The main idea is the most important one!
If you have fewer pigeon holes than pigeons and you put every pigeon in a pigeon hole, then there must result at least one pigeon hole with more than one pigeon.
Expand the idea perhaps using a finite example.
21 pigeons and 10 holes.
Try and see how at least one hole must have more than 2 pigeons.
Then think about generalizing it.
The real power comes from seeing this all in action.
A commonly used example is:
Among any N positive integers, there exists 2 whose difference is divisible by N-1.
Chat with our AI personalities
3546
The easiest way might to answer that might be to show you an example. Let's look at a chess board with two of the diagonally opposite corners removed. Is it possible to cover the board with pieces of domino whose size is exactly two board squares? The reason this is a pigeonhole problem is because the two diagonal square on a chess board are the same color. So when you remove them you have 2 more square of one color than you do of the other. So assume by contradiction that you can cover the board with pieces of domino whose size is exactly two board squares. Now every piece of domino must cover exactly two squares and these will be squares of different colors because adjacent square on the chess board are different colors. So for every domino piece I place, I set up a 1 to 1 correspondence between the set of one color square and the set of the other color squares. We now know the cardinality of the two sets is different since we removed those corners. So the pigeonhole principle tells us we can not have a 1 to 1 correspondence between two sets with different cardinalities. We conclude that it can't be done. The idea in all cases where you want to use the pigeonhole principle and prove by contradiction is to assume it works and then let the pigeonhole principle prove it can't work.
Only vector addition is applicable
That is the principle of a sextant.
It depends on what you're learning.