exact computations with algebraic numbers

asked 2020-07-06 21:18:17 +0100

b897069
21 ●1 ●2 ●4

updated 2020-07-12 03:10:09 +0100

slelievre
17839 ●22 ●164 ●354 http://carva.org/samue...

I'm trying to do exact computations with algebraic numbers. In particular, I know to expect integer answers like 1, 0, and 3 at the end of my computations, but I'm getting something slightly off.

I noticed that if I run

sage: r = sqrt(2)
sage: a = AA(r)
sage: b = AA(1/r)
sage: c = a*b

Then I get:

sage: c
1.000000000000000?

Is this being handled in the computer as exactly 1? Otherwise, how can I do exact computations with algebraic numbers in Sage? I obtain the algebraic numbers I'm working with using algdep() to find a polynomial and .roots(QQbar) to find the roots of that polynomial.

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answered 2020-07-06 22:23:25 +0100

rburing

11084 ●6 ●81 ●222 https://www.rburing.nl/

Yes, it is exact. To check:

sage: c.minpoly() # minimal polynomial over QQ
x - 1
sage: c.radical_expression()
1

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answered 2020-07-12 03:12:28 +0100

slelievre
17839 ●22 ●164 ●354 http://carva.org/samue...

Computations in algebraic numbers are costly so Sage computes the product of a and b "lazily".

Some computations force it to realise the exact value of c.

Here are the one I would favour.

sage: c = AA(r) * AA(1/r)
sage: c
1.000000000000000?
sage: c.exactify()
sage: c
1

For the sake of completeness I also include those from @rburing's answer:

sage: c = AA(r) * AA(1/r)
sage: c
1.000000000000000?
sage: c.radical_expression()
1

sage: c = AA(r) * AA(1/r)
sage: c
1.000000000000000?
sage: c.minpoly()
x - 1
sage: c
1

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