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Sqrt(2) is algebraic, and can therefore be constructed fairly explocitly. Starting with integers, you can construct the rationals as an equivelence class of ord
by gizmo686 8y ago
Sqrt(2) is algebraic, and can therefore be constructed fairly explocitly. Starting with integers, you can construct the rationals as an equivelence class of ordered pairs of integers with a particular definition of addition and multiplication. From their you can define polynomials with rational coeficients, and from there you can define quotient fields, Q[x]/<x^2-2> which contains two elements whose square is 2, and is isomorphic to a subset of R (eg, behaves as you would expect numbers to). The most problamatic step of the above is defining polynomials.
- umanwizard 8y agoCan't you just define polynomials as finite sequences of rationals with nonzero last term? E.g. encode x^2 + 2x -1/2 as (-1/2, 2, 1), and define evaluation at a given point in the obvious way.
- gizmo686 8y agoYou generally define them as infinite sequences with a finite number of non-zero terms. Since we only care about a specific quotient field you could force it to work by working with only polynomials of bounded degree (<=2 should work. <=4 is probably the least bad option), but it would get ugly. You could also ignore polynomials entirely. Since Q(sqrt(2)) is just a two dimensional vector space over Q, we could define it as the ordered pairs Q^2 with an appropriate definition of multiplication and division (like we often define the complex numbers to highshoolers), but this also gets ugly. I guess the moral of this post is 99% of the time a finitist or constroctivist complains you can rework your theory into a more ugly one that avoids the complaint.