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I have already proved that $\sqrt{2}\cdot \sqrt{2}=2$ so I hope I can now use $\sqrt{3}\cdot \sqrt{3}=3$ and the same for 6. The exercise comes from Stillwell: Mathematics and its History. The other exercises have not been complicated so probably the solution to proving $\sqrt{2}\cdot \sqrt{2}=2$ I borrowed from Martin K was way over the top. But it cannot be adapted, I think, to this one where the roots are not of the same number. Here is my attempt: $r^2=2, s^2=3$.

$r\cdot r=2$ implies $r=\sqrt{2}$ and $s\cdot s=3$ implies $s=\sqrt{3}$

I also know that $\sqrt{6}\cdot \sqrt{6}=6$.

$r^2\cdot s^2=6$ so $(rs)^2=6$ This implies that $(\sqrt{2}\sqrt{3})^2=\sqrt{6}\cdot \sqrt{6}$

Which I am not sure about, but hope, implies that $\sqrt{2}\sqrt{3}=\sqrt{6}$

Hans-Peter Stricker
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Kang
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    It is correct . – Sonal_sqrt May 03 '18 at 11:25
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    Compare your proof with this one from this site. Then you will see it. – Dietrich Burde May 03 '18 at 11:26
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    Of course, the point is that You use commutativity and associativity while stating $r^2s^2=(rs)^2=2.3=6$ – Peter Melech May 03 '18 at 11:27
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    I should have said that the context is Dedekind's proof (which I have been unable to find online. I'd love to see it.) so the commutativity of roots might not be assumable. My problem was largely concerning how much might be assumed. – Kang May 03 '18 at 11:39
  • @Kang Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user May 31 '18 at 19:50
  • @gimusi The original problem has not been solved. I need to study Dedekind's proof to answer it myself. But I've had too much else to do. Should I just accept the best answer to close it? – Kang Jun 02 '18 at 11:06

3 Answers3

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Let $a, b > 0$ then we need to prove that $ \sqrt{a}\sqrt{b} = \sqrt{ab} $
If we do not prove axiomatically, then we can do this: Since we have positive numbers, their squares will also be equal.
$$ (\sqrt{a} \cdot \sqrt{b})^2 =_{?} (\sqrt{ab})^2 $$
and, $ (xy)^2 = x^2y^2 $, then $\sqrt{a}^2\cdot\sqrt{b}^2 = ab $ by the definition of a root, we obtain equality.

Vladislav Kharlamov
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  • I want to note that the equality of squares of positive numbers is equal if and only if these numbers are equal. (Therefore the proof is correct) – Vladislav Kharlamov May 03 '18 at 11:37
  • This is not the answer I am looking for either (see comment above) but it does clarify exactly what sort of answer would satisfy me so it, too, has helped! – Kang May 05 '18 at 11:14
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Note that

$$a=\sqrt{2}\implies a^2=2$$

$$b=\sqrt{3}\implies b^2=3$$

then

$$(ab)^2=6\implies ab=\sqrt{6}$$

user
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  • Your huge reputation emboldens me to ask if you know a source of Dedekind's proof? I may not be able to understand it but I see now (see comments either side of your answer) that only something along those lines will satisfy me. – Kang May 05 '18 at 11:18
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    Sorry it wasn’t clear by your OP, I’ll take a look! Bye – user May 05 '18 at 11:37
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Let $\sqrt{2}\sqrt{3} = x$.

Substitute $\sqrt{2} = \dfrac{2}{\sqrt{2}}$ and $\sqrt{3} = \dfrac{3}{\sqrt{3}}$. Then we should have: $$\left(\frac{2}{\sqrt{2}}\right)\left(\frac{3}{\sqrt{3}}\right) = \sqrt{6}.\tag1$$ Note that the $LHS$ (Left Hand Side of the equation) is equal to $\dfrac{6}{\sqrt{2}\sqrt{3}}$. It follows, then, that: $$\begin{align} \frac{6}{\sqrt{2}\sqrt{3}} &=\sqrt{6} \\ &\Downarrow \\ \frac{6}{x} &= \sqrt{6}.\end{align}$$ For you to have made your substitutions in Eq. $(1)$, it must follow that $x = \sqrt{6}$.

$$\therefore \sqrt{2}\sqrt{3} = \sqrt{6}.\tag*{$\bigcirc$}$$

Mr Pie
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    This does help me to pin-point exactly what I am looking for and why I am not satisfied with my proof. I see I must, after all, attempt to use Dedekind cuts. – Kang May 05 '18 at 11:11
  • @Kang your proof is fine. It is correct and quite nice, imho. I was just giving a suggestion as to another proof like the other answers :) – Mr Pie May 05 '18 at 11:51
  • At the risk of offending against protocol: thank you. All the proofs offered were helpful and enabled me to see not why I was dissatisfied with my proof in particular (sorry I wasn't clear on that) but that all these proofs seem to assume more than I suspect Stillwell wants us to assume at that point. Hence the reference to Dedekind, whose cuts are the subject of the section from which the exercise comes. – Kang May 07 '18 at 09:31