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Homesquare root of square root binomial

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# square root of square root binomial

Some people call the expressions of the form $a\!+\!b\sqrt{c}$ the square root binomials, especially when $c$ is an square-free integer greater than 1 (and $a$ and $b$ rational numbers). On the high school level one may learn to perform arithmetic operations between such binomials (see e.g. division), or also polynomials containing several square root terms. Taking the square root of a square root binomial is more difficult and usually results nested square roots. However, there are some exceptions if the numbers are appropriate. We have the formulae

$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^{2}-b}}{2}}+\sqrt{\frac{a-\sqrt{a^{2}-% b}}{2}}$ |

and

$\sqrt{a-\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^{2}-b}}{2}}-\sqrt{\frac{a-\sqrt{a^{2}-% b}}{2}}.$ |

If $a^{2}\!-\!b$ happens to be square of a rational number, then the formulae allow to convert the square roots on the left side into expressions without nested square roots.

For example, because $6^{2}\!-\!20=16=4^{2}$, we obtain

$\sqrt{6\!+\!2\sqrt{5}}=\sqrt{6\!+\!\sqrt{20}}=\sqrt{\frac{6\!+\!4}{2}}+\sqrt{% \frac{6\!-\!4}{2}}=1\!+\!\sqrt{5},$ |

and because $4^{2}\!-\!7=9=3^{2}$, we get

$\sqrt{4\!-\!\sqrt{7}}=\sqrt{\frac{4\!+\!3}{2}}-\sqrt{\frac{4\!-\!3}{2}}=\frac{% \sqrt{7}\!-\!1}{\sqrt{2}}=\frac{\sqrt{14}\!-\!\sqrt{2}}{2}.$ |

# References

- 1 K. Väisälä: Algebran oppi- ja esimerkkikirja I. – Werner Söderström osakeyhtiö, Porvoo & Helsinki (1952).

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## Comments

## Proof

This was very helpful...

Could anybody show a proof of how you would derive this formula? I would really appreciate it.

Thanks

## Re: Proof

There may be an elementary approach, but one could probably do this by putting it into polar coordinates, and then using something like De Moivre's formula.

Cam

## Re: Proof

Thank you so much...

Could you help me out with the "elementary approach" because I think that is what my professor is looking for...

DK

## Re: Proof

Rewrite the binomial formula as follows:

sqrt (x^2 + y^2 + 2xy) = x + y

Let x be one of the square roots appearing on the

right-hand-side of the formula you are trying

to prove and let y be the other square root.

After a bit of simplification, you will see

that x^2 + y^2 + 2xy = a + sqrt(b).

## Re: Proof

There is no cause to use imaginary numbers in such a formula concerning real numbers.

## Re: Proof

I disagree -- if you want to find the three real roots of certain cubic polynomials with real coefficients, you're naturally led to a calculation that invokes complex numbers.

Cam

## Re: Proof

I agree, Cam, it is the casus irreducibilis :)

Jussi

## Re: Proof

Well, that certainly is a case that requires complex numbers, but I was referring to the *reducible* case, i.e., polynomials with real coefficients and three *real* roots, the roots of which can be found using complex numbers (e.g., via Cardano's formulas). A quick google search gave me the example

x^3 - 15x - 4 = 0,

which if you were to use Cardano's formulas, would invoke some complex analysis, but has three real roots.

Cam

## Re: Proof

This is an example of casus irreducibilis, mentioned e.g. in the case 3 of http://planetmath.org/encyclopedia/CardanosFormulae.html

(The original thing was a square root formula.)

Jussi

## Re: Proof

Something's wrong there -- if a polynomial f with real coefficients has all real roots (say a_1,...,a_n) then it's necessarily reducible -- it factors into linear factors as

f(x) = a(x-a_1)(x-a_2)...(x-a_n)

for some constant a.

Ah, after doing some reading -- it appears that the "irreducibilis" in this phrase is not related to the irreducibility of the polynomial. The page casus irreducibilis entry should remove the irreducibility hypothesis.

Cam

## Re: Proof

The term "casus irreducibilis" refers to the fact that the roots cannot be expressed in terms of real radicals - that imaginary numbers are required. For example, x^3-3x-1 has real roots are 2cos(k\pi/9) for k=2,8,14, but is expressible in terms of radicals only using e^{2\pi/9} (for example, 2\cos 2\pi/9 = e^{2\pi/9}+e^{-2\pi/9}).

## Re: Proof

Dear Cam, your example equation really has three real roots, but they are not rational. And since the trinomial of the left side is in Q[x], it cannot be split in that ring into factors of less degree -- so it is irreducible.

BTW, the roots are the sums of the cube roots of 2+11i and 2-11i multiplied by a third root of unity and its square, resp.

Jussi

## Re: Proof

No. The polynomial may well be irreducible (e.g. x^3-15x-4; can you factorise it in Q[x]?).

It's true that f can be always be factorised as you write, but the real numbers a_j are not rational unless n is not a power of 2.

Jussi

## Re: Proof

> No. The polynomial may well be irreducible (e.g. x^3-15x-4;

> can you factorise it in Q[x]?).

> It's true that f can be always be factorised as you write,

> but the real numbers a_j are not rational unless n is not a

> power of 2.

> Jussi

No, but you can factor it in R[x]. The point is that the term "irreducible" is relative to a given ring. When you say that a polynomial has real coefficients (as it does in the entry casus irreducibilis), the most natural place to consider its irreducibility is in the ring R[x]. If I gave you the polynomial

pi * x^2 - e*x +3,

it's essentially nonsense to talk about whether it's irreducible or not in Q[x], since it's not an element of Q[x]! Thus the only way the term "irreducible" in that entry can be taken is to mean "irreducible in R[x]", which can't be the case if it has three real roots.

There are a couple of resolutions to this. You could insist that the coefficients be rational, in which case your use of the word irreducible is fine. This is perhaps the historically best solution for the "casus irredcubilis" entry since presumably this terminology was invented only for the case of integer/rational coefficients. On the other hand, the phenomenon extends to real coefficients, so it would be a slight disservice to unnecessarily leave them out. The other option would be to remove the word irreducible.

Cam

## Re: Proof

Please look the definition of "irreducible polynomial". In algebra, the irreducibility refers to the "field of the coefficients" of the polynomial. Your example \pi{xx}-ex+3 is irreducible since it does not split in Q[\pi, e] (this is not R). If the coefficients were all in Q, then the irreducibility is checked in Q.

## Re: Proof

I think cam has this one right. The entry on casus irreducibilis talks about polynomials over the reals, not over the rationals. This is incorrect, as the classical meaning of the term refers to the case where a (monic) cubic with integral coefficients has three real roots but is irreducible (over the rationals).

The reason for the term is that, in this case, the roots, while real, cannot be expressed in terms of real radicals since, as you point out, the roots would then be constructible. The casus irreducibilis occurs when the discriminant is positive.

When the discriminant is negative, the cubic has only one real root, so of course there is not any question of being able to represent the roots using real radicals.

Roger

## Re: Proof

Roger, you and cam are partially right -- maybe it were better to not speak of _real_ coefficients a_j, since the theorem is true without making this restrictive presumption. Essential is that P(x) is irreductible, i.e. irreductible in the field

Q(a_1, ..., a_n)

of its coefficients (this practice in using the concept "irreductible" is normal in algebra and algebraic number theory). Of course, the theorem if more concrete when the a_j's are are real.

Jussi

## Re: Proof

The theorem I mean in the preceding message is

http://planetmath.org/encyclopedia/CasusIrreducibilis.html

Jussi