Saturday, August 23, 2008


JSH: Distributive property and key result

It is worth reminding yet again that I discovered a flaw in "core" mathematics which oddly enough is easily shown to be there using the distributive property and some simple quadratics.

But that is so controversial because the people who are supposed to evaluate such a discovery are number theorists who have their dreams of successful research shattered by the result, so despite the ease of proof and, of course the absoluteness of the proof, they have chosen instead to live in error, and to teach it to young minds.

The key to understanding what follows is accepting the distributive property with functions, for example,

a*(f(x) + b) = a*f(x) + a*b

If you accept that then all follows easily. I have seen arguments on the sci.math newsgroup where people dispute that and then come back to claim that they are not disputing it as they dodge on the basis of a slight difference in form as you will see

c*P(x)=(f_1(x) + c)*(f_2(x) + c)

so there are TWO cases where the distributive property is used, but it doesn't change how it works, but in that deuce, posters have tried to convince against mathematical proof.

Now then, consider in an integral ring

c*P(x)=(f_1(x) + c)*(f_2(x) + c)

where P(x) is a polynomial with integer coefficients, and c is a non-zero integer, where also P(0) is nonzero and coprime to c in Z. That is, c is coprime to P(0) in the ring of integers.

Then it cannot be true that both f_1(0) = 0 and f_2(0)=0.

Proof: Assume both DO equal 0 when x=0, then you have

c*P(0) = c^2 which contradicts with P(0) being coprime to c in Z.

Since they both cannot equal 0 when x=0, let's normalize with

f_1(x) = g_1(x) + b_1


f_2(x) = g_2(x) + b_2

where g_1(0) = 0 and g_2(0) = 0, where b_1 and b_2 are not functions of x and both cannot equal 0 as proven above.

Then making the substitutions I have

c*P(x)=(g_1(x) + b_1 + c)*(g_2(x) + b_2 + c)

and notice that at x=0 I have

c*P(0)=(b_1 + c)*(b_2 + c).

Now let d_1 be any factors in common between (b_1 + c) and c, and let d_2 be any in common between c and (b_2 + c), then I have that

c = d_1*d_2.

Then it follows from the distributive property that g_1(x) and g_2(x) must be products as well so let

g_1(x) = d_1*h_1(x)


g_2(x) = d_2*h_2(x)

and making those substitutions gives

d_1*d_2*P(x)=(d_1*h_1(x) + b_1 + d_1*d_2)*(d_2*h_2(x) + b_2 + d_1*d_2)

and now for reference return to our original expression

c*P(x)=(f_1(x) + c)*(f_2(x) + c)


f_1(x) + c = d_1*h_1(x) + b_1 + d_1*d_2

and since c = d_1*d_2, that is

f_1(x) = d_1*h_1(x) + b_1

so f_1(x) must have d_1 as a factor since b_1 does.

Proof complete.

So we now have proven the result that d_1 has distributed through f_1(x) + c and d_2 has distributed through f_2(x) + c.

To get some grasp of what that result means consider that if the ring is the ring of integers and f_1(x) and f_2(x) are polynomials then you have that d_1 is a factor of f_1(x) and d_2 is a factor of f_2(x).

But it is possible in the ring of algebraic integers for f_1(x) to NOT have d_1 as a factor so all of those steps above are not always valid in the ring of algebraic integers.

Since you can calculate d_1 and d_2, it is easy to check the result with actual polynomials.

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