Theorem 2.10: When n + 1 pigeons roost in n holes, there must be some hole containing at least two pigeons.
Prove the Pigeonhole Principle using mathematical induction.
Step 1:IF n=1 then there are two pigeons and one hole,so both pigeons roost in one hole.
Step 2:Lets take n=2 then there are 3 pigeons and 2 holes , so one pigeon roost in one hole and other 2 roost in one hole.
Step 3:taking some x , assume that if x+1 pigeons roost in x holes one hole will atleast contain 2 pigeons.
STEP 4:consider the case of x+2 pigeons among x+1 holes,choose one of the holes. It will atleast contains zero pigeons,one pigeon or two or more pigeon.IF it has two or more pigeons we proved it.If it has one pigeon then remaining x+1 pigeons roost in x holes.so by assumption one of these holes contain atleast 2 pigeons.IF it has zero pigeons then,remaining x+1 of the x+2 pigeons roost among x holes so as previously one of these x holes contains atleast two pigeons.
Hence Proved When n + 1 pigeons roost in n holes, there must be some hole containing at least two pigeons.
Theorem 2.10: When n + 1 pigeons roost in n holes, there must be some hole...
Proposition PHP2. (The Pigeonhole Principle.) If n or more pigeons are distributed among k 0 pigeonholes, then at least one pigeonhole contains at least 1 pigeons. Proof. Suppose each pigeonhole contains at most 1-1 pigeons. Then, the total number of pigeons is at most k(P1-1) < k㈜ = n pigeons (because R1-1( , RI). Exercises. Prove: (a) If n objects are distributed among k>0 boxes, then at least one box contains at most L objects (b) Given t > 0...
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