49 lines
1.2 KiB
Markdown
49 lines
1.2 KiB
Markdown
#Math #Algebra
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# Proof
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Let polynomial
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$$
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P(x) = \sum _{i = 0}^n c_i x^i
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$$
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where $c_i \in \mathbb{Z}$ (all values of $c$ are integers).
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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 $P$).
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$$
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\sum _{i = 0}^n c_i (\frac p q)^i = 0
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$$
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Multiplying by $q^n$:
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$$
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\sum _{i = 0}^n c_i p^n q^{n - i} = 0
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$$
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Now subtract $c_0 q^n$ from both sides and factor $p$ out to get:
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$$
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p\sum _{i = 1}^n c_i p^{n - 1} q^{n - i} = -c_0 q^n
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$$
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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 $c_0$.
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Doing the same thing as above but with the $a_n$ term and $q$:
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$$
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q\sum _{i = 0}^{n - 1} c_i p^n q^{n - i - 1} = -c_n p^n
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$$
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By the above logic, $q$ must divide $c_n$.
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## Conclusion
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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.
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## Notes
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For the curious, coprime integers $p$ and $q$ mean that $\gcd(p, q) = 1$.
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If future me or someone else is wondering about the excess definitions, this was made for a friend. |