# Showing that the Intermediate Value Theorem is false in the absence of irrational numbers

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.

• Where would this function cross the $x$-axis? Feb 28, 2019 at 2:11
• 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 $f(1)<0<f(2)$. Feb 28, 2019 at 2:14
• 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!
– Vale
Feb 28, 2019 at 2:15
• $f(1) = 1^2 -2 < 0$ 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. Feb 28, 2019 at 2:16

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 $$f(x) = 0$$.