Expert: Paul Klarreich - 11/11/2011

Question
Hello Sir,
If i have a system  of n equations with n unknowns, with NATURAL NUMBERS coefficients and for some reason(never mind how) I know the solution(IF THERE EXISTS ONE)consists of only NATURAL NUMBERS.If one can prove the system is linear independent in the NATURAL NUMBER domain.Does the general theorem expressed below,hold.
General Theorem:
a linear independent system of n equations in n unknowns has a unique solution.Thanks.

Answer
Questioner: sean
Country: Gambia
Category: Advanced Math
Private: No
Subject: linear independence
Question: Hello Sir,
If i have a system  of n equations with n unknowns, with NATURAL NUMBERS coefficients and for some reason(never mind how) I know the solution(IF THERE EXISTS ONE)consists of only NATURAL NUMBERS.If one can prove the system is linearLY independent in the NATURAL NUMBER domain.

>>> I am not sure what that term means: L.I. in NN.

L.I. usually means:  Given n equations, there exists a set of (n-1) numbers, A1..A[n-1], such that A1* EQ1 + ... A[n-1]*EQ[n-1] yields EQ[n].

These  Ai's would be rational, of course, and unlikely to be integers.

BUT, if you allow:

A1* EQ1 + ... A[n-1]*EQ[n-1] yields A[n]*EQ[n]

then they could be natural numbers.

Does the general theorem expressed below,hold.
General Theorem:
a linearLY independent system of n equations in n unknowns has a unique solution.Thanks.

>>> This seems so, but why would they have to be NN's?