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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 $\dfrac{23}{7}$?

José Carlos Santos
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MC2
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    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. – JonathanZ Jul 20 '22 at 16:10
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    Hint: $2/7 > 2/8 = 0.25$. This is all you need, assuming you know that $\pi \approx 3.14$. – Doug Jul 20 '22 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.) – Dave L. Renfro Jul 20 '22 at 16:33
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    $\int_0^1 \frac{x^4(1-x)^4}{(1+x^2)} = \frac{22}{7} - \pi$ 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. – Sarvesh Ravichandran Iyer Jul 20 '22 at 16:34
  • See this question for a proof of the perimeter bound using the theory of sequences. – Oscar Lanzi Jul 20 '22 at 20:07
  • Um.... I don't see how you can actually solve this without knowing significant information about the value of $\pi$. I was taught that $\frac {22}7$ is a very good estimation of $\pi$ which implies $\frac {22}7$ is a better estimation than $\frac {21}7$ or $\frac {23}7$. So If $\frac {22}7$ is a better estimate and $\frac {23}7$ is a worst estimate than $\frac {23}7$ is too big and $\pi -\frac {23}7 < 0$ so $|\pi -\frac {23}7| =\frac {23}7-\pi$.... but I was also taught that $\pi \approx 3.14$ so.... to be continued.... – fleablood Jul 20 '22 at 23:25
  • $\frac {23}7 = 3\frac 27$ and we have to compare $\frac 27$ to $0.14$ but $7\times 0.14 < 1 < 2$ so $\frac 27<0.14$ and $\frac {23}7 > 3.14\approx \pi$. But all of those assume I know something about the value of $\pi$. so..... I don't understand your question. If you don't know anything about the size of $\pi$ you can't answer this. If you do know something about the size of $\pi$ then you probably know $\pi < \frac {23}7$ and there isn't anything to answer. – fleablood Jul 20 '22 at 23:29
  • Meh..... I suppose this is probably good enough for pre-calculus. $3.14 < \pi < 3.15$ so $3.14\times 7 =21.98 < 7\pi < 3.15\times 7 = 22.05$ so $21 < 21.98 < 7\pi < 22.05 < 23$ so $\frac {21}7 < \pi < \frac {23}7$. ... [This also shows us that $21.98 < 7\pi < 22.05$ so $7\pi \approx 22$ and therefore $\pi \approx \frac {22}7$ is a very good estimation. (unfortunately this is not enough to tell us if $\frac {22}7$ is bigger or less than $\pi$..... Of course if we just do $3.141 < \pi < 3.142$ we can probably figure that out.) – fleablood Jul 20 '22 at 23:40

5 Answers5

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

insipidintegrator
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  • My mind? Blown. – JonathanZ Jul 20 '22 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. – insipidintegrator Jul 20 '22 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. – Brian Tung Jul 21 '22 at 00:37
  • 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? – Eric Snyder Jul 21 '22 at 03:25
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    @EricSnyder I found this on mathOverflow: https://mathoverflow.net/questions/67384/source-and-context-of-frac227-pi-int-01-x-x24-dx-1x2 – insipidintegrator Jul 21 '22 at 15:41
  • I’ll also add https://math.stackexchange.com/questions/2229847/asymptotic-quality-of-rational-approximations-to-pi, https://vixra.org/pdf/1711.0310v1.pdf, https://math.stackexchange.com/questions/1662540/interpretation-of-frac227-pi, and https://math.stackexchange.com/questions/3048678/who-was-dalzell-pi-22-7 @EricSnyder – insipidintegrator Jul 28 '22 at 17:47
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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

$$ P = 24 \tan \frac{\pi}{12} = 24(2-\sqrt3) $$

enter image description here

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.

Brian Tung
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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.

ArthD21
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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?

egreg
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We have that

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

user
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