Re: JSH: New core argument then

jstevh@xxxxxxx writes:

Given

7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1)

true for all x, where all the functions go to 0 at x=0, by the core
principles I've outlined, where the value of the functions doesn't
change the distributive property, you can look at x = 0, which gives

7(0 + 1)(0 + 1) = (0 + 7)(0 + 1)

and see that 7 multiplies through the first, where I'm using the
convention that A is the first function and B is the second, so IF THE
VALUE OF THE FUNCTIONS TO THE DISTRIBUTIVE PROPERTY IS IRRELEVANT then
I have for ALL x that

A(x) = 7A'(x) and B'(x) = B(x).

You know, everyone else has given counterexamples involving rational
functions and the like, but let's do something simpler. Tell me where
I go wrong here. You've said the only requirements are that each of
the four functions A, A', B, B':

(1) takes algebraic integers to algebraic integers
(2) evaluates to 0 at 0.

Let us define:

A'(x) = 0 if x = 0, 1 else.
B'(x) = 0 (for all x)
A(x) = 0 (for all x)
B(x) = 0 if x = 0, 1 else.

Clearly, these functions satisfy (1) and (2) above. Let us check that
7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1). If x = 0, then the
left hand side evaluates to

7 (0 + 1)(0 + 1) = 7

and the right hand side to

(0 + 7)(0 + 1) = 7.

If x is not 0, then the left hand side evaluates to

7 (1 + 1)(0 + 1) = 14

and the right hand side to

(0 + 7)(1 + 1) = 14.

So all of the assumptions seem to work out and hence we should
conclude that A(x) = 7 A'(x) and B(x) = B'(x). But clearly that's not
true. What went wrong?

Didn't I satisfy every condition you suggested? Aren't these
"algebraic integer functions" in the sense you defined them?

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