Can I have fries and a calculator with that?
Dave L. Renfro
something else. It speaks for itself.
----------------------------
http://www.vbforums.com/showthread.php?referrerid=43870&t=298609
How do i rationalize the denominator in this?
6/sqr(x) + sqr(3) I know for somthing like
7/sqr(4) i can use a rationalizing factor of
sqr(4) to end up getting 7sqr(4)/4 but then
im not sure if this ends up being reduced to
7sqr(1). So i guess im asking two quesitons.
Thanks.
----------------------------
I'm almost tempted to think this was a troll post
given the "two quesitons" at the end, but I think
doing so requires too much textual interpretation
to be realistic.
Dave L. Renfro
Virgil
If that denominator is (sqrt(x) + sqrt(3)), you can rationalize it by
multiplying the numerator and denominator by (sqrt(x) - sqrt(3)) or by
(-sqrt(x) + sqrt(3)).
To rationalize a denomonator of sqrt(4), note that sqrt(4) = 2, which is
already rational.
Dave L. Renfro
>> I came across the following while searching for
>> something else. It speaks for itself.
>>
>> ----------------------------
>>
>> http://www.vbforums.com/showthread.php?referrerid=43870&t=298609
>>
>> How do i rationalize the denominator in this?
>> 6/sqr(x) + sqr(3) I know for somthing like
>> 7/sqr(4) i can use a rationalizing factor of
>> sqr(4) to end up getting 7sqr(4)/4 but then
>> im not sure if this ends up being reduced to
>> 7sqr(1). So i guess im asking two quesitons.
>> Thanks.
>>
>> ----------------------------
>>
>> I'm almost tempted to think this was a troll post
>> given the "two quesitons" at the end, but I think
>> doing so requires too much textual interpretation
>> to be realistic.
Virgil wrote:
> If that denominator is (sqrt(x) + sqrt(3)), you can
> rationalize it by multiplying the numerator and denominator
> by (sqrt(x) - sqrt(3)) or by (-sqrt(x) + sqrt(3)).
>
> To rationalize a denomonator of sqrt(4), note that
> sqrt(4) = 2, which is already rational.
Uh ... I think we all know this, or at least I hope we do.
My point (see thread title) was that this person appears
to suffer from that malady some people get from calculator
overuse, where they aren't able to recognize sqrt(4) and
sqrt(1) as 2 and 1, and where they think sqrt(4) cancels
4 in sqrt(4)/4 to produce sqrt(1).
As for rationalizing denominators, here's a less trivial
example for you. We can rationalize the denominator of 1/b,
where
b = sqrt(2) + sqrt(10) + sqrt(12) + sqrt(56),
by multiplying the numerator and denominator by f(b), where
f(x) = x^15 - (640)*x^13 + (155,072)*x^11 - (18,296,832)*x^9
+ (1,125,983,744)*x^7 - (35,305,193,472)*x^5
+ (491,646,992,384)*x^3 - (1,840,594,812,928)*x.
After multiplying, but before reducing, the denominator will be
- 525,242,269,696.
The numerator will be an integer-linear combination of 15
rationally independent square root terms.
Dave L. Renfro
Virgil
"Kobu" <kobu....@gmail.com> wrote:
But can you in general rationalize a denominator of form
sqrt(a) + sqrt(b) + sqrt(c) + sqrt(d) + sqrt(e),
for natural numbers a,b,c,d, and e?
Bill Dubuque
> example for you. We can rationalize the denominator of 1/b,
>
> where b = sqrt(2) + sqrt(10) + sqrt(12) + sqrt(56),
>
> by multiplying the numerator and denominator by f(b), where
>
> f(x) = x^15 - (640)*x^13 + (155,072)*x^11 - (18,296,832)*x^9
>
> + (1,125,983,744)*x^7 - (35,305,193,472)*x^5
>
> + (491,646,992,384)*x^3 - (1,840,594,812,928)*x.
>
> After multiplying, but before reducing, the denominator will be
>
>
> The numerator will be an integer-linear combination of 15
> rationally independent square root terms.
I.e. g(x) := x f(x) - d is the minimum polynomial of b over Q.
Of course one may employ the minimum poly of any algebraic number
w in a similar manner to rationalize w in a denominator. This is
essentially the same as multiplying w by all of its conjugates.
This property characterizes algebraic (vs. transcendental) numbers,
w is algebraic iff w f(w) is integral for some poly f(w), i.e.
w algebraic over domain D <-> w|d in D[w] for some d in D
NOTES
1) The above method of rationalizing denominators is actually
just a special case of employing the extended euclidean algorithm
to compute inverses in Q[x]/g(x). i.e. Bezout gcd of x, g(x) is
- g(x) + f(x) x = d
hence within Q[x]/g(x): f(x)/d = 1/x, so inverting x there.
This should be familiar to anyone who has used the extended
euclidean algorithm to compute inverses in Z/m = integers (mod m).
There are essentially equivalent ways to do this in terms of
norms via, resultants, Grobner bases, etc, e.g. see
----------------------------------------------------------------------------
91m:13011 13F20 (13B99)
Myerson, G. (5-MCQR)
Norms in polynomial rings.
Bull. Austral. Math. Soc. 41 (1990), no. 3, 381-386.
----------------------------------------------------------------------------
Let A be a ring, and B a ring containing A such that B is a finitely
generated A-module. The norm of b in B, N_A^B b, is the determinant
of multiplication by b in B as a linear operator on B. The following
THEOREM is proved: Let A be an integral domain and B = A[X_1,...,X_n]/I;
then for f in A[X_1,..,X_n] we have N_A^B ~f = prod_{P in Z(I)} f(P)^{m_P},
where m_P is the multiplicity of P in Z(I).
COROLLARY: If g in A[X] is monic, B = A[X]/(g), then for all f in A[X],
R(f,g) = N_A^B ~f, where R(f,g) is the resultant.
Reviewed by Ralf Friberg
----------------------------------------------------------------------------
2) Simpler than b is b/2 = (1 + /5 + /6 + 2 /7 )/ /2, with min poly
16 14 12 10 8 6
X - 160 X + 9692 X - 285888 X + 4398374 X - 34477728 X
4 2 2 2
+ 120031004 X - 112340992 X + 19 149
--Bill Dubuque
Peter Webb
"Bill Dubuque" <w...@nestle.csail.mit.edu> wrote in message
news:y8zwtid...@nestle.csail.mit.edu...
> "Dave L. Renfro" <renf...@cmich.edu> wrote:
>>
>> As for rationalizing denominators, here's a less trivial
>> example for you. We can rationalize the denominator of 1/b,
>>
>> where b = sqrt(2) + sqrt(10) + sqrt(12) + sqrt(56),
>>
>> by multiplying the numerator and denominator by f(b), where
>>
>> f(x) = x^15 - (640)*x^13 + (155,072)*x^11 - (18,296,832)*x^9
>>
>> + (1,125,983,744)*x^7 - (35,305,193,472)*x^5
>>
>> + (491,646,992,384)*x^3 - (1,840,594,812,928)*x.
>>
>> After multiplying, but before reducing, the denominator will be
>>
>> d = -525,242,269,696.
>>
>> The numerator will be an integer-linear combination of 15
>> rationally independent square root terms.
>
> I.e. g(x) := x f(x) - d is the minimum polynomial of b over Q.
> Of course one may employ the minimum poly of any algebraic number
> w in a similar manner to rationalize w in a denominator. This is
> essentially the same as multiplying w by all of its conjugates.
Indulge me here a little, this is a hobby (not a good start, I know).
We know that sin (90 degrees) = 1, and we know that the value of sin(45
degrees) satisfies the equation 2x^2 = 1.
We further know that the value of sin(1 degree) satisfies some equation
sigma (choose(90,i)x^i) = 1 (or similar).
Therefore sin(1 degree) is algebraic, and so is sin(1+1).
We can therefore construct sin(n) or indeed sin(m/n) as algebraic.
Therefore sin maps all rational degrees to algebraic numbers.
The interesting (for me) is the fact that this doesn't occur for sin in
radians, where sin(1) seems unobtainable within algebraics.
Dave L. Renfro
> But can you in general rationalize a denominator of
> form sqrt(a) + sqrt(b) + sqrt(c) + sqrt(d) + sqrt(e),
> for natural numbers a,b,c,d, and e?
Do you remember asking essentially the same question back
on October 26, 2002 in the sci.math thread "rationalize
this denominator"? (I mean the question, of course, not the
exact date and thread title.) There were several follow-ups
in the thread that discussed your question.
-----------------------
sci.math "rationalize this denominator" (October 26, 2002)
http://groups.google.com/group/sci.math/msg/5cdc035e33c3edaf
> But how would you do it for, say,
> 1/(sqrt(2) + sqrt(3) + sqrt(5) + sqrt(7) + sqrt(11))?
-----------------------
Here's what I did with sqrt(2) + sqrt(10) + sqrt(12) + sqrt(56),
which should sufficiently illustrate what to do for any rational
linear combination of square roots of rational numbers. Sums of
n many m'th roots can be handled in a similar way (but the work
is much more involved), and any sum of different roots can be
rewritten as a sum of m'th roots for some fixed m by taking m
to be the least common multiple of the various root indices.
The essential idea is to find a polynomial P(x) with integer
coefficients (rational is sufficient, but integer is what
we'll usually get) such that the number in question is a
zero of P(x). By bringing out factors of x, we can assume
that P(0) is not zero. For example, if we somehow came up with
8x^5 - 2x^3 + x^2 that has the number as a zero, then we can
rewrite this as
(x^2)*(8x^3 - 2x + 1),
and hence we'd use 8x^3 - 2x + 1 for P(x).
Now note that P(x) - P(0) is divisible by x. Therefore, we have
P(x) - P(0) = x*f(x)
(put in x=b) 0 - P(0) = b*f(b)
- P(0) = b*f(b),
where f(x) is a polynomial with integer coefficients and b is
the number in question. We can use this last equation to
rationalize the denominator in the following way.
Multiplying the numerator and denominator of 1/b by f(b) gives
f(b) / [b*f(b)] = f(b) / [-P(0)],
which rationalizes the denominator. Note that b, b^2, b^3, etc.
will each be an integer-linear combination of radicals, and
hence f(b) will also be an integer-linear combination of radicals.
So it comes down to finding a polynomial P(x) such that P(b) = 0.
Algebraists will usually try to get the minimal polynomial for b,
but since I'm not an algebraist, I'm not going to worry whether
the polynomial P(x) I get is minimal or not. (Note that the
argument I gave above didn't require P(x) to be minimal.)
In the case of rational-linear combinations of roots of various
orders (not necessarily square roots) of rational numbers,
there is a specific constructive way of doing this, which I'll
illustrate with
b = sqrt(2) + sqrt(10) + sqrt(12) + sqrt(56).
I don't know whether a _constructive_ method exists for any
explicit algebraic number, however. There are plenty of numbers
of the form sqrt[integer + sqrt(integer)] that can't be expressed
as a rational-linear combination of roots of rational numbers,
to say nothing of much more complicated nested radicals.
The idea will be to do a bit of reverse engineering to find a
polynomial P(x) such that P(b) = 0. Thus, we'll start with what's
usually the last step when solving an equation:
x = sqrt(2) + sqrt(10) + sqrt(12) + sqrt(56).
Rewrite each square root term so that every number under
a radical sign is square free.
x = sqrt(2) + sqrt(10) + 2*sqrt(3) + 2*sqrt(14).
Now consider the various prime factors of the numbers left under
any of the radical signs. In our case, these are 2, 3, 5, and 7.
What we'll do is to rewrite and square, on four separate occasions,
in such a way that we'll successively eliminate all numbers under
a radical sign that are divisible by 2, 3, 5, and 7.
To take care of 2, put all the radicals of numbers divisible
by 2 on one side and factor out sqrt(2):
sqrt(2) + sqrt(10) + 2*sqrt(14) = x - 2*sqrt(3)
[sqrt(2)]*[1 + sqrt(5) + 2*sqrt(7)] = x - 2*sqrt(3)
Squaring both sides gives
[sqrt(2)]^2 * [1 + sqrt(5) + 2*sqrt(7)]^2 = [x - 2*sqrt(3)]^2
68 + 4*sqrt(5) + 8*sqrt(7) + 8*sqrt(35) = x^2 - (4x)*sqrt(3) + 12.
To take care of 3, put all the radicals of numbers divisible
by 3 on one side, factor out sqrt(3) (this step not needed here),
and then square both sides:
[4x*sqrt(3)]^2 = [x^2 - 56 - 4*sqrt(5) - 8*sqrt(7) - 8*sqrt(35)]^2
.......... expand, combine terms, rearrange terms, square .........
This will leave us with only radicals of numbers whose prime
factors are 5 and/or 7. Repeat this process for 5, and then for 7:
[sqrt(5)]^2 * [terms with no radicals]^2
= [all other terms, including radicands divisible by 7]^2
.......... expand, combine terms, rearrange terms, square .........
[sqrt(7)]^2 * [terms with no radicals]^2
= [all other terms, none of which involve radicals]^2
Finally, putting all the terms on one side of the equals
sign and combining like terms will give us the following
16'th degree polynomial P(x) such that P(b) = 0.
P(x) = x^16 - (640)*x^14 + (155,072)*x^12 - (18,296,832)*x^10
+ (1,125,983,744)*x^8 - (35,305,193,472)*x^6
+ (491,646,992,384)*x^4 - (1,840,594,812,928)*x^2
- 525,242,269,696.
Thus, using the remarks I made just before I began the calculations
for finding P(x), we get the results that I stated in an earlier
post:
We can rationalize the denominator of 1/b, where
b = sqrt(2) + sqrt(10) + sqrt(12) + sqrt(56),
by multiplying the numerator and denominator by f(b), where
f(x) = x^15 - (640)*x^13 + (155,072)*x^11 - (18,296,832)*x^9
+ (1,125,983,744)*x^7 - (35,305,193,472)*x^5
+ (491,646,992,384)*x^3 - (1,840,594,812,928)*x.
After multiplying, but before reducing, the denominator will be
- 525,242,269,696.
The numerator will be an integer-linear combination of 15
rationally independent square root terms. In general, if you're
dealing with a sum of square roots and there are k many prime
factors of the radicands involved, then at the end you'll have
at most (2^k) - 1 many "different" square root terms showing up.
This is because the number of different square free integers
that could show up under a radical sign is the number of
different singleton to k-fold products of the prime numbers,
and this will be C(k,1) + C(k,2) + ... + C(k,k), where C(k,j)
is the binomial coefficient for the number of j-element subsets
of a set with k elements. This sum is 2^k - 1, since C(K,0) = 1
and C(K,0) + C(k,1) + ... + C(k,k) = 2^k (look at what happens
when you expand (1+1)^k to see this last part).
Dave L. Renfro
Dave L. Renfro
> Indulge me here a little, this is a hobby (not a good start,
> I know).
>
> We know that sin (90 degrees) = 1, and we know that the value
> of sin(45 degrees) satisfies the equation 2x^2 = 1.
>
> We further know that the value of sin(1 degree) satisfies some
> equation sigma (choose(90,i)x^i) = 1 (or similar).
>
> Therefore sin(1 degree) is algebraic, and so is sin(1+1).
>
> We can therefore construct sin(n) or indeed sin(m/n) as algebraic.
>
> Therefore sin maps all rational degrees to algebraic numbers.
>
> The interesting (for me) is the fact that this doesn't occur
> for sin in radians, where sin(1) seems unobtainable within
> algebraics.
You might be interested in these two July 1999
alt.math.undergraduate posts of mine:
http://groups.google.com/group/alt.math.undergrad/msg/097f0dbd69a82f71
http://mathforum.org/kb/message.jspa?messageID=684452
http://groups.google.com/group/alt.math.undergrad/msg/077c6d6c8c71ccbf
http://mathforum.org/kb/message.jspa?messageID=684453
I now know of probably about 3 times as many references in
the literature for this result than I gave in the second post,
but I'm not going to look them up and type them into a usenet
post right now, not after the essay on rationalizing denominators
that I just wrote and posted.
Dave L. Renfro
Gerry Myerson
Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
> This should be familiar to anyone who has used the extended
> euclidean algorithm to compute inverses in Z/m = integers (mod m).
> There are essentially equivalent ways to do this in terms of
> norms via, resultants, Grobner bases, etc, e.g. see
> ----------------------------------------------------------------------------
> 91m:13011 13F20 (13B99)
> Myerson, G. (5-MCQR)
> Norms in polynomial rings.
> Bull. Austral. Math. Soc. 41 (1990), no. 3, 381-386.
> ----------------------------------------------------------------------------
> Let A be a ring, and B a ring containing A such that B is a finitely
> generated A-module. The norm of b in B, N_A^B b, is the determinant
> of multiplication by b in B as a linear operator on B. The following
>
> THEOREM is proved: Let A be an integral domain and B = A[X_1,...,X_n]/I;
> then for f in A[X_1,..,X_n] we have N_A^B ~f = prod_{P in Z(I)} f(P)^{m_P},
> where m_P is the multiplicity of P in Z(I).
>
> COROLLARY: If g in A[X] is monic, B = A[X]/(g), then for all f in A[X],
> R(f,g) = N_A^B ~f, where R(f,g) is the resultant.
> Reviewed by Ralf Friberg
> ----------------------------------------------------------------------------
The reviewer is Froberg, umlaut over the o.
I never noticed before that the reviewer uses I without telling
the reader what it stands for. But I suppose it is clear from
the context that I is an ideal in A[X_1, ..., X_n].
Also, Z(I) means the set of all zeros of I over an algebraically
closed field k containing A, and in the paper B is required to be
finitely-generated and free.
For the corollary, A need not be an integral domain; any commutative
ring with unity will do. The corollary has appeared in one form or
another at least half a dozen times in the literature.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Dave L. Renfro
> We can rationalize the denominator of 1/b, where
>
> b = sqrt(2) + sqrt(10) + sqrt(12) + sqrt(56),
>
> by multiplying the numerator and denominator by f(b), where
>
> f(x) = x^15 - (640)*x^13 + (155,072)*x^11 - (18,296,832)*x^9
>
> + (1,125,983,744)*x^7 - (35,305,193,472)*x^5
>
> + (491,646,992,384)*x^3 - (1,840,594,812,928)*x.
>
> After multiplying, but before reducing, the denominator will be
>
> - 525,242,269,696.
>
> The numerator will be an integer-linear combination of 15
> rationally independent square root terms. In general, if you're
> dealing with a sum of square roots and there are k many prime
> factors of the radicands involved, then at the end you'll have
> at most (2^k) - 1 many "different" square root terms showing up.
I was so confident that all 15 square root terms would show up
when one rationalizes the denominator of
1 / [ sqrt(2) + sqrt(10) + sqrt(12) + sqrt(56) ]
that I didn't even bother to qualify my statement by saying
something like: "The numerator will be an integer-linear
combination of _at_most_ 15 rationally independent square
root terms." Well, I should have! This past weekend I
carried out the rationalization with a little help from
Scientific Workplace (MAPLE based) and it turns out that
only 8 square roots appear (i.e. 7 of the square root terms
that could have appeared have coefficients of zero).
A rationalization for this expression is:
(2117/11324) * sqrt(2) + (301/2831) * sqrt(3)
+ (681/11324) * sqrt(10) - (62/2831) * sqrt(14)
+ (425/5662) * sqrt(15) - (173/2831) * sqrt(21)
- (2117/11324) * sqrt(70) - (61/2831) * sqrt(105).
Dave L. Renfro