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I read a little proposition that any quantity obtainable from rational numbers by a finite number of operations $+$, $-$, $\cdot$, $\div$, $\sqrt{{}}$ can be put in a standard form $\frac{p}{q}\cdot A$, where $\frac{p}{q}\in\mathbb{Q}$ and $A$ is an expression of only integers and $+$, $-$, $\cdot$, and $\sqrt{{}}$.

I don't think it's difficult to believe this is the case, but I don't know of a nice, clear way to prove it.

I tried to simplify the argument first by saying that any operation $+$, $-$, $\cdot$, $\div$ can be rewritten entirely in terms of $+$, since $-$ is the addition of a the additive inverse, $\cdot$ is repeated addition, and $\div$ is multiplication of the reciprocal.

I also decided that given a radicand consisting of rationals and operations $+$, $-$, $\cdot$, $\div$, you could then rewrite the radicand as a sum of rationals, and multiply the radical by $\frac{\sqrt{\Delta^2}}{\Delta}$, where $\Delta$ is some multiple of the denominators. This would put the radicand in standard form, with a fraction $\frac{1}{\Delta}$ outside the radical. For example, $$ \sqrt{\frac{a}{b}+\frac{c}{d}}=\sqrt{\frac{ad+cb}{bd}}=\frac{\sqrt{(bd)^2}}{bd}\sqrt{\frac{ad+cb}{bd}}=\frac{1}{bd}\sqrt{(bd)(ad+cb)}. $$

I suppose for nested radicals, the process could be repeated. Then any sequence of these operations can be put into a sum of terms with rational denominators by rationalizing any denominator, and then one could do the same procedure of pulling a common multiple of all the terms out in front to put the expression in standard form.

However, I wanted to formalize this idea, but it seems like there are so many different things to consider, that I don't know how to go about it. Is there a way to make basic induction work, or perhaps some other method? Thanks.

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up vote 9 down vote accepted

This sort of thing is proved by induction over the number of operations by treating the "outermost" operation:

Case $+$: $\left(\frac{p}{q}A\right)+\left(\frac{r}{s}B\right)=\frac{1}{qs}(psA+qrB)$

Case $-$: $\left(\frac{p}{q}A\right)-\left(\frac{r}{s}B\right)=\frac{1}{qs}(psA-qrB)$

Case $\cdot$: $\left(\frac{p}{q}A\right)\cdot\left(\frac{r}{s}B\right)=\frac{pr}{qs}(AB)$

Case $\div$: $\left(\frac{p}{q}A\right)\div\left(\frac{r}{s}B\right)=\frac{ps}{qr}\frac{A}{B}$

Case $\sqrt{}$: $\sqrt{\frac{p}{q}A}=\frac{1}{q}\left(\sqrt{pqA}\right)$

In the first version of this answer, I claimed: "Now all that remains to be shown is that you can deal with the $B$ in the denominator in the $\div$ case, and this you can show with the same kind of induction by cases over the number of operations in $B$."

It turns out this was overly optimistic, and actually the more subtle part of the proof lies in dealing with that case. So here's a proper proof:

The proof is by induction over $(n,m)$, where $m$ is the number of operations in $B$ and $n$ is the number of different radicands that appear in $B$. For given $(n,m)$, assume that we can move all roots to the numerator if the denominator has $(n',m') < (n,m)$ in lexicographical order, i.e. either $n' < n$, or $n'=n$ and $m'<m$.

To simplify things, let's replace each "$-$" by "$+(-1)\cdot$", so only the operations $+$, $\cdot$ and $\sqrt{}$ remain.

Case $\sqrt{}$: This is easy:


and we've reduced the denominator's $n$ by $1$.

Case $\cdot$: This is also easy:

$$\frac{1}{B}=\frac{1}{C\cdot D}=\frac{1}{C}\frac{1}{D}\;,$$

and we can apply the induction hypothesis separately to $C$ and $D$ (both of which have lower $m$ than $B$ and at most equal $n$) to move all roots in both fractions to the numerator.

Case $+$: This is the harder part. We multiply out any products of sums that are not inside a root. This may increase $m$ but leaves $n$ unchanged, and it yields a sum of products of one or more factors, each of which is an integer or a root. If the same root appears twice in a product, we replace the two occurrences by the radicand (which doesn't increase $n$) and then again multiply out all products of sums, iterating until there are no more roots occurring twice in the same term of the outer sum. (This process must terminate since each radicand inserted when resolving a duplicate root contains fewer operations than the root, and only roots in this radicand can lead to new duplicates.) Now we choose a root among the roots in the products that contains the greatest number of operations (picking one arbitrarily if there are several with the same number of operations), and we factor it out of any terms in which it occurs. Since it occurs at most once in each term (since we resolved double occurrences) we get an expression of the form $C + D \cdot \sqrt{E}$, where $\sqrt{E}$ occurs nowhere else in the expression. (It cannot occur outside of roots since we factored it out, and it cannot occur inside roots since it is the root containing the greatest number of operations.) Now we expand the fraction with $C - D\sqrt{E}$, yielding

$$\frac{1}{B}=\frac{1}{C+D\cdot\sqrt{E}}=\frac{C-D\cdot\sqrt{E}}{C\cdot C - D\cdot D \cdot E}\;.$$

We may have greatly increased $m$, but we have decreased $n$, since $E$ no longer occurs as a radicand in the denominator and no new radicands have been introduced. That completes the induction.

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Thanks joriki. I'm not quite sure what you mean by dealing with $B$. For example, suppose $+$ is the outermost operation in $B$. Then $$B=\frac{a}{b}C+\frac{c}{d}D=\frac{1}{bd}(adC+bcD)$$ which implies $$\frac{ps}{qr}\frac{A}{B}=\frac{psbd}{qr}\frac{A}{adC+bcD}?$$ Is this the way to go? What can I conclude from that? –  yunone Mar 7 '11 at 10:46
@yunone: You used the wrong form for $B$ -- the form for the overall expressions instead of the integer expressions, which are supposed to use only $+$, $-$, $\cdot$ and $\sqrt{}$. The case for $+$ would be $B=C+D$, $1/B=1/(C+D)=(C-D)/(C^2-D^2)$, so in fact I was wrong and you can't simply do induction on the number of operations in $B$, but have to use some more complicated ordering like the lexical order of the number of radicals and the number of other operations to show that the recursion terminates -- I'll think about that a bit and correct the answer. –  joriki Mar 7 '11 at 10:58
@yunone: Actually, the induction has to be on the number of radicals in $B$ -- I'll post a proper proof in the answer in a bit. –  joriki Mar 7 '11 at 11:16
@joriki: Is this normal form guaranteed to be unique? That is, for $A,B$ of different expression but $A=B$, are reduced($A$) and reduced($B$) guaranteed to be equivalent expressions? (I think √ poses a problem) –  Mitch Mar 7 '11 at 14:27
@Mitch: I don't think any of this is remotely unique. For instance, every integer in the integer part of the expression can be expressed as a sum of other integers to get a different expression of the same form with the same value. If uniqueness is desired (there was no mention of it in the question), one would have to first define what "reduced(A)" is to be -- no such concept had occurred either in the question or in my (still incomplete) proof. –  joriki Mar 7 '11 at 14:48
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