Rationalize the denominator in $\frac{1}{1+ \sqrt2+ \sqrt3}$

Question

I found this method in a book.

To rationalize the denominator in $\frac{1}{1+\sqrt2 +\sqrt3}$,

we multiply denominator and numerator so that we get the denominator $$(1+\sqrt2 +\sqrt3)(1+\sqrt2 -\sqrt3)(1-\sqrt2 +\sqrt3)(1-\sqrt2 -\sqrt3)$$.

The mapping $\sqrt2 \mapsto -\sqrt2$ and $\sqrt3 \mapsto -\sqrt3$ leaves the term unchanged. Thus the term is rational.

My questions are: 1) What does mapping mean? Does it mean the transformation of $\sqrt 3 $ to $-\sqrt 3$ leaves the term unchanged?

2) Why does that imply that the term(denominator) is rational?

3) In general, if we have any any expression involving any number of such radical symbols that does not change under all the possible transformations of irrational terms then does it always imply that the expression is rational? If so, Why?

Note: I would really appreciate it if you could prove/explain this without using Galois theory as I haven't learnt it yet.

Your question is actually about Galois theory. The maps are authomorphisms of certain field extentions and there is a result that the only elements unchanged under all authomorphisms are in the ground field (rational numbers in this case). — Invincible, Apr 21 '20 at 14:33
I can explain it in more details if what I just said makes any sense to you. — Invincible, Apr 21 '20 at 14:34
I'm sorry. I don't know Galois Theory. Is there any other way you could derive this result? — Aditya, Apr 21 '20 at 14:43
Someone please tell me if the following argument is correct: — Aditya, Apr 22 '20 at 12:54
Since the expression is invariant under the transformation $x \mapsto -x$ and $y \mapsto -y$, the final expression must have even power of $x$ and $y$ ( power is at least 2) and hence it must be rational. — Aditya, Apr 22 '20 at 12:57

score 2 · Answer 1 · answered Apr 21 '20 at 15:50

You are over-thinking the problem. If you don't want to get into Galois theory, but just want an algorithm to rationalize the denominator, all you need is the slogan "Get rid of one square root at a time." In your example, first multiply numerator and denominator by $(1+\sqrt 2)-\sqrt 3 $ But if you would like a deeper understanding, I suggest reading Zariski and Samuel's book on Commutative algebra, especially the chapter on field extensions/Galois theory.

Will Jagy · Answer 2 · 2020-04-22T15:26:36.830

This is the identity from Heron's formula for the area of a triangle, given all three sides.

$$ (x+y+z) (-x+y+z)(x-y+z)(x+y-z) \; \; \color{magenta}{=} \color{red}{ \; \; 2 \left( y^2 z^2 + z^2 x^2 + x^2 y^2 \right) - x^4 - y^4 - z^4 } $$

When two or three of $x,y,z$ are square roots, this product clears them all because every term is squared. Multiply it out in this $x,y,z$ version, it will help you remember

Apparently this works with more variables: keep the first variable with no minus sign, and multiply the $2^{n-1}$ expressions with the other signs taking all combinations of $\pm$ signs. It is probably not too hard to prove that all exponents in the resulting polynomial are even, thus clearing all the square roots. I had it write out $n=4.$ Note that the coefficient of $w^2 x^2 y^2 z^2$ is $-40$

? f = (w + x + y + z ) * (w - x + y + z ) * (w + x - y + z ) * (w + x + y - z ) * 
  (w - x - y + z ) * (w - x + y - z ) * (w + x - y - z ) * (w - x - y - z )
%1 = x^8 + (-4*w^2 + (-4*y^2 - 4*z^2))*x^6 +
 (6*w^4 + (4*y^2 + 4*z^2)*w^2 + (6*y^4 + 4*z^2*y^2 + 6*z^4))*x^4 +
 (-4*w^6 + (4*y^2 + 4*z^2)*w^4 + (4*y^4 - 40*z^2*y^2 + 4*z^4)*w^2 + (-4*y^6 + 4*z^2*y^4 + 4*z^4*y^2 - 4*z^6))*x^2 +
 (w^8 + (-4*y^2 - 4*z^2)*w^6 + (6*y^4 + 4*z^2*y^2 + 6*z^4)*w^4 + (-4*y^6 + 4*z^2*y^4 + 4*z^4*y^2 - 4*z^6)*w^2 + (y^8 - 4*z^2*y^6 + 6*z^4*y^4 - 4*z^6*y^2 + z^8))
? 

? polcoeff(f, 8,x)
%2 = 1
? polcoeff(f, 7,x)
%3 = 0
? polcoeff(f, 6,x)
%4 = -4*w^2 + (-4*y^2 - 4*z^2)
? polcoeff(f, 5,x)
%5 = 0
? polcoeff(f, 4,x)
%6 = 6*w^4 + (4*y^2 + 4*z^2)*w^2 + (6*y^4 + 4*z^2*y^2 + 6*z^4)
? polcoeff(f, 3,x)
%7 = 0
? polcoeff(f, 2,x)
%8 = -4*w^6 + (4*y^2 + 4*z^2)*w^4 + (4*y^4 - 40*z^2*y^2 + 4*z^4)*w^2 + (-4*y^6 + 4*z^2*y^4 + 4*z^4*y^2 - 4*z^6)
? polcoeff(f, 1,x)
%9 = 0
? polcoeff(f, 0,x)
%10 = w^8 + (-4*y^2 - 4*z^2)*w^6 + (6*y^4 + 4*z^2*y^2 + 6*z^4)*w^4 + (-4*y^6 + 4*z^2*y^4 + 4*z^4*y^2 - 4*z^6)*w^2 + (y^8 - 4*z^2*y^6 + 6*z^4*y^4 - 4*z^6*y^2 + z^8)
?

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What about the third question that I posed? This identity will be true only for an expression having 3 numbers under the radical sign. What about $n$ terms, each under the radical symbol? — Aditya, Apr 22 '20 at 04:44
For example, $\sqrt a_1 +\sqrt a_2 + \cdots +\sqrt a_n$ multiplied with all possible permutations of signs within that expression will be rational. How would you prove this? — Aditya, Apr 22 '20 at 04:46

Rationalize the denominator in $\frac{1}{1+ \sqrt2+ \sqrt3}$

2 Answers2