Why does this hold?

[pallid field]

This proof is trying to show that n choose p is always a natural number while p <= n

Why does the idea n+1 choose p hold for all p <= n+1 specifically because n+1 choose n+1 holds? is there a property that could be pointed to for this idea that I am overlooking?

Any videos doing this proof would be helpful too. I couldn't find any.

[tame quarry]

It's proof by induction. Are you familiar with that, @pallid field?

[pallid field]

I did a few problems by proof by induction before this. But those were problems where i followed a pretty linear three steps of induction:

1. Prove that k=1 is true.
2. Assume k=n is true
3. So it should follow that k+1 =n is true (Prove k+1 = n)
   It was basically just swapping out arguments and equal expressions in step by step changes until n = k+1. But this seems to be a more general use of the idea than what I'd read and understood. Seemingly the second step is. Suppose that n choose p is a.. and the third step is it follows that. But it is confusing to me because I don't really see the steps to show that the it follows that is true and I do not really know the properties that assure it is true.

[grim marlin]

nCp and nCp-1 (p-1 is still <=n) are natural numbers per the hypothesis, so their sum n+1 choose p is also a natural number (the sum of natural numbers will always be a natural number, this is also pascal's identity). n+1Cn+1 is a natural number is trivial, anything choose itself is one. So it holds that our p, when n=n+1, can also be equal to n+1, so n+1Cp is a natural number for all p<=n+1

[pallid field]

mm i think i kind of understand it now. hmmm

hesitantly

thank you

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