Simplifying reciprocals of sums of square roots
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Consider the following expression: $1/(sqrt a_1 ± sqrt a_2 ± ... sqrt a_n)$ where $a_1 ... a_n$ are positive integers
I would like a general algorithm by which I can convert expressions of this form into expression of the following form: exp/k
where k is some positive integer and exp is an expression consisting of additions, subtractions, multiplications and square roots but no divisions.
Here is an example of a simple case:
$1/(sqrt 2 + sqrt 3) = (sqrt 3 - sqrt 2)/(3 - 2) = sqrt 3 - sqrt 2$
Here I multiplied the top and bottom by the conjugate. This technique can be used in the general problem for n≤4 by breaking the sum up into pairs of roots. I will illustrate this below in the case where all square root terms are added rather than having subtraction mixed in.
consider this denominator: $(sqrt a + sqrt b) + (sqrt c + sqrt d)$
multiplying by its conjugate: $(sqrt a + sqrt b) - (sqrt c + sqrt d)$ yields:
$(sqrt a + sqrt b)^2 - (sqrt c + sqrt d)^2$
$a+b+2sqrt (ab) - c - d - 2sqrt (cd) = $
$ A + (sqrt B - sqrt C)$ where $A= a+b-c-d$ and $B=4ab$ and $C=4cd$
multiplying by its conjugate $ A - (sqrt B - sqrt C)$ yields:
$ A^2 - (sqrt B - sqrt C)^2 =$
$ A^2 - B - C + 2sqrt (BC) =$
$ D + sqrt E =$ where $D=A^2-B-C$ and $E=4BC$
now multiply by its conjugate $ D - sqrt E$ yields:
$ D^2 - E$ which is an integer
I would like to know how to generalize this to n=5 and beyond, or if this is not possible, why that is the case. Thanks
@ancientmathemetician:
I am not sure how to do this for the case n=8. Consider the following expression: $(sqrt a_1 + sqrt a_2 + sqrt a_3 + sqrt a_4) + (sqrt a_5 + sqrt a_6 + sqrt a_7 + sqrt a_8)$
I could multiply by its conjugate which would yield the following
$(sqrt a_1 + sqrt a_2 + sqrt a_3 + sqrt a_4)^2 - (sqrt a_5 + sqrt a_6 + sqrt a_7 + sqrt a_8)^2$
However once each squared term is distributed out there will be 6 new square root terms introduced rather than just the original 4. Below I will do the first term:
$(sqrt a_1 + sqrt a_2 + sqrt a_3 + sqrt a_4)^2 =$
$a_1 + a_2 + a_3 + a_4 + 2sqrt (a_1a_2) + 2sqrt (a_1a_3) + 2sqrt (a_1a_4) + 2sqrt (a_2a_3) + 2sqrt (a_2a_4) + 2sqrt (a_3a_4)$
as a result, I go from 8 roots in the denominator to 12 and a single non root term
If instead I broke up the expression into a group of 5 square roots and a group of 3 square roots I would end up with 10 roots from the group of 5 and 3 from the group of 3 totalling 13.
in either case the number of roots does not go down, which is the trick I have been exploiting. It seems to me a new trick of sorts is required to find the general solution.
Thanks
algebra-precalculus
add a comment |
up vote
1
down vote
favorite
Consider the following expression: $1/(sqrt a_1 ± sqrt a_2 ± ... sqrt a_n)$ where $a_1 ... a_n$ are positive integers
I would like a general algorithm by which I can convert expressions of this form into expression of the following form: exp/k
where k is some positive integer and exp is an expression consisting of additions, subtractions, multiplications and square roots but no divisions.
Here is an example of a simple case:
$1/(sqrt 2 + sqrt 3) = (sqrt 3 - sqrt 2)/(3 - 2) = sqrt 3 - sqrt 2$
Here I multiplied the top and bottom by the conjugate. This technique can be used in the general problem for n≤4 by breaking the sum up into pairs of roots. I will illustrate this below in the case where all square root terms are added rather than having subtraction mixed in.
consider this denominator: $(sqrt a + sqrt b) + (sqrt c + sqrt d)$
multiplying by its conjugate: $(sqrt a + sqrt b) - (sqrt c + sqrt d)$ yields:
$(sqrt a + sqrt b)^2 - (sqrt c + sqrt d)^2$
$a+b+2sqrt (ab) - c - d - 2sqrt (cd) = $
$ A + (sqrt B - sqrt C)$ where $A= a+b-c-d$ and $B=4ab$ and $C=4cd$
multiplying by its conjugate $ A - (sqrt B - sqrt C)$ yields:
$ A^2 - (sqrt B - sqrt C)^2 =$
$ A^2 - B - C + 2sqrt (BC) =$
$ D + sqrt E =$ where $D=A^2-B-C$ and $E=4BC$
now multiply by its conjugate $ D - sqrt E$ yields:
$ D^2 - E$ which is an integer
I would like to know how to generalize this to n=5 and beyond, or if this is not possible, why that is the case. Thanks
@ancientmathemetician:
I am not sure how to do this for the case n=8. Consider the following expression: $(sqrt a_1 + sqrt a_2 + sqrt a_3 + sqrt a_4) + (sqrt a_5 + sqrt a_6 + sqrt a_7 + sqrt a_8)$
I could multiply by its conjugate which would yield the following
$(sqrt a_1 + sqrt a_2 + sqrt a_3 + sqrt a_4)^2 - (sqrt a_5 + sqrt a_6 + sqrt a_7 + sqrt a_8)^2$
However once each squared term is distributed out there will be 6 new square root terms introduced rather than just the original 4. Below I will do the first term:
$(sqrt a_1 + sqrt a_2 + sqrt a_3 + sqrt a_4)^2 =$
$a_1 + a_2 + a_3 + a_4 + 2sqrt (a_1a_2) + 2sqrt (a_1a_3) + 2sqrt (a_1a_4) + 2sqrt (a_2a_3) + 2sqrt (a_2a_4) + 2sqrt (a_3a_4)$
as a result, I go from 8 roots in the denominator to 12 and a single non root term
If instead I broke up the expression into a group of 5 square roots and a group of 3 square roots I would end up with 10 roots from the group of 5 and 3 from the group of 3 totalling 13.
in either case the number of roots does not go down, which is the trick I have been exploiting. It seems to me a new trick of sorts is required to find the general solution.
Thanks
algebra-precalculus
One small comment: you mustn't overdo it. For example, if $D=sqrt{E}$ you stop there and don't multiply by $D-sqrt{E}=0$.
– ancientmathematician
Nov 23 at 7:52
Hints: If I wanted to go along this path I'd do two things. (i) I'd persuade myself that the process applies equally well to $(b_1 sqrt a_1 ± b_2 sqrt a_2 ± ... b_n sqrt a_n)$ with integral $b_i$, some possibly $0$; and (ii) I'd show I could do it for $n=2,4,8,16, dots $. Then with the right number of zero $b_i$ I'd be done for all $n$.
– ancientmathematician
Nov 23 at 7:57
Thanks for pointing out the case where D=√E, didn't think of that. but ultimately if such a case arizes then one can just end the algorithm early. I agree it would be worth doing the more general case with b√a instead of just √a because it captures the changes in sign. As for showing it can be done for the cases n=2,4,8,16... I am not sure how to do it for the n=8 case so if you want to demonstrate that I'll accept that as an answer
– mathew
Nov 23 at 8:29
You are right to point out the way in which it won't work. So to save it I think you have to deal with the fully general case. That is, the general case of $b_0$, $b_0+b_1 sqrt{a_1}$, $b_0+b_1 sqrt{a_1}+b_2 sqrt{a_2} + b_{1,2}sqrt{a_1 a_2}$, $dots$ - in other words, allow for all the cross terms from the outset.
– ancientmathematician
Nov 23 at 10:15
add a comment |
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1
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up vote
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down vote
favorite
Consider the following expression: $1/(sqrt a_1 ± sqrt a_2 ± ... sqrt a_n)$ where $a_1 ... a_n$ are positive integers
I would like a general algorithm by which I can convert expressions of this form into expression of the following form: exp/k
where k is some positive integer and exp is an expression consisting of additions, subtractions, multiplications and square roots but no divisions.
Here is an example of a simple case:
$1/(sqrt 2 + sqrt 3) = (sqrt 3 - sqrt 2)/(3 - 2) = sqrt 3 - sqrt 2$
Here I multiplied the top and bottom by the conjugate. This technique can be used in the general problem for n≤4 by breaking the sum up into pairs of roots. I will illustrate this below in the case where all square root terms are added rather than having subtraction mixed in.
consider this denominator: $(sqrt a + sqrt b) + (sqrt c + sqrt d)$
multiplying by its conjugate: $(sqrt a + sqrt b) - (sqrt c + sqrt d)$ yields:
$(sqrt a + sqrt b)^2 - (sqrt c + sqrt d)^2$
$a+b+2sqrt (ab) - c - d - 2sqrt (cd) = $
$ A + (sqrt B - sqrt C)$ where $A= a+b-c-d$ and $B=4ab$ and $C=4cd$
multiplying by its conjugate $ A - (sqrt B - sqrt C)$ yields:
$ A^2 - (sqrt B - sqrt C)^2 =$
$ A^2 - B - C + 2sqrt (BC) =$
$ D + sqrt E =$ where $D=A^2-B-C$ and $E=4BC$
now multiply by its conjugate $ D - sqrt E$ yields:
$ D^2 - E$ which is an integer
I would like to know how to generalize this to n=5 and beyond, or if this is not possible, why that is the case. Thanks
@ancientmathemetician:
I am not sure how to do this for the case n=8. Consider the following expression: $(sqrt a_1 + sqrt a_2 + sqrt a_3 + sqrt a_4) + (sqrt a_5 + sqrt a_6 + sqrt a_7 + sqrt a_8)$
I could multiply by its conjugate which would yield the following
$(sqrt a_1 + sqrt a_2 + sqrt a_3 + sqrt a_4)^2 - (sqrt a_5 + sqrt a_6 + sqrt a_7 + sqrt a_8)^2$
However once each squared term is distributed out there will be 6 new square root terms introduced rather than just the original 4. Below I will do the first term:
$(sqrt a_1 + sqrt a_2 + sqrt a_3 + sqrt a_4)^2 =$
$a_1 + a_2 + a_3 + a_4 + 2sqrt (a_1a_2) + 2sqrt (a_1a_3) + 2sqrt (a_1a_4) + 2sqrt (a_2a_3) + 2sqrt (a_2a_4) + 2sqrt (a_3a_4)$
as a result, I go from 8 roots in the denominator to 12 and a single non root term
If instead I broke up the expression into a group of 5 square roots and a group of 3 square roots I would end up with 10 roots from the group of 5 and 3 from the group of 3 totalling 13.
in either case the number of roots does not go down, which is the trick I have been exploiting. It seems to me a new trick of sorts is required to find the general solution.
Thanks
algebra-precalculus
Consider the following expression: $1/(sqrt a_1 ± sqrt a_2 ± ... sqrt a_n)$ where $a_1 ... a_n$ are positive integers
I would like a general algorithm by which I can convert expressions of this form into expression of the following form: exp/k
where k is some positive integer and exp is an expression consisting of additions, subtractions, multiplications and square roots but no divisions.
Here is an example of a simple case:
$1/(sqrt 2 + sqrt 3) = (sqrt 3 - sqrt 2)/(3 - 2) = sqrt 3 - sqrt 2$
Here I multiplied the top and bottom by the conjugate. This technique can be used in the general problem for n≤4 by breaking the sum up into pairs of roots. I will illustrate this below in the case where all square root terms are added rather than having subtraction mixed in.
consider this denominator: $(sqrt a + sqrt b) + (sqrt c + sqrt d)$
multiplying by its conjugate: $(sqrt a + sqrt b) - (sqrt c + sqrt d)$ yields:
$(sqrt a + sqrt b)^2 - (sqrt c + sqrt d)^2$
$a+b+2sqrt (ab) - c - d - 2sqrt (cd) = $
$ A + (sqrt B - sqrt C)$ where $A= a+b-c-d$ and $B=4ab$ and $C=4cd$
multiplying by its conjugate $ A - (sqrt B - sqrt C)$ yields:
$ A^2 - (sqrt B - sqrt C)^2 =$
$ A^2 - B - C + 2sqrt (BC) =$
$ D + sqrt E =$ where $D=A^2-B-C$ and $E=4BC$
now multiply by its conjugate $ D - sqrt E$ yields:
$ D^2 - E$ which is an integer
I would like to know how to generalize this to n=5 and beyond, or if this is not possible, why that is the case. Thanks
@ancientmathemetician:
I am not sure how to do this for the case n=8. Consider the following expression: $(sqrt a_1 + sqrt a_2 + sqrt a_3 + sqrt a_4) + (sqrt a_5 + sqrt a_6 + sqrt a_7 + sqrt a_8)$
I could multiply by its conjugate which would yield the following
$(sqrt a_1 + sqrt a_2 + sqrt a_3 + sqrt a_4)^2 - (sqrt a_5 + sqrt a_6 + sqrt a_7 + sqrt a_8)^2$
However once each squared term is distributed out there will be 6 new square root terms introduced rather than just the original 4. Below I will do the first term:
$(sqrt a_1 + sqrt a_2 + sqrt a_3 + sqrt a_4)^2 =$
$a_1 + a_2 + a_3 + a_4 + 2sqrt (a_1a_2) + 2sqrt (a_1a_3) + 2sqrt (a_1a_4) + 2sqrt (a_2a_3) + 2sqrt (a_2a_4) + 2sqrt (a_3a_4)$
as a result, I go from 8 roots in the denominator to 12 and a single non root term
If instead I broke up the expression into a group of 5 square roots and a group of 3 square roots I would end up with 10 roots from the group of 5 and 3 from the group of 3 totalling 13.
in either case the number of roots does not go down, which is the trick I have been exploiting. It seems to me a new trick of sorts is required to find the general solution.
Thanks
algebra-precalculus
algebra-precalculus
edited Nov 23 at 8:39
asked Nov 23 at 7:18
mathew
408215
408215
One small comment: you mustn't overdo it. For example, if $D=sqrt{E}$ you stop there and don't multiply by $D-sqrt{E}=0$.
– ancientmathematician
Nov 23 at 7:52
Hints: If I wanted to go along this path I'd do two things. (i) I'd persuade myself that the process applies equally well to $(b_1 sqrt a_1 ± b_2 sqrt a_2 ± ... b_n sqrt a_n)$ with integral $b_i$, some possibly $0$; and (ii) I'd show I could do it for $n=2,4,8,16, dots $. Then with the right number of zero $b_i$ I'd be done for all $n$.
– ancientmathematician
Nov 23 at 7:57
Thanks for pointing out the case where D=√E, didn't think of that. but ultimately if such a case arizes then one can just end the algorithm early. I agree it would be worth doing the more general case with b√a instead of just √a because it captures the changes in sign. As for showing it can be done for the cases n=2,4,8,16... I am not sure how to do it for the n=8 case so if you want to demonstrate that I'll accept that as an answer
– mathew
Nov 23 at 8:29
You are right to point out the way in which it won't work. So to save it I think you have to deal with the fully general case. That is, the general case of $b_0$, $b_0+b_1 sqrt{a_1}$, $b_0+b_1 sqrt{a_1}+b_2 sqrt{a_2} + b_{1,2}sqrt{a_1 a_2}$, $dots$ - in other words, allow for all the cross terms from the outset.
– ancientmathematician
Nov 23 at 10:15
add a comment |
One small comment: you mustn't overdo it. For example, if $D=sqrt{E}$ you stop there and don't multiply by $D-sqrt{E}=0$.
– ancientmathematician
Nov 23 at 7:52
Hints: If I wanted to go along this path I'd do two things. (i) I'd persuade myself that the process applies equally well to $(b_1 sqrt a_1 ± b_2 sqrt a_2 ± ... b_n sqrt a_n)$ with integral $b_i$, some possibly $0$; and (ii) I'd show I could do it for $n=2,4,8,16, dots $. Then with the right number of zero $b_i$ I'd be done for all $n$.
– ancientmathematician
Nov 23 at 7:57
Thanks for pointing out the case where D=√E, didn't think of that. but ultimately if such a case arizes then one can just end the algorithm early. I agree it would be worth doing the more general case with b√a instead of just √a because it captures the changes in sign. As for showing it can be done for the cases n=2,4,8,16... I am not sure how to do it for the n=8 case so if you want to demonstrate that I'll accept that as an answer
– mathew
Nov 23 at 8:29
You are right to point out the way in which it won't work. So to save it I think you have to deal with the fully general case. That is, the general case of $b_0$, $b_0+b_1 sqrt{a_1}$, $b_0+b_1 sqrt{a_1}+b_2 sqrt{a_2} + b_{1,2}sqrt{a_1 a_2}$, $dots$ - in other words, allow for all the cross terms from the outset.
– ancientmathematician
Nov 23 at 10:15
One small comment: you mustn't overdo it. For example, if $D=sqrt{E}$ you stop there and don't multiply by $D-sqrt{E}=0$.
– ancientmathematician
Nov 23 at 7:52
One small comment: you mustn't overdo it. For example, if $D=sqrt{E}$ you stop there and don't multiply by $D-sqrt{E}=0$.
– ancientmathematician
Nov 23 at 7:52
Hints: If I wanted to go along this path I'd do two things. (i) I'd persuade myself that the process applies equally well to $(b_1 sqrt a_1 ± b_2 sqrt a_2 ± ... b_n sqrt a_n)$ with integral $b_i$, some possibly $0$; and (ii) I'd show I could do it for $n=2,4,8,16, dots $. Then with the right number of zero $b_i$ I'd be done for all $n$.
– ancientmathematician
Nov 23 at 7:57
Hints: If I wanted to go along this path I'd do two things. (i) I'd persuade myself that the process applies equally well to $(b_1 sqrt a_1 ± b_2 sqrt a_2 ± ... b_n sqrt a_n)$ with integral $b_i$, some possibly $0$; and (ii) I'd show I could do it for $n=2,4,8,16, dots $. Then with the right number of zero $b_i$ I'd be done for all $n$.
– ancientmathematician
Nov 23 at 7:57
Thanks for pointing out the case where D=√E, didn't think of that. but ultimately if such a case arizes then one can just end the algorithm early. I agree it would be worth doing the more general case with b√a instead of just √a because it captures the changes in sign. As for showing it can be done for the cases n=2,4,8,16... I am not sure how to do it for the n=8 case so if you want to demonstrate that I'll accept that as an answer
– mathew
Nov 23 at 8:29
Thanks for pointing out the case where D=√E, didn't think of that. but ultimately if such a case arizes then one can just end the algorithm early. I agree it would be worth doing the more general case with b√a instead of just √a because it captures the changes in sign. As for showing it can be done for the cases n=2,4,8,16... I am not sure how to do it for the n=8 case so if you want to demonstrate that I'll accept that as an answer
– mathew
Nov 23 at 8:29
You are right to point out the way in which it won't work. So to save it I think you have to deal with the fully general case. That is, the general case of $b_0$, $b_0+b_1 sqrt{a_1}$, $b_0+b_1 sqrt{a_1}+b_2 sqrt{a_2} + b_{1,2}sqrt{a_1 a_2}$, $dots$ - in other words, allow for all the cross terms from the outset.
– ancientmathematician
Nov 23 at 10:15
You are right to point out the way in which it won't work. So to save it I think you have to deal with the fully general case. That is, the general case of $b_0$, $b_0+b_1 sqrt{a_1}$, $b_0+b_1 sqrt{a_1}+b_2 sqrt{a_2} + b_{1,2}sqrt{a_1 a_2}$, $dots$ - in other words, allow for all the cross terms from the outset.
– ancientmathematician
Nov 23 at 10:15
add a comment |
1 Answer
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$$(sqrt a+sqrt b+sqrt c)(s+tsqrt a+usqrt b+vsqrt c+wsqrt{ab}+xsqrt{ac}+ysqrt{bc}+zsqrt{abc})=1$$
The rational coefficients of the various square-roots give you eight equations in eight unknowns $s$ to $z$.
It looks like it works for any number of square-roots. A $2^Ntimes2^N$ matrix has to be inverted. I don't know how to tell if it's going to be singular.
The matrix is very sparse - mostly zeros - so this isn't the most efficient method. I think it shows there is an answer.
add a comment |
1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
up vote
0
down vote
$$(sqrt a+sqrt b+sqrt c)(s+tsqrt a+usqrt b+vsqrt c+wsqrt{ab}+xsqrt{ac}+ysqrt{bc}+zsqrt{abc})=1$$
The rational coefficients of the various square-roots give you eight equations in eight unknowns $s$ to $z$.
It looks like it works for any number of square-roots. A $2^Ntimes2^N$ matrix has to be inverted. I don't know how to tell if it's going to be singular.
The matrix is very sparse - mostly zeros - so this isn't the most efficient method. I think it shows there is an answer.
add a comment |
up vote
0
down vote
$$(sqrt a+sqrt b+sqrt c)(s+tsqrt a+usqrt b+vsqrt c+wsqrt{ab}+xsqrt{ac}+ysqrt{bc}+zsqrt{abc})=1$$
The rational coefficients of the various square-roots give you eight equations in eight unknowns $s$ to $z$.
It looks like it works for any number of square-roots. A $2^Ntimes2^N$ matrix has to be inverted. I don't know how to tell if it's going to be singular.
The matrix is very sparse - mostly zeros - so this isn't the most efficient method. I think it shows there is an answer.
add a comment |
up vote
0
down vote
up vote
0
down vote
$$(sqrt a+sqrt b+sqrt c)(s+tsqrt a+usqrt b+vsqrt c+wsqrt{ab}+xsqrt{ac}+ysqrt{bc}+zsqrt{abc})=1$$
The rational coefficients of the various square-roots give you eight equations in eight unknowns $s$ to $z$.
It looks like it works for any number of square-roots. A $2^Ntimes2^N$ matrix has to be inverted. I don't know how to tell if it's going to be singular.
The matrix is very sparse - mostly zeros - so this isn't the most efficient method. I think it shows there is an answer.
$$(sqrt a+sqrt b+sqrt c)(s+tsqrt a+usqrt b+vsqrt c+wsqrt{ab}+xsqrt{ac}+ysqrt{bc}+zsqrt{abc})=1$$
The rational coefficients of the various square-roots give you eight equations in eight unknowns $s$ to $z$.
It looks like it works for any number of square-roots. A $2^Ntimes2^N$ matrix has to be inverted. I don't know how to tell if it's going to be singular.
The matrix is very sparse - mostly zeros - so this isn't the most efficient method. I think it shows there is an answer.
edited Nov 23 at 10:31
answered Nov 23 at 10:17
Empy2
33k12159
33k12159
add a comment |
add a comment |
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One small comment: you mustn't overdo it. For example, if $D=sqrt{E}$ you stop there and don't multiply by $D-sqrt{E}=0$.
– ancientmathematician
Nov 23 at 7:52
Hints: If I wanted to go along this path I'd do two things. (i) I'd persuade myself that the process applies equally well to $(b_1 sqrt a_1 ± b_2 sqrt a_2 ± ... b_n sqrt a_n)$ with integral $b_i$, some possibly $0$; and (ii) I'd show I could do it for $n=2,4,8,16, dots $. Then with the right number of zero $b_i$ I'd be done for all $n$.
– ancientmathematician
Nov 23 at 7:57
Thanks for pointing out the case where D=√E, didn't think of that. but ultimately if such a case arizes then one can just end the algorithm early. I agree it would be worth doing the more general case with b√a instead of just √a because it captures the changes in sign. As for showing it can be done for the cases n=2,4,8,16... I am not sure how to do it for the n=8 case so if you want to demonstrate that I'll accept that as an answer
– mathew
Nov 23 at 8:29
You are right to point out the way in which it won't work. So to save it I think you have to deal with the fully general case. That is, the general case of $b_0$, $b_0+b_1 sqrt{a_1}$, $b_0+b_1 sqrt{a_1}+b_2 sqrt{a_2} + b_{1,2}sqrt{a_1 a_2}$, $dots$ - in other words, allow for all the cross terms from the outset.
– ancientmathematician
Nov 23 at 10:15