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0 pointsmannerheim4y ago0 comments

An inductive proof has two steps:

1) Prove H(0)

2) Prove that if H(n), then H(n+1)

Then, by the axiom of induction, this is proven for all positive integer values of n. Because if those two conditions hold, we would have

H(0) is true

H(1) is true because H(0) -> H(1) ((2) with n=0) and H(0)

H(2) is true because H(1) -> H(2) ((2) with n=1) and H(1)

and so on.

The assumption isn't what's being proven, it's proving the inductive hypothesis for the next integer.

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2 comments · 1 top-level

phkahler4y ago· 1 in thread

Yes, I was being dumb.

As the article says there is a problem talking about equality among members of a set with size 1. So saying "all the horses in a set of size 1 are the same color" begs the question "same as what?" and that's where the real problem is, and the difficulty in pinning down that kind of thing is why I'm not a mathematician ;-)

mannerheimOP4y ago

There are several 'mathematical' ways you could phrase it (glossing over how to define the colour of a horse):

1) In a set of horses with size n, if horses A and B being both members of this set implies horse A has the same colour as horse B, then all horses in the set have the same colour.

2) In a set of horses with size n, and let f be a function on this set that produces each horse's colour. Then all horses in this set have the same colour if and only if the codomain of f on this set has size 1.

You could probably prove these are equivalent definitions.

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