0

I am confused but I'd think the answer is no because that being the case I could, for instance, in an induction proof suppose $p(k)$ for any (an arbitrary) $k$ (let's say a natural number) and then, if a found another natural number like $k+1$ I could conclude, in particular for $k+1$ what I assumed for $k$. Because after all that's a property of an arbitrary object. I mean: If I know that "for all object with a certain property something happens" whenever I found "a particular object the certain property" I can, then, claim that "the something happens", right? Yet I've given proofs arguing so because I chose $k$ to be ARBITRARY and apparently, along with good reason, these proofs are wrong given that they were based in such a property of arbitrary objects.

If they don't mean the same, when should I use each? and What does it mean "arbitrary but fixed" in the context of a proof by induction?

  • 1
    This is all very hard to follow. What's the question? Can you cite some examples where the phrase "arbitrary but fixed" was used? – lulu Jul 05 '21 at 03:04
  • @lulu We know that $\forall r\in \mathbb{R}: r>0 \veebar r=0 \veebar r<0$ so since $2\in \mathbb{R}$ then one of these options (we know which one) is true. We conclude something in particular for $2$ for being a real numbers. My question is: Are those phrases really interchangeable because if that's so in an induction proof I may assume $p(k)$ for an ARBITRARY $k\in \mathbb{N}$ and try to prove $p(k+1)$. But since $k+1$ would be natural for being the succesor of $k$, Could I conclude IN PARTICULAR what was assume for $k$ for $k+1$ because the latter is also a natural number? Like in example. –  Jul 05 '21 at 03:23
  • If not, I would think the words "arbitrary" and "arbitrary but fixed" would not be interchangeable. –  Jul 05 '21 at 03:24

0 Answers0