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?

of irrational numbersbeen proved?" $\endgroup$ – Greg Martin Sep 22 '18 at 0:14Dedekind's theorem:$\sqrt{2}\times\sqrt{3}=\sqrt{6}$, The American Mathematical Monthly,99no 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. $\endgroup$ – David Roberts Sep 22 '18 at 0:26