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#26 Re: Help Me ! » Matrix Multiplications and Determinants » 2012-10-09 04:42:16
Nevertheless, I did finish the question with all of your help and now it's onto this week's assignment! I'll be back if I get stuck.
I hope so (not that you get stuck... that you post more interesting questions 
) since i'm studying the same subject right now. Good work!
#27 Re: Help Me ! » Matrix Multiplications and Determinants » 2012-10-06 02:10:36
Yes, it's exactly the same argument but wrote in a less pedant manner
In my opinion, your demonstration in post#1 is not incomplete: i've read lots of (basic level) demonstration where "formal" things
are left to intuition and they're still considered good demonstration
For example consider the classical demonstration that sqrt2 is irrational:
we start with two proposition which are:
p: (a/b)^2=2
q: a and b are irreducible
and we prove that p AND q => NOT q;
but we have to assume q to be true, because in each case it's false we get back to a case in which is true.
so NOT q is false... so p AND q is false, but since q is true, p must be false.
But i've never (for God's sake) read a math's book explaining that in proposition in these terms.
#28 Re: Help Me ! » Matrix Multiplications and Determinants » 2012-10-05 07:19:42
ok thanks...
seems to me your (Anakin) third demonstration is right
call
p: AA=A;
p': A!=I;
q: detA=0
q': detA=1
we have that p=>(q OR q')
and q'=>NOT p' (or, equivalent, p'=>NOT q')
so p AND p' => (q OR q') AND (NOT q') = q
Do you agree?
EDIT: to be correct i think i should say:
q' AND p => not p' which is equivalent to p' => (NOT q' or NOT p) => NOT q'
because we assume p is true
#29 Re: Help Me ! » Matrix Multiplications and Determinants » 2012-10-05 00:16:28
sorry anakin
what is "A!" ?
#30 Re: Help Me ! » Interesting square problem (and a way to solve it) » 2012-09-27 04:18:00
Hi bobbym, thank you for your answer...
Wow! This is a nice formula! Thank you.
Have some idea about how to show EF=FG just using triangles properties?
i.e. we can show PFC=QFA in fact
(1) PD=QD;
(2) DA=DC;
(3) angle ADC is common
(1-2-3)=>PAD=DCQ, subtracting PFQD from both we get PFC=QFA.
I'm sure it can be shown in similar ways that FG=FE.
#31 Help Me ! » Interesting square problem (and a way to solve it) » 2012-09-26 23:17:53
- Fistfiz
- Replies: 3
Hi guys! Yesterday I found a cool problem:
in the square ABCD, Q and P cut AD and DC in half. Calculate the ratio of areas BEFG/ABCD.
Now, despite this is correct, I find it a little complicated, and I think that some calculations could be avoided. For example, could you show that GF=FE without explicitly calculating it as I did? Or, how would you show that BF lies on BD?
In general, how would you solve the whole problem?
edit: maybe a mod could put a spoiler on my solution... i don't know how to do that
Hope you like this :-)
#32 Re: Help Me ! » Can you solve this series? » 2012-07-21 04:28:15
Thanks :-)
Unfortunately I have no idea of what that psi function is...
What do you think is not correct? My solution to the original problem or your closed form?
#33 Help Me ! » Can you solve this series? » 2012-07-20 22:27:27
- Fistfiz
- Replies: 3
Hi guys, can you help me with this? (i'm sorry, i don't know how to write in code, maybe a mod can edit this :-) )
∑ (from k=1 to j) of : m/(m+1-k)
can you find a general formula for this series?
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OT:
The problem which led me to this other problem is this:
I have a bag with 100 blue balls;
to do one exchange means to take one random ball from the bag and put a red ball in;
how many average exchange does it take to have 50 red balls in the bag??
my reasoning is:
the average number of exchanges to add a red ball is given by
n=1/P, where P is the probability of taking a blue ball from the bag (P=#blueballs/#allballs)
since a missed tentative does not affect the number of blue balls in the bag, our number of exchanges is given by:
n=1+100/99+100/98+...+100/51 that you can obtain from the series above by putting m=100 and j=50
what do you think about it?
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sorry for the english, i'm from Italy... hope i've made it clear