rational cubic polynomial (2001 Moldova MO Grade 7 P5)

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jasperE3

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jasperE3

#1 h Apr 12, 2021, 8:33 PM • 1 Y

Y by Mango247

Let $a,b,c,d$ be real numbers. Prove that the set $M=\left\{ax^3+bx^2+cx+d|x\in\mathbb R\right\}$ contains no irrational numbers if and only if $a=b=c=0$ and $d$ is rational.

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laffytaffy

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laffytaffy

#2 h Apr 12, 2021, 10:21 PM

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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?

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jasperE3

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jasperE3

#3 h Apr 12, 2021, 10:31 PM

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#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?

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laffytaffy

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laffytaffy

#4 h Apr 12, 2021, 11:16 PM

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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.

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bluelinfish

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bluelinfish

#5 h Apr 13, 2021, 1:00 AM • 2 Y

Y by jasperE3, laffytaffy

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.

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jasperE3

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#6 h Apr 13, 2021, 1:21 AM

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#5

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

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bluelinfish

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#7 h Apr 13, 2021, 2:01 AM

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What I mean by WLOG let $n>d$ is that the $n<d$ case is analogous.

@below the like button exists for a reason

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jasperE3

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#8 h Apr 13, 2021, 2:08 AM

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I like this solution.

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laffytaffy

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laffytaffy

#9 h Apr 13, 2021, 1:22 PM

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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?

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jasperE3

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#10 h Apr 13, 2021, 1:27 PM • 1 Y

Y by laffytaffy

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

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