1.2 KiB
#Math #Algebra
Proof
Let polynomial
P(x) = \sum _{i = 0}^n c_i x^i
where c_i \in \mathbb{Z} (all values of c are integers).
Now let P(\frac p q) = 0, where p and q are coprime integers (let a fraction \frac p q be in simplest form and be a root of P).
\sum _{i = 0}^n c_i (\frac p q)^i = 0
Multiplying by q^n:
\sum _{i = 0}^n c_i p^n q^{n - i} = 0
Now subtract c_0 q^n from both sides and factor p out to get:
p\sum _{i = 1}^n c_i p^{n - 1} q^{n - i} = -c_0 q^n
Now p must divide -c_0q^n. However, we know p cannot divide q^n (since \frac p q is in simplest form / p and q are coprime), so p must divide c_0.
Doing the same thing as above but with the a_n term and q:
q\sum _{i = 0}^{n - 1} c_i p^n q^{n - i - 1} = -c_n p^n
By the above logic, q must divide c_n.
Conclusion
For all rational roots in simplest form (\frac p q where p and q are coprime integers), p must be a factor of the last coefficient while q must be a factor of the first coefficient.
Notes
For the curious, coprime integers p and q mean that \gcd(p, q) = 1.
If future me or someone else is wondering about the excess definitions, this was made for a friend.