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 $$\dfrac{23}{7}$$?

• What facts about $\pi$ do you know, or are allowed to use, in answering this question? Because if you know that the first few digits of its decimal expansion are $3.14$, it's pretty easy. Jul 20, 2022 at 16:10
• Hint: $2/7 > 2/8 = 0.25$. This is all you need, assuming you know that $\pi \approx 3.14$.
– Doug
Jul 20, 2022 at 16:11
• If an explanation is not required in your answer (e.g. the question is "answer only" or "multiple choice"), then it follows from the (essentially) "common knowledge fact" that $22/7$ is the best approximation for $\pi$ using one- and two-digit integers that $\pi$ has to be between $21/7$ and $23/7.$ (Why best? Common sense -- if another such approximation was better, then everyone would be learning the other approximation and not the $22/7$ approximation.) Jul 20, 2022 at 16:33
• $\int_0^1 \frac{x^4(1-x)^4}{(1+x^2)} = \frac{22}{7} - \pi$ is the integral of a positive function, hence a positive quantity. Therefore $\frac {22}7 > \pi$. The same follows for $\frac {23}7$. I'm not sure this is what you're looking for, it is "analytically" showing what you need. Jul 20, 2022 at 16:34
• See this question for a proof of the perimeter bound using the theory of sequences. Jul 20, 2022 at 20:07

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 $$\Bigg |\pi-\dfrac{23}{7}\Bigg|=\dfrac{23}{7}-\pi$$.

• My mind? Blown. Jul 20, 2022 at 17:12
• Thanks @JonathanZsupportsMonicaC! Actually where I am from, “Prove that $\int_0^1\frac{t^4(1-t)^4}{1+t^2}dt=\frac{22}{7}-\pi$” is a very popular question in worksheets. So all I had to do was to connect the OP’s question to the integral and write the last paragraph by myself. Jul 20, 2022 at 19:27
• Not to detract from @insipidintegrator's fine solution (which I upvoted), but it can also be found elsewhere on Math.SE. It is a clever approach to the approximation, and a clearly non-Archimedean one. Jul 21, 2022 at 0:37
• Where does this integral come from? One of the series expansions for $\pi$? It seems too simple to have been conjured from scratch just for an exercise? Jul 21, 2022 at 3:25
• @EricSnyder I found this on mathOverflow: mathoverflow.net/questions/67384/… Jul 21, 2022 at 15:41

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.

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.

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?

We have that

$$\pi - \dfrac{23}{7} <\frac{315}{100}-\dfrac{23}{7}=\frac{2205}{700}-\dfrac{2300}{700}<0$$