Wikipedia:Reference desk/Archives/Mathematics/2022 April 19

= April 19 =

Proof by induction for real numbers?
Is there a parallel for proof by induction which works with real numbers, replacing the induction step "if the statement holds for any given case n = k, then it must also hold for the next case n = k + 1" with "if the statement holds for any given case n = k, then it must also hold for the next case n = k + 1 / r" and letting r become arbitrarily large? 2A01:E34:EF5E:4640:44FE:12D:9E52:1C3D (talk) 11:00, 19 April 2022 (UTC)


 * There is no proof by induction for real numbers, as there is no such thing as the "next case". I don't know if you'd consider it a "parallel", but there are ways to reduce proofs about all real numbers to smaller steps. For example, you can use the fact that the real numbers are connected to prove that a statement is true for all real numbers by showing that the set where it holds is nonempty, open and closed. So you can show that (i) your statement holds for 0 (ii) if it holds for a real number r, then there is s>0 such that it holds for all numbers in the interval (r-s,r+s), (iii) if it holds for a convergent sequence of real numbers, it holds for their limit. Then (i)+(ii)+(iii) imply your statement holds for all real numbers. —Kusma (talk) 11:13, 19 April 2022 (UTC)


 * This sounds a bit like Proof by exhaustion. -- Jayron 32 11:59, 19 April 2022 (UTC)


 * ZFC implies the existence of a well-order on the set of real numbers, which would seem to imply the possibility of using transfinite induction. Unfortunately, the existence proof is completely non-constructive; we have no way of actually defining this order, so there is no way to prove any of the three cases (zero, successor and limit) other than by first proving the conclusion being sought. --Lambiam 13:40, 19 April 2022 (UTC)
 * Anyway, extending a proof to n = k + 1/r would only cover rational numbers, not all the reals (assuming r is an integer). -- Verbarson talkedits 08:48, 21 April 2022 (UTC)
 * And proofs by induction for the rationals are feasible. The well-known proof by infinite descent of the irrationality of $\sqrt{2}$ is a proof by mathematical induction in disguise. --Lambiam 11:15, 21 April 2022 (UTC)
 * Wouldn't Cantor's diagonal argument also qualify as inductive? -- Jayron 32 12:26, 21 April 2022 (UTC)
 * It is a famous example of a proof by contradiction. --Lambiam 13:23, 21 April 2022 (UTC)