Find all polynomial $p(x)$ with rational coefficients so that for all irrational number $x$, $p(x)$ is also irrational.
If it is useful, this exercise was found in one of my linear algebra textbook.
Finding polynomial p
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$\begingroup$
polynomials
vector-spaces
irrational-numbers
1 Answers
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Polynomials of degree $0$ and $1$ with rational coefficients will work. But polynomials of higher degree will not. Namely, suppose $p$ is a prime that does not divide the numerator or denominator of any of the coefficients of your polynomial $P(x)$. Then there is no rational $x$ such that $P(x)$ has denominator $p$: if $x$ does not have denominator divisible by $p$, $P(x)$ does not either, while if $x$ has denominator divisible by $p$, $P(x)$ has denominator divisible by $p^d$ where $d$ is the degree of $P$.
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0Thank you but I still don't see where this is leading us – 2017-02-22
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0I think it would be better to prove that for every polynomial $p(x)$ with degree greater than $1$, there is always a irrational number $x$ so that $p(x)$ is rational – 2017-02-22
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0That's exactly what it does prove. Take a rational $y$ in the range of $P$ whose denominator is $p$. There is no rational $x$ such that $P(x)=y$, so there is an irrational $x$. – 2017-02-22
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0Polynomials of degree 0 are not solutions (their value is rational for all arguments). – 2017-02-22
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0Oh I got it, thank you very much – 2017-02-22