Re: JSH: A simple error

On 27-05-2007 14:45, jstevh@xxxxxxxxx wrote:

Posters in attacking my proofs showing inconsistency with the ring of
algebraic integers routinely move outside the ring. Here is a post
meant to show you how they do it.

In the ring of integers, consider x^2 + 3x + 2 = 0, which of course
factors as

x^2 + 3x + 2 = (x+2)(x+1)

and now solve for it using the quadratic formula, but kind of weird by
NOT resolving the square root then

x = (-3 +/- sqrt(1))/2

and now make the substitution, x=2y, so you get

How can I make that substitution? You wrote that we are working in the
ring of integers. And, given an integer _x_ it is often false that I
can write an integer _x_ as 2y for an integer _y_. In a ring, you cannot
divide by arbitrary elements.

4y^2 + 6y + 2 = 0, so you can divide by 2 to get

2y^2 + 3y + 1 = 0,

and your solution now becomes

y = (-3+/- sqrt(1))/4

which is two solutions where one is not an integer.

So you moved outside the ring of integers.

So what's the trick?

The trick consists in stating that we are working in the ring of
integers and then doing some operation which cannot be performed on that
ring.

Well, with integer solutions you can resolve the square root and throw
away one solution, which is what most people routinely do, so they say
that sqrt(4) = 2.

With integer solutions, the solutions are (-3 + sqrt(1))/2 = -1 and
(-3 - sqrt(1))/2 = -2. That makes two solutions, not one.

Best regards,

Jose Carlos Santos
.

Relevant Pages

  • RIT Spring Juggle-In 2007 HLGCB Thread...
    ... i had soooooo much fun! ... Sean Blue's ring juggling! ... bruising my back on the SOLID floor doing my forward roll trick in the ... Laura and Linka for being super cute ...
    (rec.juggling)
  • Re: Abstract Algebra
    ... If I remember correctly, your ring was the ... ring of all rational numbers that can be written as a/b, ... not divide b. ... You'll have to figure out what the ideals are first. ...
    (sci.math)
  • Re: Timed Profiles? Can it be done?
    ... is there a trick to it? ... vol 0%, ring vol 100%, respectively), added their schedule times ... to use night mode instead and set calls off ...
    (microsoft.public.pocketpc.phone_edition)
  • Re: Dumb arguments, and social stuff
    ... > that they do in the ring of algebraic integers. ... the reader and the editor was supposed to GUESS what ring ... pulled a nasty trick on you, ...
    (sci.math)
  • Re: Ring with Postitive & Finite Number of Primes?
    ... I think any finite commutative ring with 1 which is not a field will ... Each of 0, y, and x+y divide x^2=0 but do not divide x, so they are ... Note, by the by, that each nonzero nonunit is irreducible: ... to have no primes at all, I sense it would have to be infinite. ...
    (sci.math)