Proving an onto function
How do I prove that the function defined as is an onto function by mathematical induction? The bracket meant "n choose r".
I started off this way:
So f(1,1) is true since f(1,1)=1, the element "1" in the codomain has its preimages m=1 and n=1.
Then assume true for .
Then there exists a such that
But how do I continue to show that this will be true for all k+1 and that this is an onto function?
Are there any "easier tricks/process" when doing induction? I am trying to keep doing induction until I get the hang of it but I am always stuck after the basis step. (Headbang)
Doesn't feel like a friendly technique.
Thanks for any help.
If f(m,n)=k then what is f(m+1,n-1) ?
Thanks but how did you figure out that f(m+1,n-1)=k+1 ? Do I have to keep doing trial and error until I get this?
Originally Posted by running-gag
One way to get k+1 is to have m+1 as last term of the sum but then you change 2 choose m+n-1 into 2 choose m+n. Coming back to m+n-1 can be done by changing n into n-1
Originally Posted by xEnOn
I know this is already solved, but if you want to find the ordered pair that maps to some number k from scratch, it helps to realize the connection to triangular numbers here. Any k can be written as n(n-1)/2+j, where 0<=j<n, or in other words k lies between 2 successive triangular numbers. If you find this n and j then you can find your ordered pair.
For example, if you want to know how to get 43,276, you just need to find that 43,276=293(294)/2+205, and f(205,294-205+1)=43,276. It's not induction, but it might be good to know.
edit: Also, I was just thinking that if you're still going to use induction you need to be careful with the triangular numbers. If you do the m+1, n-1 thing at a certain point you'll end up with something like f(5,1) and you can't have f(6,0).