<s>[INST] Express $|\pi - \frac{23}{7}|$ without the absolute value symbol 
Express $\left|\pi - \dfrac{23}{7}\right|$ without the absolute value symbol.

I know I have to check if $\pi - \dfrac{23}{7}$ is greater than (or equal to) zero, but how can I do it analytically (without a calculator)?
I know that $\pi \gt 3=\dfrac{21}{7}$ but how to compare $\pi$ with $\dfrac{23}{7}$?
 [/INST] I will not use any approximations in this answer.
Consider the definite integral $$\int_0^1\frac{t^4(1-t)^4}{1+t^2}dt.$$ Simply expand the numerator using binomial formula and reduce the numerator in terms of the denominator. I’ll skip a few steps for the sake of brevity: $$\int_0^1\frac{t^4(1-t)^4}{1+t^2}dt=\int_0^1\left(-4t^5+t^6+t^4+\frac{4t^6}{1+t^2} \right)dt$$$$= \int_0^1\left(-4t^5+t^6+5t^4-\frac{4t^4}{1+t^2} \right)dt= \int_0^1\left(-4t^5+t^6+5t^4-4t^2+\frac{4t^2}{1+t^2} \right)dt $$$$=  \int_0^1\left(-4t^5+t^6+5t^4-4t^2+4-\frac{4}{1+t^2} \right)dt$$$$=\bbox[5px, border:2px solid red]{\frac{22}{7}-\pi.}$$ This is a very nice expression containing both $\dfrac{22}{7}$ and $\pi$.
Now, note that the function $\displaystyle f(t)= \frac{t^4(1-t)^4}{1+t^2}$ is ALWAYS positive for all $t\in (0,1)$. This means that the integral $\displaystyle\int_0^1 \frac{t^4(1-t)^4}{1+t^2} dt$ is also strictly positive. Thus, we get, $$\bbox[5px, border:2px solid gold]{\frac{22}{7}-\pi>0\implies \frac{22}{7}>\pi.}$$
Hence we finally arrive at the desired conclusion: $$\pi<\frac{22}{7}<\frac{23}{7}.$$

Thus, we can write $\Bigg |\pi-\dfrac{23}{7}\Bigg|=\dfrac{23}{7}-\pi$.
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