Timeline for Why did Dedekind claim that $\sqrt{2}\cdot\sqrt{3}=\sqrt{6}$ hadn't been proved before?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Sep 28, 2018 at 17:28 | comment | added | Peter Heinig | (Minute detail: it would have been better had Corry translated "Verstand" as 'mind' or 'intellect'.) | |
| Sep 27, 2018 at 14:13 | comment | added | Andrej Bauer | For rationals we know that $(m \cdot n)^2 = m^2 \cdot n^2$, so let's just use that for irrationals too. And while we are at it, let's also postulate the identity $(m \cdot n)^2 = m^2 \cdot n^2$ for complex numbers as well, and for quaternions too. What could go wrong? It's just numbers after all. | |
| Sep 27, 2018 at 13:40 | comment | added | YCor | Dedekind was probably aware that using carelessly the formula $\sqrt{z_1z_2}=\sqrt{z_1}\sqrt{z_2}$ (for complex numbers, with $z_1=z_2=-1$) leads to a contradiction (still very common, insofar as many people still teach the notation $\sqrt{-1}$). | |
| Sep 27, 2018 at 13:31 | answer | added | user21349 | timeline score: 7 | |
| Sep 27, 2018 at 10:58 | answer | added | zvonimir šikić | timeline score: 3 | |
| Sep 22, 2018 at 11:00 | comment | added | Alec Rhea | @LaurentMoret-Bailly Good call, much appreciated. | |
| Sep 22, 2018 at 9:12 | comment | added | Hans-Peter Stricker | I shoud have given stronger emphasis to these facts: (i) Dedekind (esp. in Continuity and irrational numbers) refers to "geometrical evidence" and relates his number theoretic consideratons to (Euclidean?) geometry, (ii) multiplication can be done and the associativity and commutativity of multiplication can be shown geometrically for arbitrary lengths (not only rational lengths), (iii) $\sqrt{a}\sqrt{b} = \sqrt{ab}$ can - presumably - be shown geometrically for arbitrary lengths. From the valuable comments above I still don't see, why Dedekind did not "trust" these constructions. | |
| Sep 22, 2018 at 6:48 | comment | added | Laurent Moret-Bailly | @AlecRhea: Dedekind's text doesn't say he couldn't prove the equality. | |
| Sep 22, 2018 at 4:50 | comment | added | Alec Rhea | Huh, this would seem to indicate that Dedekind cuts hadn't been discovered by Dedekind yet (at this juncture at least), as they offer a very simple proof: $\sqrt{2}\times\sqrt{3}=\{q\in\mathbb{Q}:q^2<2\vee q<0\}\times\{q\in\mathbb{Q}:q^2<3\vee q<0\}=\{p\times q\in\mathbb{Q}:p^2\times q^2=(p\times q)^2<2\times 3=6\vee p\times q<0\}=\sqrt{6}$. Is this a Cauchy-type situation where his name ended up on the construction as a homage rather than because he discovered it? | |
| Sep 22, 2018 at 0:38 | comment | added | ex0du5 | It takes a bit more apparatus, I think, to conclude an isomorphism between (rationals + irrationals) and points on a geometric line that respect these distance constructions than it does to specify this multiplication property of irrationals. To me it seems the multiplication property will always be prior to proving any such geometric representation rigorously across the union of number types. | |
| Sep 22, 2018 at 0:26 | comment | added | David Roberts♦ | Fowler treats this in Dedekind's theorem:$\sqrt{2}\times\sqrt{3}=\sqrt{6}$, The American Mathematical Monthly, 99 no 8 (1992) pp 725-733, doi.org/10.1080/00029890.1992.11995919 He considers how one might prove the equation using other, pre-rigorous definitions of real numbers, like infinite decimal expansions or continued fractions etc. | |
| Sep 22, 2018 at 0:14 | comment | added | Greg Martin | Dedekind might well have responded to your post "Have associativity and commutativity of multiplication of irrational numbers been proved?" | |
| Sep 22, 2018 at 0:13 | comment | added | user44191 | As for your question - couldn't it be justified - the point seems not that it couldn't be, but that the teachers didn't do so when teaching. | |
| Sep 22, 2018 at 0:11 | comment | added | user44191 | A large piece of an answer might come from the sentence: "Not even a minor explanation of the product of two irrational numbers is provided in advance ..." In other words, to proceed, a student working from the axioms would have to assume that multiplication can be made to make sense on irrational numbers. Further compounding the problem, a theorem that's easy over integers (or rationals) is then stated without relating it to the definition of multiplication. It looks like a complaint about the fact that theorems on irrational arithmetic assume "obvious" things from integers without justifying | |
| Sep 21, 2018 at 23:45 | history | asked | Hans-Peter Stricker | CC BY-SA 4.0 |