Monday, August 06, 2007
JSH: Simple construct
Let
7x^2 + 5x - 8 = 0
and
7y^2 + 5y - 1 = 0
so subtracting the second from the first I have
7(x^2 - y^2) + 5(x-y) - 7 = 0
so
7(x+y) + 5 = 7/(x-y).
Therefore 1/y has 7 as a factor, right?
Wrong. For one root of 7y^2 + 5y - 1 = 0, 1/y has 7 as a factor in a proper ring, while for the other root it is coprime to 7.
However, in the ring of algebraic integers for neither root does 1/y have 7 as a factor.
What's nice about this example is that it is trivially easy for you to understand if you mastered adding fractions together.
If you didn't then yeah, you may still have problems.
Odd that such a simple way to show the ring of algebraic integers is worthless for analysis was always there, eh? I wonder why I just thought of it now.
[A reply to someone who explained James why he was wrong.]
Sigh. So you have problems adding fractions?
Look again at
7(x+y) + 5 = 7/(x-y)
and notice that x+y must divide off all factors of 7 on the left side so that you can add with 5 which is coprime to 7, to get a result that has factors in common with 7.
On the right side you are just multiplying 7 by more factors.
It is a fascinatingly simple example that you should see quickly if you can handle fractions in your head, or if you have trouble, solve out for x and y and try again.
7x^2 + 5x - 8 = 0
and
7y^2 + 5y - 1 = 0
so subtracting the second from the first I have
7(x^2 - y^2) + 5(x-y) - 7 = 0
so
7(x+y) + 5 = 7/(x-y).
Therefore 1/y has 7 as a factor, right?
Wrong. For one root of 7y^2 + 5y - 1 = 0, 1/y has 7 as a factor in a proper ring, while for the other root it is coprime to 7.
However, in the ring of algebraic integers for neither root does 1/y have 7 as a factor.
What's nice about this example is that it is trivially easy for you to understand if you mastered adding fractions together.
If you didn't then yeah, you may still have problems.
Odd that such a simple way to show the ring of algebraic integers is worthless for analysis was always there, eh? I wonder why I just thought of it now.
[A reply to someone who explained James why he was wrong.]
Sigh. So you have problems adding fractions?
Look again at
7(x+y) + 5 = 7/(x-y)
and notice that x+y must divide off all factors of 7 on the left side so that you can add with 5 which is coprime to 7, to get a result that has factors in common with 7.
On the right side you are just multiplying 7 by more factors.
It is a fascinatingly simple example that you should see quickly if you can handle fractions in your head, or if you have trouble, solve out for x and y and try again.
# posted by James Harris @ 20:43 
