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The following question is located on Page 80 of Calculus with Analytic Geometry by George Simmons.

Assume for a moment that the rational numbers are the only numbers that exist. Under this assumption, show that the Intermediate Value Theorem is false by considering the function $y=f(x)=x^2 - 2$ on the inteval $[1,2]$.

The only way I have been able to interpret this is to solve $f(x)$ for some irrational value. As the number would not exist in this hypothetical, the IVT would fail. However, I feel like that's an oversimplification and I'm missing what this question is actually asking me.

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    $\begingroup$ Where would this function cross the $x$-axis? $\endgroup$
    – Randall
    Feb 28, 2019 at 2:11
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    $\begingroup$ To elaborate, no, you don't want to solve $f(x)=c$ for an irrational value of $c$. You want to solve it for $c=0$, noting that $f(1)<0<f(2)$. $\endgroup$ Feb 28, 2019 at 2:14
  • $\begingroup$ Ah, that is exactly what I was looking for. Crosses at sqrt(2), but f(1) is negative, f(2) is positive. As IVT says f(x)=0 at some point for this to happen, but this point does not exist, IVT fails. Thank you! $\endgroup$
    – Vale
    Feb 28, 2019 at 2:15
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    $\begingroup$ $f(1) = 1^2 -2 < 0$ and $f(2) = 2^2 - 2 > 0$ so if the Intermediate Value Theorem holds then there exists a $\psi \in (1, 2)$ such that $f(\psi) = 0$. But then $\psi = \sqrt{2}$ a contradiction. $\endgroup$ Feb 28, 2019 at 2:16

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In general you need to find $c$ between $f(1)$ and $f(2)$ such that no rational $x$ satisfies $f(x) = c$.

In particular, as mentioned in the comment, you could show that there's no rational $x$ such that $f(x) = 0$.

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