to the solution of the general problem.
13h
comment give a group that is isomorphic to the figure.
Yes, that is exactly correct.
14h
comment A basic question of basic writing
I don't think it would be wise to use $\equiv$, which almost always means modular equivalence. Simple $=$ is correct, and Grigory Ilizirov gives several other options.
14h
comment A basic question of basic writing
\stackrel can be useful here. For example \stackrel\Delta= is $\stackrel\Delta=$.
14h
revised Definition of the sum of natural numbers
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14h
comment Definition of the sum of natural numbers
@Trevor Feel free to edit.
15h
revised Definition of the sum of natural numbers
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15h
answered Definition of the sum of natural numbers
18h
comment Give me an idea
Post the question in the original language, and let someone else translate it.
20h
comment Find orbit of $1$ for $\sigma$
@Travis: It's standard to talk about the orbit of some element of a set $X$ under $g$ where $g$ is an element of a group that acts on $X$. I checked Fraleigh A First Course in Abstract Algebra to confirm and found that this question is Exercise 8.10 from that book.
1d
revised Is the $\varphi \to \varphi$ axiom in Hilbert calculus redundant?
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1d
answered Is the $\varphi \to \varphi$ axiom in Hilbert calculus redundant?
1d
revised transformation of $y=3(4-x)^3-6$
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1d
revised How can one know every Cauchy sequence in a complete metric space converges?
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1d
comment Let $G$ be a group and $u \in G$ be a fixed element. By the following, prove that $(G,\bullet)$ is a group.
There is a general theorem that if $\bullet$ is an associative operation on a set $G$ for which $u$ is a left and right identity, and $b'$ is a left inverse for $b$, then $b'$ is also a right inverse for $b$. In general, any three of the four properties (left identity, right identity, left inverse, right inverse) imply the fourth, so that $\langle G, \bullet\rangle$ is a group, and it is a well-known fact that left inverses in groups are the same as right inverses.
1d
comment Prove $\overline{B} - (A-\overline{B}) \subseteq \overline{B}$
It couldn't hurt, but my sense is that OP doesn't need the tutorial as much as he needed to be told to stop putting $ $ between every two symbols.
1d
comment Prove $\overline{B} - (A-\overline{B}) \subseteq \overline{B}$
Even so, there is no reason to write x $\in$ $\overline{B}$ instead of simply $x\in\overline{B}$.
1d
comment Prove $\overline{B} - (A-\overline{B}) \subseteq \overline{B}$
You should not write x $\in$ $\overline{B}$ $\,$^$\,$x$\,$$\in$$\,$($\overline{A}$$\cup$$\overline{B}$), which comes out as «x $\in$ $\overline{B}$ $\,$^$\,$x$\,$$\in$$\,$($\overline{A}$$\cup$$\overline{B}$)». Instead, it looks better and is easier to write simply $x \in \overline{B}\land x\in(\overline{A}\cup \overline{B})$ which renders as «$x \in \overline{B}\land x\in(\overline{A}\cup \overline{B})$». Everyone wins.
1d
revised Prove $\overline{B} - (A-\overline{B}) \subseteq \overline{B}$
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