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Here is something fuzzy to think about while going to sleep:
Math Koan: "How is adding the numbers (1, 2, ... n) together, the same as counting up the number of ways to choose two objects out of (n+1) objects?"
Here is one way:
1. The sum of numbers from 1 to n can be written as n*(n+1)/2.
It's an old trick, you group the outer pairs together and walk inward. So if n=100, you get (1,100), (2,99), (3,98), ..., (50,51). Each of these pairs adds up to 101=(n+1). And there are n/2 of these pairs, so you get n*(n+1)/2.
2. The ways to choose two objects out of n objects is n*(n-1)/2.
Again, it's straightforward. There are n objects to pick for the first one, and for each pick, you get one of the (n-1) leftovers for the second, giving you n*(n-1) choices. Since it doesn't matter which one you pick first or second, it's n*(n-1)/2.
But it's just a numerical relation, and it doesn't explain WHY... does it? It is fun to meditate on whether there is a connection beside the formulas - could someone suddenly realize one day that the adding is the same as the choosing, without working out the numbers?
Are there other kinds of adding, which might be just as natural, that are connected to picking three objects out of (n+2) or something?
Can you think of an addition algorithm which reduces to choosing two objects, and that you could teach to your child?
- rw
Math Koan: "How is adding the numbers (1, 2, ... n) together, the same as counting up the number of ways to choose two objects out of (n+1) objects?"
Here is one way:
1. The sum of numbers from 1 to n can be written as n*(n+1)/2.
It's an old trick, you group the outer pairs together and walk inward. So if n=100, you get (1,100), (2,99), (3,98), ..., (50,51). Each of these pairs adds up to 101=(n+1). And there are n/2 of these pairs, so you get n*(n+1)/2.
2. The ways to choose two objects out of n objects is n*(n-1)/2.
Again, it's straightforward. There are n objects to pick for the first one, and for each pick, you get one of the (n-1) leftovers for the second, giving you n*(n-1) choices. Since it doesn't matter which one you pick first or second, it's n*(n-1)/2.
But it's just a numerical relation, and it doesn't explain WHY... does it? It is fun to meditate on whether there is a connection beside the formulas - could someone suddenly realize one day that the adding is the same as the choosing, without working out the numbers?
Are there other kinds of adding, which might be just as natural, that are connected to picking three objects out of (n+2) or something?
Can you think of an addition algorithm which reduces to choosing two objects, and that you could teach to your child?
- rw
