Life In 19x19

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They're more studious and respect the idea there could be a fast lane to mastery, which may not be to their spontaneous liking. Whether their genes or culture is the primary cause for this, I cannot tell.
I tried my best to just, go to sleep, but this sentence somehow hit my mind so hard that ...

Japan is sure part of the (far) east. It is located in the east of Eurasia; the American continent was unknown in the mainstream history for a very long time.

The original question was, how the knight's move is called in the 'east', not in Japan.

Your answer 'It is called keima (in the east)' was ...

Unusedname wrote:http://senseis.xmp.net/?Keima

It's called a keima. I prefer knights move.

Japan is not 'East', neither 'East Asia'

Just curious, do they call it a knight's move in the east? I doubt it seeing as it refers to chess and so it seems like an American go term.
In Korea, it is called 날일자행마(Narilja-haengma) or just 날일자(Narilja). It means to move as the Chinese charactor 日 (see the squares as on the go board). In the ...

schawipp wrote:Just a question on the syntax - does _nC_k stand for the number of combinations, where k elements are selected out of n (i.e. n!/(k!(n-k)!)?

Sure, as you can see in the code.

I'm quite sorry to keep using this forum this way.. but I just feel so uncomfortable unanswered, so.

6236754_1402771389.jpg
This is everything.

Mathics, a freeware alternative to Mathematica, claims the above true.

(Input)
C := (#1)!/((#2)!*(#1-#2)!) &
Sum[C[n, k], {k, 0, s-1 ...

How about this. Let me make a suggestion.

1. Not every single file in your computer needs to be protected or be unseen by others.
2. Suppose there is an important text file 'haha.txt' which has your bank card PIN number and email password written down.
3. You don't want others to be able to access ...

Yes I am trying to clearly define a set of a five-dimensional space.

As RBerenguel pointed out, it is currently not well defined.

I will need to think more to do this.

Perhaps some clues of the properties of set X will occur after being clearly defined.

And in his example the intervall A/3<A+B<-A/2 is empty (if A is positive). I'm pretty sure, the misunderstanding lies here.
The interval is not empty, and n({(A, B)|A/3 < A+B < -A/2}) = ∞
(n(X) means number of the elements of set X)
I never stated the numbers should be positive but rather {A, B ...

You are confusing yourself because you did not word your problem correctly to yourself.

Is the following implication true FOR EVERY 5-tuple (a1,a2,b1,b2,x) of real numbers?

If (a1<x<a2 if-and-only-if b1<x<b2) then (a1=b1 AND a2=b2)

To disprove it, you need to find that FOR SOME 5-tuple (a1,a2 ...

Let me make things clear...

First, I am still not much familiar with math in English. What I meant by 'given' is actually 'in a situation that', so 'when' I think is a better word there.

So here is (probably) the final version of this problem.

Prove that below is not true&#46;

when,
{a, b, c, d ...

MJK, just curious: what's the background of this question ? Are you trying to understand it for fun ?
Or, is it a homework assignment ?

If it's homework, which school level ? Which subject ? (Mathematics, or physics, or something else, like economics ? )
Background is a situation I faced during ...

My question can be summarized as,

The thing is, inequalities don't work like this. Even though a+b (alpha+beta) satisfy these two inequalities, it doesn't mean the extremes have to coincide. You wouldn't say:

5<d<7
2<d<9

And conclude that 5=2 and 7=9.
Thank you for your such long reply with iPad.

But in your example
5<d<7 ...