# Why did Dedekind claim that $\sqrt{2}\cdot\sqrt{3}=\sqrt{6}$ hadn't been proved before?

In a letter to Lipschitz (1876) Dedekind doubts that $$\sqrt{2}\cdot\sqrt{3}=\sqrt{6}$$ had been proved before:

quoted from Leo Corry, Modern algebra, German original:

Why did Dedekind doubt that $$(\sqrt{a}\cdot\sqrt{b})^2=(\sqrt{a})^2\cdot(\sqrt{b})^2$$ had been proved (which lies at the heart of his vicious circle)? What's wrong with this identity which easily follows from the associativity and commutativity of multiplication:

$$(m\cdot n)\cdot(m\cdot n) = (m\cdot m)\cdot(n\cdot n)$$

He argues that rational and irrational numbers are of a different kind, and thus $$(m\cdot n)^2 = m^2\cdot n^2$$ (which had been proved for the rationals) may not be "applied without scruples" to the irrationals.

But why – from his point of view – had $$(m\cdot n)^2 = m^2\cdot n^2$$ been proved only for the rationals? Couldn't it – for example – have been proved already in the system of Euclid's Elements by elementary geometrical considerations, regardless of rational or irrational, i.e. for arbitrary lengths:

The proof that the fat green circle intersects the horizontal line at the same point as the fat red line (which is constructed as the parallel to the thin red line going through 1) may have been complicated for Euclid, but it seems possible.

If not so: Why couldn't it have been proved by Euclid?

And finally: What was Dedekind's alternative proof, eventually?

• 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 – user44191 Sep 22 '18 at 0:11
• 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. – user44191 Sep 22 '18 at 0:13
• Dedekind might well have responded to your post "Have associativity and commutativity of multiplication of irrational numbers been proved?" – Greg Martin Sep 22 '18 at 0:14
• 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. – David Roberts Sep 22 '18 at 0:26
• 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. – ex0du5 Sep 22 '18 at 0:38

I don't think Dedekind's complaint can be that a proof in an axiomatic system must be invalid if the objects in the system have not been constructed from simpler objects. This obviously leads to an infinite regress. Today we typically take "set" to be undefined, but Dedekind would surely have been familiar with treatments of Euclidean geometry in which "point" is an undefined term.

I suspect that Dedekind was complaining that people were not working in a full axiomatic system for the reals, but rather were using only the elementary axioms and did not realize that the completeness axiom was needed. Note that in the quote, the argument he criticizes uses only elementary properties such as commutativity and associativity.

Without completeness, we don't have any way of proving the existence of square roots, but the textbook treatments he complains about presumably don't admit that the theorem $$\sqrt{2}\sqrt{3}=\sqrt{6}$$ is conditional on the unproved existence of these roots.

Why couldn't it have been proved by Euclid?

Euclid couldn't have proved it to modern standards of rigor because Euclid couldn't have proved that the circles in your diagram intersected. This is equivalent to lacking the completeness property of the reals.

• Euclid couldn't have proved that the circles in your diagram intersected. <-- sick burn bro :-D – David Roberts Sep 27 '18 at 21:54

Dedekind point was that nobody before him has defined reals, their roots and their multiplication; hence (before him) there was no proof.

• But roots of rational numbers had been defined. – Hans-Peter Stricker Sep 27 '18 at 11:34