(I assume trivial) floor function problem.

Question

Question. What is the floor of $$x=\frac{5(\sqrt{13}+2)}{9}.$$

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My attempt. For brevity; I will denote $f(x)$ as the floor of $x$. Now, observe $3<\sqrt{13}<4$. Then, my naive approach has $$f\Big(\frac{5(\sqrt{13}+2)}{9}\Big)=f\Big(\frac{5\cdot 3+10}{9}\Big)=f\Big(\frac{25}{9}\Big)=2.$$ But having looked this value up in a calculator, it is rougly $3.1141...$ which gives the floor of $3$ of course.

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So where have I gone wrong?

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Reflection. Thank you to all that have helped me with this. I will be wary on improving my bound next time.

Answer 1

$$ 325 > 289 $$ $$ \sqrt{325} > \sqrt{289} $$ $$ \sqrt{5^2 \cdot 13} > \sqrt{17^2} $$ $$ 5 \sqrt {13} > 17 $$ $$ 5 \sqrt {13} + 10 > 27 $$ $$ 5 ( \sqrt {13} + 2) > 27 $$ $$ \frac{ 5 ( \sqrt {13} + 2)}{9} > 3 $$

Answer 2

You've gone wrong in assuming that you could just substitute the $3=\lfloor\sqrt{13}\rfloor$ when caluculating the floor of an expression containing $\sqrt{13}$. Just to show you how wrong that is $\lfloor 5\sqrt{13}\rfloor=\lfloor\sqrt{325}\rfloor\approx\lfloor 18.027756\rfloor=18$ while $5*\lfloor\sqrt{13}\rfloor=15$.