Radicalar equalities ad libitum...

[alainv...@gmail.com]

Bonjour,

Building radicalar equalities seems very easy.
We may lay upon simple things like
(u^n)^(1/n) = u and a+b+(d-a-b) = d

What about, say ,
(108 - 40sqrt(2))^(1/2) +(1800+434sqrt(2))^(1/3)
+(827175+257629sqrt(2))^(1/5)

=

Amicalement,

Alain

[quasi]

It equals 37.

Letting

A = 108 - 40*sqrt(2)
B = 1800 + 434*sqrt(2)
C = 827175 + 257629*sqrt(2)

we get

A = a^2
B = b^3
C = c^5

where

a = 10 - 2*sqrt(2)
b = 12 + sqrt(2)
c = 15 + sqrt(2)

Thus

A^(1/2) + B^(1/3) + C^(1/5) = a + b + c = 37

However, letting

K = {x + y*sqrt(2) | x,y in Q}

I dare you to find A,B,C in K such that

A^(1/2) + B^(1/3) + C^(1/5)

is in K, but at least one of

A^(1/2), B^(1/3), C^(1/5)

is not in K.

quasi

[alainv...@gmail.com]

On 17 fév, 10:54, quasi <qu...@null.set> wrote:
> On Thu, 17 Feb 2011 00:57:50 -0800 (PST), "alainvergh...@gmail.com"

Bonjour Quasi,

Being in K doesn't matter.
We might have chosen
a = 10 - 2*sqrt(3)
b = 12 + sqrt(3)
c = 15 + sqrt(3)

Alain

[quasi]

Sure, but then you have a new field K, namely

K = {x + y*sqrt(3) | x,y in Q}

The point is, you've made sure that the expressions
inside the outer radicals are identically perfect
powers of the right type. Thus, each radical
simplifies separately.

quasi

[alainv...@gmail.com]

On Feb 17, 11:49 am, quasi <qu...@null.set> wrote:
> On Thu, 17 Feb 2011 02:29:53 -0800 (PST), "alainvergh...@gmail.com"

> Sure, but then you have a new field K, namely
>
>    K = {x + y*sqrt(3) | x,y in Q}
>
>>

> - Show quoted text -

Well,

I do catch your meaning.

I will work something else,

Amicalement,

Alain

[achille]

On Feb 17, 5:54 pm, quasi <qu...@null.set> wrote:

> However, letting
>
>    K = {x + y*sqrt(2) | x,y in Q}
>
> I dare you to find A,B,C in K such that
>
>    A^(1/2) + B^(1/3) + C^(1/5)
>
> is in K, but at least one of
>
>    A^(1/2), B^(1/3), C^(1/5)
>
> is not in K.
>
> quasi

I don't think there are such A,B,C.

Let's consider the case none of A^(1/2), B^(1/3), and
C^(1/5) are in K as an example. Since char(K) = 0, we
know for any prime p and D in K, polynomials of the
form x^p - D is either irreducible or has a root in K.
This implies the three polynomials

x^2 - A, x^3 - B and x^5 - C

are all irreducible. Let a, b, c be roots of these
polynomials in some splitting field of K, we have:

[K(a):K] = 2, [K(b):K] = 3, [K(c):K] = 5.

Since gcd(2,3) = 1
=> [K(a,b):K] = 2*3 = 6.
=> [K(c):K] = 5 not a divisor of [K(a,b):K]
=> c not in K(a,b)
=> a + b + c not in K.

[quasi]

That's why I dared Alain to find such A,B,C -- I knew there was no
chance of success (an evil trap, I admit).

quasi