Ok, correct me if I am wrong.
I think I see now what's going on here.
1) we build up that trick I couldn't see to get rid of everything under the root in the GM side;
2) we have a 3 in the LHS cause it comes from the AM in the RHS;
3) yeah, we do need three elements to implement that trick on...
Jeez. That's really bad... completely mathematically blind.
I am quite ashamed of myself. :redface:
Now that I see that 2+2=4, I have some problems that I am afraid will be challenging like 3+3=?.
There are some questions I have related with other things I don't understand.
1) Why do...
Homework Statement
Let ##a## and ##b## real numbers such that ##a>b>0##.
Determine the least possible value of ##a+ \frac{1}{b(a-b)}##
I took this example from page 3 of this paper
Homework Equations
In the article previously linked, explaining the example, the author writes...
Theorem (Dilworth, 1950):
Every poset whose width is ##n## is equal to the union of ##n## chains.
Proof:
We let ##R## be a partial order relation over a set and we let ##P## be an arbitrary poset with width ##w(P)## equal to ##n##.
We proceed on induction over ##w(P)##.
i) Base case:
We...
It is the very first time I try such a long shot, attempting to prove a well-known theorem (small one, but still not completely obvious).
I would be really grateful to anyone ready to give a feedback (contents or style doesn't matter - they are both important).
PS: Obviously I am not sure...
My two cents: Jim McNamara logic is sound.
The problem with your logic is that - if you assume that the tricky side of your problem is not in the way in which it is written, but in some mathematical content - then you have to consider the possibility that you take the 16 face cards from the...
I don't enter in the discussion about "same size" because my mathematical skills are not that advanced, even if basing on my limited knowledge I would agree with Michael Redei.
Btw, going back to my original Case 3 of the proof, I think that this example based on fractions supports exactly my...
Is it not a bit redundant?
My line of reasoning is more or less the same of that I wrote down in my last reply to Michael Redei.
I add an element to ##B## and I come out with ##B'##. Now, I assume that ##B## has a minimal element. So, if the new element and the minimal element of ##B## bear...
Well, don't worry. I appreciate the fact that you had this thought, but it's really not a problem and I didn't feel it. Actually, to be honest, when I started to read your post and I saw "base case" with the inverted commas, at the beginning I thought you wanted to be sarchastic and I kinda...
The server stop gave me enough time to get something hopefully decent that should work. :smile:
Still, I have some doubts.
Can I prove the result without going backward to the singleton cases?
Or, in other words, is there a way to prove it having as a base case the two elements one I...
Statement:
Suppose R is a partial order on a set A. Then every finite, nonempty set B \subseteq A has an R-minimal element.
Proof:
Let B be an arbitrary subset of A. We prove the statement by induction on the cardinality of B.
i) Base step:
Assume that B is a singleton. Then the only...
Huge mistake..I completely forgot that the notion of set implies that.
For a second, I thought that I was implicity assuming that it was a loset and not a poset. So, here we are: indeed I assumed completeness.
I don't see why it is the case.
This. I was looking for that word! :smile...