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General Forum: Education | | ******MATHEMATICS****** | |
| ala kaadu .. hmmm lemme put it like this
1 1/2 1/3 1/4 ... ila vunna sequence ni nenu kanaka. S = {1/n for all n )1 } annamu anuko .. explicit ..as in clean cut realtion
ala nee solution ni three sets S1,S2,S3 ni explicit ga define cheyyagalva ani ?? hope I make sense :-)
Posted by: Mr. Frank Abagnale Jr. At: 16, Nov 2003 11:44:31 AM IST
infinite solutions,i mean just arranging the subets in R+ acc to my def.....
like.....
instead of starting from [0.4,1) and going to left and right......
we can assume {........,[0.1,0.25) , [0.25,0.625),...........}
so on as far as i know.....
Posted by: Raj Sekhar At: 16, Nov 2003 11:40:16 AM IST
Hi Raj,
I saw this pattern earlier. But, wats causing a hitch is when we define a set it shd have an explicit definition. A maximum and an minimum ( if they exist ) shd be apparent. If u can put these sets in mathematical set notation. I wud agree that this is the solution.
Ok, so u agree that there are infinite solutions in R3 except that a trick in arranging the sets, is wat is making it unsolvable in each set ?
Posted by: Mr. Frank Abagnale Jr. At: 16, Nov 2003 10:46:45 AM IST
hmmm.....ok krishna,u have done it almost....anyhow i am posting the answer....the credit of answering the question goes to u although....
R+ = {............., [0.16,0.4), [0.4,1), [1,2.5), [2.5,6.25) , [6.25, 15.625) , [15.625,39.0625),.................}
hope u understand the above representation....the entire R+ is included..........
same as ur arguement,each sub interval in that set doesnt have any solution.......
and now
S1 = set containing the 1st,4th,7th,10th subsets from the above representation
S2 = 2nd,5th,8th....
S3 = 3rd,6th,9th,.....so on....
see the subintervals.....left closed,right open.....and i have taken proper care of no mixing of one subset into one another to get a solution......
like that there can be many(infinite) solutions........coz we ca divide the subintervals in infinite number of ways....
hoep u got it now....
Posted by: Raj Sekhar At: 16, Nov 2003 10:34:16 AM IST
hi Raj..I am including the previous solution on which u agreed, before I proceed further.
Lets define function S(x,y):R+ ----) R+ such that S(x,y) = x+y .lets consider an interval (c.d). Since the function is S is monotonously increasing and binear with both x and y (You can check this result by taking second derivates .. ). the maxima and the minima of the function occur at the extremes . Hence, the maximum of the function would be 2d and the minima would be 2c.
Now for the function f(z) = 5z. for z belonging to the same interval the maxima and the minima would be (5c , 5d) . For the equality not to hold good , (5c,5d) should lie outside (2c , 2d), this would result only in one imposing condition d ( (5/2) c . So as long as u divide the real line into such intervals it will hold good trivially. But wat I dont understand how can u give specific intervals ? yes , you can define intervals in recursive fashion. ( Wats bothering me if u say just three sets .. then max value on of the set in unbounded :O not gud news buddy ... coz we want it to be less than something else ..
Now we wud like to see find three sets S1,S2,S3 such that S1 U S2 U S3 = R+ .
I will define S1 , S2 , S3 in this fashion
S1 = { (x y) such that y (5x/2 and x)0 , y(2 }
S2 = (2 5)
S3 = { (x y) such that y (5x/2 and x )5 }
we can chose any interval S2 which satisfies the (x 5x/2) condition and then choose corresponding S1,S3. Probably this is wat u are luking in the answer. Let me add that this is not the answer.I also have a strong feeling that your question cannot have a solution . I am just posting an solution ( much debatable??) ..which I think is wat u r thinking or luking for.
Newayz , post the answer.. we will argue further .
bye
Krishna
Posted by: Mr. Frank Abagnale Jr. At: 15, Nov 2003 2:54:01 AM IST
ohh i c
Posted by: Mr. Frank Abagnale Jr. At: 13, Nov 2003 11:11:29 PM IST
not like that.......chinnappati nundi enthu in mathematics.........
Posted by: Raj Sekhar At: 13, Nov 2003 11:06:42 PM IST
no no .. lemme try .. newayz we can continue as long as the thread goes.. lemme try once more ... adhey I was wondering distinctive sets vundakoodadu vundalevu ..obviously .. andukE koncham bulb anna .. btw ..Raj enti idanta mee IIT preparation selava ? ;-).. mean u r enthu in mathematics ?
Posted by: Mr. Frank Abagnale Jr. At: 13, Nov 2003 10:56:16 PM IST
aa three sets union R+ kaavaali....
aa three sets lo okko set lo union of infinite subsets undacchukadaa.....
ok then,i am posting the answer f u wish....
Posted by: Raj Sekhar At: 13, Nov 2003 10:52:33 PM IST
ok .. thats fine.. aa three sets union full R+ line kaavala ? avasaram ledha ?
Posted by: Mr. Frank Abagnale Jr. At: 13, Nov 2003 10:47:42 PM IST
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