If $a=b=c=d$ and $d$ is rational, then the set $M$ becomes $\left\{d \vert x \in \mathbb R\right\} = \left\{d\right\}.$ Thus, since $d$ is rational, the set contains no irrational numbers. Next, we need to prove that if set $M$ contains no irrational numbers, then $a=b=c=0$ and $d$ is rational. If $a, b,$ or $c$ are irrational, then $M$ would contain one or more irrational numbers. Furthermore, if $a,b,$ and $c$ are rational, there is still the possibility that $x$ is irrational since $\mathbb P \subset \mathbb R$. In order to rule out this possibility, we need to eliminate all terms with $x$, and make sure that $d \in \mathbb R$. This happens only when $a=b=c=0$ and $d$ is rational. Therefore, set $M$ contains no irrational numbers if and only if $a=b=c=0$ and $d$ is rational. Is this right?

#2

This isn't obvious, since for instance if $(a,b,c,d)=\left(\sqrt2,0,0,0\right)$ and $x=\sqrt[6]2$, we get a rational number. Also, if $a=\sqrt2$ and $b=2-\sqrt2$ and $c=1$ and $d=1$, the substitution $x=1$ yields a rational number, so you also need to justify the next sentence. Also not sure what $\mathbb P$ is. The primes?

jasperE3 wrote:

#2

This isn't obvious, since for instance if $(a,b,c,d)=\left(\sqrt2,0,0,0\right)$ and $x=\sqrt[6]2$, we get a rational number. Also, if $a=\sqrt2$ and $b=2-\sqrt2$ and $c=1$ and $d=1$, the substitution $x=1$ yields a rational number, so you also need to justify the next sentence. Aren't those just specific cases? Since for your first example, inputting $x=1$ would result in an irrational number? (You could find an $x$ value like that for your second example too.) And, since set $M$ contains numbers of the form $ax^3 + bx^2 + cx + d$ evaluated for all $x \in \mathbb R$ (given values of $a, b, c,$ and $d$), wouldn't there be irrational numbers in $M$ when the values of $a,b,c,$ and $d$ are what they are in your examples? However, I do get your point, so maybe a better statement would be "If $a, b,$ or $c$ are irrational, then $M$ could contain one or more irrational numbers?" jasperE3 wrote: Also not sure what $\mathbb P$ is. The primes? It's for irrational numbers.

Suppose $P(x)=ax^3+bx^2+cx+d$ is nonconstant. Clearly $P(0)=d$. Since $P(x)$ is nonconstant, there must exist $i$ such that $P(i)=n\ne d$. WLOG let $n>d$. Then since $P(x)$ is continuous, by IVT there must exist $x$ such that $P(x)=j$ for all $d\le j\le n$. Since this is a nontrivial range, this must contain at least one irrational number.

#5

But what is $a=0$ and $P$ is bounded (ie $P(x)<d\forall d$)?

What I mean by WLOG let $n>d$ is that the $n<d$ case is analogous. @below the like button exists for a reason

I like this solution.

bluelinfish wrote: Suppose $P(x)=ax^3+bx^2+cx+d$ is nonconstant. Clearly $P(0)=d$. Since $P(x)$ is nonconstant, there must exist $i$ such that $P(i)=n\ne d$. WLOG let $n>d$. Then since $P(x)$ is continuous, by IVT there must exist $x$ such that $P(x)=j$ for all $d\le j\le n$. Since this is a nontrivial range, this must contain at least one irrational number. Does a nontrivial range mean that it doesn't contain 0?

A nontrivial range means that it is a superset of a closed interval. It's true since the irrationals are dense in the reals.