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15m
comment K is the least positive integer divisible by all positive integers less than or equal to 10. Find the total number of factors of K?
+1: Looks good!
3h
comment K is the least positive integer divisible by all positive integers less than or equal to 10. Find the total number of factors of K?
It may be an explicit part of the text's definition of "factor" (though that seems like a bad idea to me).
3h
comment K is the least positive integer divisible by all positive integers less than or equal to 10. Find the total number of factors of K?
I suspect they may be looking for proper factors only (excluding $1$ and $K$), in which case subtracting $2$ from your answer gives the desired one.
23h
comment What is the fastest, most correct way to solve this simultaneous of two linears?
Clear the denominators (don't forget to distribute), gather like terms, then solve for one variable, and then the other. Finally, check your solution(s) in both equations.
1d
comment What is $0\div0\cdot0$?
@johannesvalks: Rather, $0^{-1}$ has meaning (though not much) if and only if we're talking about the trivial ring, which has exactly one member, serving as both additive and multiplicative identity (i.e.: $0=1$).
2d
revised Counting combinations with a restriction of the form “either … or …, but not both”
typo fix
2d
comment Max Euclidean Distance between two points in a set
@Jeppe: You're quite right!
Jul
29
comment Find the number of flags of different types using induction
As for your answer, you're off to a good start!
Jul
29
comment Find the number of flags of different types using induction
That violates the conditions. The height of the pole must equal the sum of the heights of the displayed flags.
Jul
29
comment Find the number of flags of different types using induction
First of all, $f_1$ is the number of flag arrangements available for a $1$-foot flagpole. Second of all, you can't say that $$f_n=f_{n-1}+2f_{n-2}$$ when $n=1,$ since $1\not\geq 3.$ Third of all, you're miscounting $f_3$, because (for example) $a red flag over a blue flag is a different arrangement than a blue flag over a red flag.
Jul
29
comment Prove that if f is a continuous strictly monotone function defined on an interval, then its inverse is also a continuous function.
You are exactly right about the reason for strict monotonicity.
Jul
28
answered What is meaning of this question and how to solve it?
Jul
26
comment Can one construct a “Cayley diagram” that lacks only an inverse?
I just updated my answer. I'm sure you no longer need the help, but I've recently better acquainted myself with (how to use TeX to make) Cayley-like diagrams. I suspect you're even better at it than I am, at this point, so if you have any suggestions for how to clean up the formatting, let me know. (I also included a reference to the portion of Carter's text that explains determinism, for future users' reference.)
Jul
26
revised Can one construct a “Cayley diagram” that lacks only an inverse?
Improved formatting, included textual references.
Jul
26
comment How to solve “$4\sqrt5$ is the same as which square root?”?
+1: This is the best approach, but (unfortunately, the most opaquely put. @BigSwede: The idea is to show that $$\left(4\sqrt5\right)^2=4^2\sqrt5^2=16\cdot5=80.$$ Hence, since $4\sqrt5$ is positive (why?), and since $4\sqrt5$ is a square root of $80$ by definition, then $4\sqrt5=\sqrt{80}.$
Jul
24
comment Can I choose the larger of 2 arbitrary numbers with P>0.5 after only seeing 1?
@z5h: But once you choose your distribution to pick $G$, the probability isn't undefined at all. Using Hagen's example of the standard normal distribution, for any given real $A,B$ with $A\lt B,$ there is a positive probability that your $G$ lies strictly between $A$ and $B$. You don't know what the probability is, but you do know that it's positive.
Jul
22
comment Ordering $2n$ numbers
@Turbo: True, but David still raises a point worth clarifying.
Jul
22
comment Ordering $2n$ numbers
What do you mean by "group" and "ordering" in your last sentence?
Jul
22
comment Why do we show that structures aren't isomorphic by exhibiting a property not shared by one of them?
+1: And if you did want to seek a contradiction, you could show that $$\text{[isomorphism]}\implies\text{[property]}$$ and that $$\text{[property]}\implies\text{[contradiction]},$$ so that, indeed, $$\text{[isomorphism]}\implies\text{[contradiction]}.$$ For example, we suppose that $f:\Bbb N\to\Bbb Z$ is an order isomorphism. Thus, $f(0)$ is the least element of $\Bbb Z.$ But $f(0)-1$ is then not less than $f(0),$ whence $0\le-1,$ which is absurd.
Jul
22
comment Suppose every convergent Sequence has a unique limit point in space $X$, then $X$ is Hausdorff
@Struggler: Don't try to prove that it's Hausdorff. It isn't Hausdorff.