Proof by induction: nCr always an integer

[app_oos]

since nCr and nC{r-1} are both integers by assumption.

This is the only part I don't get. I know nCr was assumed to be an integer (induction hypothesis) but where was nC(r-1) assumed to be an integer? Is this some sort of axiom, that if nCr is an integer, then nC(r-1) is one too? I understand the other parts of the proof and I know what induction is, by the way.

 

[D H]

This is the only part I don't get. I know nCr was assumed to be an integer (induction hypothesis) but where was nC(r-1) assumed to be an integer?

If nCr is an integer for all real r, then so is nC{r-1} because r-1 is a real.

 

[HallsofIvy]

Well, for all positive integers r. nCr is only defined for n and r positive integers.

 

[D H]

Well, for all positive integers r. nCr is only defined for n and r positive integers.

The point of this thread is to extend to the concept nCr to real r. The constraint that n be a non-negative integer, true.