How do you rationalise the denominator of #1/(sqrt(4)+sqrt(5)+sqrt(6)+sqrt(8))# ?

2 Answers
Nov 19, 2016

Answer:

#1/(sqrt(4)+sqrt(5)+sqrt(6)+sqrt(8))#

#=(270+382 sqrt(2)-208 sqrt(3)-139 sqrt(5)-157 sqrt(6)-136 sqrt(10)+56 sqrt(15)+92 sqrt(30))/431#

Explanation:

In the given example, note that #sqrt(4) = 2# is already rational.

That does not actually affect the process however.

Choose one of the terms in the denominator to retain its sign, then multiply both numerator and denominator by all the other possible combinations of signs of the other #3# terms.

So in our example, that means multiplying by:

#(2+sqrt(5)+sqrt(6)-sqrt(8)) * #
#(2+sqrt(5)-sqrt(6)+sqrt(8)) * #
#(2+sqrt(5)-sqrt(6)-sqrt(8)) * #
#(2-sqrt(5)+sqrt(6)+sqrt(8)) * #
#(2-sqrt(5)+sqrt(6)-sqrt(8)) * #
#(2-sqrt(5)-sqrt(6)+sqrt(8)) * #
#(2-sqrt(5)-sqrt(6)-sqrt(8))#

This is a pretty nasty thing to multiply, but note that every partial product along the way can be written in the form:

#a+bsqrt(2)+csqrt(3)+dsqrt(5)+esqrt(6)+fsqrt(10)+gsqrt(15)+hsqrt(30)#

for some integers #a, b, c, d, e, f, g, h#.

I used a computer to help with the arithmetic to find the resulting numerator:

#-270-382 sqrt(2)+208 sqrt(3)+139 sqrt(5)+157 sqrt(6)+136 sqrt(10)-56 sqrt(15)-92 sqrt(30)#

and the denominator:

#-431#

Reversing the signs, we have:

#1/(sqrt(4)+sqrt(5)+sqrt(6)+sqrt(8))#

#=(270+382 sqrt(2)-208 sqrt(3)-139 sqrt(5)-157 sqrt(6)-136 sqrt(10)+56 sqrt(15)+92 sqrt(30))/431#

Nov 20, 2016

Answer:

#1/(sqrt(4)+sqrt(5)+sqrt(6)+sqrt(8))#

#=(270+382 sqrt(2)-208 sqrt(3)-157 sqrt(6)-139 sqrt(5)-136 sqrt(10)+56 sqrt(15)+92 sqrt(30))/431#

Explanation:

Another way to approach this problem is using matrices.

We can represent any number of the form #a+bsqrt(2)# where #a, b in QQ# using a #2xx2# matrix of the form:

#((a, b),(2b,a))#

Arithmetic of matrices of this form corresponds precisely to arithmetic of expressions of the form #a+bsqrt(2)#.

Similarly, we can represent any number of the form:

#a+bsqrt(2)+csqrt(3)+dsqrt(6)+esqrt(5)+fsqrt(10)+gsqrt(15)+hsqrt(30)#

where #a, b, c, d, e, f, g, h in QQ# using an #8xx8# matrix of the form:

#((a, b, c, d, e, f, g, h),(2b, a, 2d, c, 2f, e, 2h, g),(3c, 3d, a, b, 3g, 3h, e, f),(6d, 3c, 2b, a, 6h, 3g, 2f, e),(5e, 5f, 5g, 5h, a, b, c, d),(10f, 5e, 10h, 5g, 2b, a, 2d, c),(15g, 15h, 5e, 5f, 3c, 3d, a, b), (30h, 15g, 10f, 5e, 6d, 3c, 2b, a))#

In our example, we have:

#sqrt(4)+sqrt(5)+sqrt(6)+sqrt(8)#

#=2+2sqrt(2)+0sqrt(3)+1sqrt(6)+1sqrt(5)+0sqrt(10)+0sqrt(15)+0sqrt(30)#

which is represented by the matrix:

#((2, 2, 0, 1, 1, 0, 0, 0),(4, 2, 2, 0, 0, 1, 0, 0),(0, 3, 2, 2, 0, 0, 1, 0),(6, 0, 4, 2, 0, 0, 0, 1),(5, 0, 0, 0, 2, 2, 0, 1),(0, 5, 0, 0, 4, 2, 2, 0),(0, 0, 5, 0, 0, 3, 2, 2), (0, 0, 0, 5, 6, 0, 4, 2))#

Use your favourite matrix inversion method to find the inverse of this matrix, namely:

#1/431 ((270, 382, -208, -157, -139, -136, 56, 92), (764, 270, -314, -208, -272, -139, 184, 56), (-624, -471, 270, 382, 168, 276, -139, -136), (-942, -624, 764, 270, 552, 168, -272, -139), (-695, -680, 280, 460, 270, 382, -208, -157), (-1360, -695, 920, 280, 764, 270, -314, -208), (840, 1380, -695, -680, -624, -471, 270, 382), (2760, 840, -1360, -695, -942, -624, 764, 270))#

This inverse matrix will be the representation of the reciprocal of the original number.

Then you can read off the coefficients from the top line of the matrix to find:

#1/(sqrt(4)+sqrt(5)+sqrt(6)+sqrt(8))#

#=(270+382 sqrt(2)-208 sqrt(3)-157 sqrt(6)-139 sqrt(5)-136 sqrt(10)+56 sqrt(15)+92 sqrt(30))/431#