Pigeon Hole Principle: Let a={1,2,3,4,5,6,7,8} Show that if

[Guest]

Let a={1,2,3,4,5,6,7,8} Show that if you choose any five distinct members of A, then there will be two integers such that their sum is 9.

I understand the principle and I can see if you choose 5 distinct members of A that there will be 2 integers whose sum is 9.

How do you show this?

 

[stapel]

How does "A" relate? (You've given a set "a", but no information is given about "A", is why I ask.)

Thank you.

Eliz.

 

[pka]

Re: Pigeon Hole Principle: Let a={1,2,3,4,5,6,7,8} Show that

Let A={1,2,3,4,5,6,7,8} Show that if you choose any five distinct members of A, then there will be two integers such that their sum is 9.

Take note that {1,8},{2,7},{3,6},{4,5} are the only pairs where the sum is 9. Then there are only four such subsets and they partition A. Thus if we pick five numbers from A two of them must come from one of those four sets.

 

[Guest]

stapel, I forgot to capitalize my initial a. sorry for the confusion.

pka, thanks for the help.

Because I have 4 possibilities I have 4 'holes' but I have 5 numbers to pick so the 5th must pair with one of the other numbers to add up to 9. Makes sense.