Fun with algebraic number theory
Dave Rusin
The following example was pointed out to me by John Wolfskill.
(1) Show that sqrt( 8 + 3 sqrt(7) ) is the sum of the square roots of
two integers.
(2) Generalize.
This came up as an illustration of some topics in Galois theory -- what
looks like it ought to involve a dihedral Galois group lies in a field
with group Z/2 x Z/2 -- but the topic expands a bit upon generalization,
e.g., just which two quadratic extensions of Q have a compositum big
enough to hold the required algebraic numbers.
dave
Robin Chapman
Do you really need algebraic nuumber theory to solve this sort of
problem? Let's consider when sqrt(a + sqrt(b)) with a and b rational is
a sum of two square roots of rationals. We want to solve
a + sqrt(b) = (sqrt(c) + sqrt(d))^2
with c, d rational, and dismissing trivial cases we have to have
a = c + d, and sqrt(b) = 2 sqrt(c)sqrt(d) and so b/4 = cd. Thus
c and d are roots of a quadratic and can be found.
In this example, c + d = 8 and cd = 63/4 giving, c = 9/2 and
d = 7/2 so that sqrt(8 + 3sqrt(7)) = sqrt(9/2) + sqrt(7/2).
But, Dave, did you really want square roots of *integers"?
--
Robin Chapman
http://www.maths.ex.ac.uk/~rjc/rjc.html
"`Well, I'd already done a PhD in X-Files Theory at UCLA, ...'"
Greg Egan, _Teranesia_
Sent via Deja.com http://www.deja.com/
Before you buy.
Kurt Foster
<ru...@vesuvius.math.niu.edu> said:
> The following example was pointed out to me by John Wolfskill.
> (1) Show that sqrt( 8 + 3 sqrt(7) ) is the sum of the square roots of
> two integers.
Hmmm. (sqrt(a) + sqrt(b))^2 = (a+b) + 2*sqrt(a*b). It's kind of hard
to get 8 + 3*sqrt(7) if a and b are integers.
However, 8 + 3*sqrt(7) = [(sqrt(18) + sqrt(14))/2]^2.
. (2) Generalize.
Well, in the above, I tried to get a*b to be 7 times a square, and there
had to be a factor of 3 in the root of that square. I made a try using
sqrt(9) + sqrt(7) and was able to get the above from that. The only other
prime I had to contend with was 2.
Note that 8 + 3*sqrt(7) is a unit u in the quadratic field K = Q(u). In
this situation, if L = K(sqrt(u)), about the only possibilities for primes
that ramify in L/K are 2 and (if u < 0) the infinite prime. This limits
the possibilities for expressions of the form sqrt(a) + sqrt(b) being
equal to u, or perhaps to 4u.
Pertti Lounesto
> The following example was pointed out to me by John Wolfskill.
>
> Show that sqrt(8+3sqrt(7)) is the sum of the square roots of
> two half-integers.
The square-root of a complex number x+iy is sqrt(x+iy) =
sqrt((sqrt(x^2+y^2)+x)/2)+i*sqrt((sqrt(x^2+y^2)-x)/2).
Slightly modifying this, we have sqrt(x+sqrt(y)) =
sqrt((x+sqrt(x^2-y))/2)+sqrt((x-sqrt(x^2-y))/2), which
gives sqrt(8+sqrt(3^2*7)) = sqrt((8+sqrt(64-63))/2)+
sqrt((8-sqrt(64-63))/2) = sqrt(9/2)+sqrt(7/2).
I was a bit astonished to learn that this al-gebraic
simplification, reducing a square-root from below
a square-root, known already to Arabs, was not
implemented into Mathematica or Maple.
So disappears the mathematical knowledge, at the
altar of advancing technology, put forward by math.
Lee Rudolph
>The square-root of a complex number x+iy is sqrt(x+iy) =
>sqrt((sqrt(x^2+y^2)+x)/2)+i*sqrt((sqrt(x^2+y^2)-x)/2).
>Slightly modifying this, we have sqrt(x+sqrt(y)) =
>sqrt((x+sqrt(x^2-y))/2)+sqrt((x-sqrt(x^2-y))/2), which
>gives sqrt(8+sqrt(3^2*7)) = sqrt((8+sqrt(64-63))/2)+
>sqrt((8-sqrt(64-63))/2) = sqrt(9/2)+sqrt(7/2).
>
>I was a bit astonished to learn that this al-gebraic
>simplification, reducing a square-root from below
>a square-root, known already to Arabs, was not
>implemented into Mathematica or Maple.
>
>So disappears the mathematical knowledge, at the
>altar of advancing technology, put forward by math.
Is this simplification implemented into CLICAL?
Lee Rudolph
denis.feldmann
Pertti Lounesto a écrit dans le message <38492F3E...@hit.fi>...
>ru...@vesuvius.math.niu.edu (Dave Rusin) wrote:
>
>> The following example was pointed out to me by John Wolfskill.
>>
>> Show that sqrt(8+3sqrt(7)) is the sum of the square roots of
>> two half-integers.
>
>sqrt((sqrt(x^2+y^2)+x)/2)+i*sqrt((sqrt(x^2+y^2)-x)/2).
>Slightly modifying this, we have sqrt(x+sqrt(y)) =
>sqrt((x+sqrt(x^2-y))/2)+sqrt((x-sqrt(x^2-y))/2), which
>gives sqrt(8+sqrt(3^2*7)) = sqrt((8+sqrt(64-63))/2)+
>sqrt((8-sqrt(64-63))/2) = sqrt(9/2)+sqrt(7/2).
>
>I was a bit astonished to learn that this al-gebraic
>simplification, reducing a square-root from below
>a square-root, known already to Arabs, was not
>implemented into Mathematica or Maple.
>
This is Maple (version 4) output:
sqrt(8+sqrt(3^2*7));
1/2 1/2
3/2 2 + 1/2 14
I didn't even had to use radnormal...
Pertti Lounesto
> Is this simplification implemented into CLICAL?
No. CLICAL's design goals, in 1970's and 1980's,
were to develop an easy-to-use, calculator-type,
semi-symbolic computer program for non-degenerate
real Clifford algebras, the part of Clifford algebras
which I estimated could be explored during my
active life in science-making. Semi-symbolic
excluded such implementations.
denis.feldmann wrote:
> This is Maple (version 4) output:
>
> sqrt(8+sqrt(3^2*7));
>
> 1/2 1/2
> 3/2 2 + 1/2 14
Why don't you ask Maple, who informed them
about lack of this feature, in earlier versions?
Virgil
<Pertti....@hit.fi> wrote:
>The square-root of a complex number x+iy is sqrt(x+iy) =
>sqrt((sqrt(x^2+y^2)+x)/2)+i*sqrt((sqrt(x^2+y^2)-x)/2).
I believe that you mean that ONE of the square roots is given by the
above. The other square root is the negative of the one above.
--
Virgil
vm...@frii.com
Dave L. Renfro
[sci.math 3 Dec 1999 21:49:12 GMT]
<http://forum.swarthmore.edu/epigone/sci.math/foinulstix>
wrote
>The following example was pointed out to me by John Wolfskill.
>
>(1) Show that sqrt( 8 + 3 sqrt(7) ) is the sum of the square
> roots of
> two integers.
>
>(2) Generalize.
>
>This came up as an illustration of some topics in Galois
>theory -- what looks like it ought to involve a dihedral
>Galois group lies in a field with group Z/2 x Z/2 -- but
>the topic expands a bit upon generalization, e.g., just
>which two quadratic extensions of Q have a compositum big
>enough to hold the required algebraic numbers.
>
>dave
You might want be interested in
Webster Wells, "Advanced Course in Algebra",
D. C. Heath and Co., 1904 [pp. 235-237]
<< BEGIN "QUOTE" ---->>>>>
ARTICLE 390: If sqrt[a + sqrt(b)] = sqrt(x) + sqrt(y),
where a, b, x, and y are rational expressions, and
a greater than sqrt(b), then
sqrt[a - sqrt(b)] = sqrt(x) - sqrt(y).
[proof omitted]
ARTICLE 391: The preceding principles may be used to find
the square root of certain expressions which are in the
form of the sum of a rational expression and a quadratic
surd.
Example: Find the square root of 13 - sqrt(160).
Assume, sqrt[13 - sqrt(160)] = sqrt(x) - sqrt(y) (1).
Then by #390, sqrt[13 + sqrt(160)] = sqrt(x) + sqrt(y).
Multiplying the equations gives sqrt(169 - 160) = x - y.
Hence, x - y = 3. Squaring (1), 13 - sqrt(160) =
x - 2*sqrt(xy) + y. Hence, x + y = 13. [Citing the
previously proved result that a + sqrt(b) = c + sqrt(d),
where a, c are rational and b, d are quadratic surds,
implies a=c and b=d.] It now easily follows that x=8
and y=5.
<<<<<<------ END "QUOTE" >>
Using this method for sqrt[8 + 3*sqrt(7)] gives
sqrt[8 + 3*sqrt(7)] = sqrt(9/2) + sqrt(7/2).
You can't write sqrt[8 + 3*sqrt(7)] as a sum
sqrt(x) + sqrt(y) with x and y both integers, however.
[There are not many pairs of integers to check. For
instance, among other restrictions, 1 <= x,y <= 8,
since the number in question is less than 4.]
Wells goes on to show that
sqrt[8 + sqrt(48)] = sqrt(6) + sqrt(2)
sqrt[22 - 3*sqrt(32)] = 3*sqrt(2) - 2
sqrt(392) + sqrt(360) = [2^(1/4)]*[3 + sqrt(5)]
ARTICLE 394: It may be proved, as in #390, that if
cube:root[a + sqrt(b)] = x + sqrt(y), where a, x are
rational expressions and sqrt(b), sqrt(y) quadratic
surds, then cube:root[a - sqrt(b)] = x - sqrt(y).
***************************************************
***************************************************
Some more exotic surd transformation techniques can
be found in:
Salvatore Composto, "Sulla trasformaxione del radicale
sqrt{a + sqrt[b + sqrt(c)]}", Peridico di Matematica
(3) 4 (1907), 32-36.
Salvatore Composto, "Sulla trasformazione dei radicali
sqrt[a +/- fourth:root(b)], sqrt[sqrt(a) +/- fourth:root(b)],
sqrt[fourth:root(a) +/- fourth:root(b)]", Peridico di
Matematica (3) 4 (1907), 75-83.
Here are some examples from these papers:
1. sqrt{12 + sqrt[38 + 2*sqrt(105)]}
= (1/2)*[sqrt(15) + sqrt(7) + sqrt(21) - sqrt(5)]
2. sqrt{3 + sqrt[-20 + 12*sqrt(5)]}
= (1/2)*[sqrt(10) + 2*fourth:root(5) - sqrt(2)]
3. sqrt{sqrt(2) + sqrt[-6 + 4*sqrt(3)]}
= (1/2)*[fourth:root(18) + fourth:root(24) - fourth:root(2)]
4. sqrt[7 + 2*fourth:root(132)]
= (1/2)*{sqrt[6 + 2*sqrt(33)] + sqrt[22 - 2*sqrt(33)]}
There are additional examples in the papers I
cited if anyone wants to pursue this further.
Dave L. Renfro
Dave L. Renfro
[sci.math (hasn't posted yet)]
<http://forum.swarthmore.edu/epigone/sci.math/foinulstix>
wrote (first line only)
>You might want be interested in
^^^^^^^
Sigh (loudly) . . .
Dave L. Renfro
Tapio Hurme
Dave L. Renfro <dlre...@gateway.net> kirjoitti
viestissä:klea64...@forum.swarthmore.edu...
| Dave Rusin <ru...@vesuvius.math.niu.edu>
| [sci.math 3 Dec 1999 21:49:12 GMT]
| <http://forum.swarthmore.edu/epigone/sci.math/foinulstix>
| ARTICLE 394: It may be proved, as in #390, that if
| cube:root[a + sqrt(b)] = x + sqrt(y), where a, x are
| rational expressions and sqrt(b), sqrt(y) quadratic
| surds, then cube:root[a - sqrt(b)] = x - sqrt(y).
|
| ***************************************************
| ***************************************************
| Dave L. Renfro
ARTICLE 394 implies that SOME cuberoots are possible to construct with the
aid of compass (as seen earlier in this ng). :-)
Tapio
Dave Rusin
> The following example was pointed out to me by John Wolfskill.
sqrt(8+3 sqrt(7)) + sqrt(8-3 sqrt(7)) = 3 sqrt(2) .)
So let's turn this to
> (1) Show that sqrt( 8 + 3 sqrt(7) ) is the sum of the square roots of
two rational numbers.
In article <82b69o$trq$1...@nnrp1.deja.com>,
Robin Chapman <r...@maths.ex.ac.uk> wrote:
>Do you really need algebraic nuumber theory to solve this sort of
>problem? Let's consider when sqrt(a + sqrt(b)) with a and b rational is
Of course there are other ways to do this, but this made for a nice example
for Wolfskill's Galois theory class. Let alpha = sqrt( epsilon ) where
epsilon = a + b sqrt(D) is an element of a quadratic extension of Q.
Clearly alpha satisfies the rational polynomial (X^2 - a)^2 - b^2 D = 0,
and except for some special cases this is its minimal polynomial. (For
example, we need epsilon itself not to be a square!) Now,
both alpha and -alpha lie in the field K = Q(alpha), of course, but
the other two roots beta = sqrt(a-b sqrt(D)) and -beta do not in
general lie in K -- in the generic case the normal closure of K is
a degree-8 extension of Q with dihedral Galois group. On the other hand,
alpha*beta = sqrt( Norm(epsilon) ) [the norm being from the quadratic
field down to Q], so in the special case that epsilon is a unit with
norm = 1, then alpha and beta (or -beta) are inverses, and in
particular, the minimal polynomial of alpha splits completely in K,
K/Q is Galois, and Gal(K/4) is a group of order 4. Clearly the
action of the Galois group on the set of these four roots is limited to
the obvious sign changes, from which we conclude Gal(K/Q) is the four-group
Z2 x Z2 rather than the cyclic group Z4.
But now there are three intermediate subgroups of index 2 and hence
K contains three quadratic extensions of Q, any two of which must then
have K as their compositum. OK, well it's clear that Q(sqrt(D)) is
one of the intermediate quadratic fields. But where are the other two?
They should be found by taking orbit-sums under the Galois group.
I note that alpha +- beta has a rational square 2(a+-1), so that
K contains the quadratic subfields sqrt(2(a+-1)). (You can get one from
the other by noting that if the three fields are Q(sqrt(d_i)) then
we must have d1 d2 = d3 * (a square); and (a+1) (a-1) = D * (b^2). )
When D is prime, there's not a lot of variation in the intermediate
quadratic fields, but for example taking epsilon = 6 + sqrt(35)
gives a little variation.
So yes, there are easier ways to finish the problem of writing some
of these numbers in the form sqrt(x)+sqrt(y). They're just not as
much fun :-)
dave
Richard Bumby
material. Interested readers should track down the original postings.
Robin Chapman <r...@maths.ex.ac.uk> writes:
>In article <829dso$s46$1...@gannett.math.niu.edu>,
> ru...@vesuvius.math.niu.edu (Dave Rusin) wrote:
>> The following example was pointed out to me by John Wolfskill.
>> (1) Show that sqrt( 8 + 3 sqrt(7) ) is the sum of the square roots
>> of two integers.
>> (2) Generalize.
>> This came up as an illustration of some topics in Galois theory --...<snip>...
>Do you really need algebraic nuumber theory to solve this sort of
>problem? ...<snip>... We want to solve
>a + sqrt(b) = (sqrt(c) + sqrt(d))^2 ...<snip>... we have to have
>a = c + d, and sqrt(b) = 2 sqrt(c)sqrt(d) and so b/4 = cd. ...<snip>...
>c and d are roots of a quadratic and can be found.
This approach to sqrt(a + sqrt(b)) goes back to Euclid, but it still
seems fresh. Modern notation makes it look easy, but it can appear in
unexpected places. In looking for some topics to lecture on while my
students were finishing the exercises assigned earlier, I was led to
reexamining something I published in volume 25 (1987, p. 62) of the
Fibonacci Quarterly, applying this to the "incredible identity" that
Dan Shanks published in vol. 12 (1974, p. 271) of the same journal.
Maybe you don't need algebraic number theory to solve the problem, but
there is a strong connection. Consider, for example, an article of
Shanks in the 7th SE conference on combinatorics, graph theory and
computing (1976).
--
R. T. Bumby ** Rutgers Math || Amer. Math. Monthly Problems Editor 1992--1996
bu...@math.rutgers.edu ||
Telephone: [USA] 732-445-0277 (full-time message line) FAX 732-445-5530