Question

can anybody explain me the pigeon hole principle using this: if a1 , a2 , a3 ... a7 are seven not necessarily distinct real numbers such that 1 14 years ago

Answer 1

i know the theorem..just donno how to apply it here.. :(

Answer 2

In mathematics and computer science, the pigeonhole principle states that if n items are put into m pigeonholes with n > m, then at least one pigeonhole must contain more than one item.

Answer 3

this is the theorem...i guess

Answer 4

yah.. wikipedia..lol :)

Answer 5

how to apply here in this problem?

Answer 6

yes i am studying from wiki.......i don't know it.............

Answer 7

see i think u have to make three groups of the ai's....

Answer 8

yes..the principle is simple..but how will we apply it here?

Answer 9

each grop will correspond to a side.............
let u take p,q,r no. of ai at a time....p+q+r=7
let sum of p ai is b, that of q is c, that of r is d.
p<b<13p,q<c<13q,r<d<13r
p+q<b+c<13(p+q)
now we can take the ai in such a way that always...b+c>d
this is the condition of a triangle.........

Answer 10

see ai s are different....trivially i can say that choose larger ai s for the grouping of p,q so that this force fully leads to b+c>d
again u can take r=1. and this element in the group of r element is such that a_r1=min(ai)

Answer 11

ai's may not be distinct..

Answer 12

so what.......say all are same then take p=3,q=2,r=2

Answer 13

p<b<13p,q<c<13q,r<d<13r
p+q<b+c<13(p+q)
now we can take the ai in such a way that always...b+c>d

i dont understand this..how do u get b<13p

Answer 14

u gave...1<ai<13...so for p ai's......1+1...p times...<b<13+13...p times..
p<b<13p

Answer 15

"now we can take the ai in such a way that always...b+c>d"

u need the 3 numbers to satisfy simultaneously the 3 inequalities..a+b>c;c+b>a;a+c>b..not any one..

Answer 16

b+c>d
13(p+q)>b+c>p
13(7-r)>b+c>7-r
b+c>d
gives
7-r>r
2r<7
similarly for other two 2p<7, 2q<7....

Answer 17

13(7-r)>b+c>7-r
b+c>d
gives
7-r>r

this is wrong i think...

Answer 18

why?
13r>d>r
b+c>d
means.......p+q>r
7-r>r.....
please explain ur thinking...

Answer 19

b+c>r and b+c>p+q doesnt mean p+q>r

Answer 20

no .......here u have to impose this....that p+p must be greater than r.

Answer 21

so u r taking p+q>r as assumption.
well then there was no need of bringing p<b<13p etc...

Sorry, if i dont understand..can u xplain me ... :)

Answer 22

this is not assumption . this we impose so that we can form the triangle...this is the condition that we have to satisfy........

Answer 23

then why do we need p+q>r to have b+c>d.?
p+q>r doesnt imply b+c>d.. !

Answer 24

why?

Answer 25

u have b+c>r and b+c>p+q and d>r...
now p+q>r implies b+c>r.
so now u have b+c>r and d>r.
u dont know if b+c>d or b+c<d or b+c=d !!!

Answer 26

this three conditions lies the ai s u have chosen to form b,c,d..

Answer 27

yah..u r right..but p+q>r does not imply b+c>d...so now u need a condition on p,q,r that will imply b+c>d,a+b>c,c+a>d

Answer 28

yes.....actually pigeon hole theorem helps us to say that we can form a group with one or more than one ai that will be our side of a triangle....but the choosing depends on the relations....

Answer 29

nice
!

Answer 30

I got a one-line solution: if 3 or more of the ai's are equal, we can easily form an equilateral triangle. so assume not more than 2 ai's to be equal.now divide the interval (1,13) as follows
(1,5] , (5,9) , [9,13)
using pigeon hole principle, atleast three of the ai's lie in the same interval which can be used to form a triangle.

i just saw 2,2,5 doesnt work!

Answer 31

satellite73 suppose u take 2 two times from interval one and then 5. this proof is wrong!
by the way:this proof was not mine:) lol

Answer 32

k. then take intervals as [2,4],[5,8],[9,12]...
now 2,2,4 gives a problem :(

Answer 33

yup...[2,3],[4,7],[8,12] gives no problems! :)