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 $\dfrac{23}{7}$?
 A: Here's an idea that, like insipidintegrator's answer, avoids an approximation. Use a circumscribing polygon to upper-bound the circumference of the circle. I use a dodecagon, whose perimeter is
$$
P = 24 \tan \frac{\pi}{12} = 24(2-\sqrt3)
$$

We want to show that this $P < 2\left(\frac{23}{7}\right) = \frac{46}{7}$.
\begin{align}
24(2-\sqrt3) < \frac{46}{7}
    & \Leftrightarrow 84(2-\sqrt{3}) < 23 \\
    & \Leftrightarrow 168-84\sqrt{3} < 23 \\
    & \Leftrightarrow 145 < 84\sqrt{3} \\
    & \Leftrightarrow 21025 < 21168
\end{align}
Then, since $2\pi < P$, we have $\pi < \frac{23}{7}$. It should be pointed out that this does require you to accept that a circumscribing polygon is longer in perimeter than the circumference of the circle.
A: You have already said that $\frac{21}{7}$ is 3, and that both $\pi$ and $\frac{23}{7}$ are greater than 3. But, $\frac{23}{7}$ would be $3\frac{2}{7}$, and we know that the first digit of $\frac{2}{7}$ after the decimal point is 2 (because 20 divided by 7 is 2.something). So, since $\pi$ starts with 3.1, it must be less than $\frac{23}{7}$, and thus the expression in question must be less than 0.
A: It has been known for more than 22 centuries that $\pi<22/7$ (Archimedes). But if you don't know it, you can do primary school division:
\begin{array}{cccc|l}
2&3&0&0&\underline{7~~~~} \\ 
 &2&0& &328 \\
 & &6&0& \\
 & & &4
\end{array}
Thus we deduce that $23/7>3.28$
Can you finish?
A: 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 $\Bigg |\pi-\dfrac{23}{7}\Bigg|=\dfrac{23}{7}-\pi$.
A: We have that
$$\pi - \dfrac{23}{7} <\frac{315}{100}-\dfrac{23}{7}=\frac{2205}{700}-\dfrac{2300}{700}<0$$
