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The following argument that '''22/7 > π''' will be readily understood by persons with no knowledge of ] beyond first-year ]. The following argument that '''22/7 > π''' will be readily understood by persons with no knowledge of ] beyond first-year ].

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pi

The following argument that 22/7 > π will be readily understood by persons with no knowledge of mathematics beyond first-year calculus.

Some non-mathematicians have suggested that it is "well-known" that π = 3.14159..., and therefore it must be less than 22/7 = 3.142857... . But in order to know that π = 3.14159..., as opposed to believing it because all of the textbooks say so, we would need to go through a process of computation much more complicated than the argument below.

The idea

0 < ∫ 0 1 x 4 ( 1 − x ) 4 1 + x 2 d x = 22 7 − π . {\displaystyle 0<\int _{0}^{1}{\frac {x^{4}(1-x)^{4}}{1+x^{2}}}\,dx={\frac {22}{7}}-\pi .}

The details

It's not hard to see the integral is positive. Observe that the integrand is a quotient whose numerator and denominator are both nonnegative, being sums or products of even powers of real numbers. Thus the integrand is nonnegative, and evaluating it at 1/2 gives a positive value, so the integral from 0 to 1 is positive.

It remains to show that the integral in fact evaluates to the desired quantity:

0 < ∫ 0 1 x 4 ( 1 − x ) 4 1 + x 2 d x {\displaystyle 0<\int _{0}^{1}{\frac {x^{4}(1-x)^{4}}{1+x^{2}}}\,dx}
= ∫ 0 1 x 4 − 4 x 5 + 6 x 6 − 4 x 7 + x 8 1 + x 2 d x {\displaystyle =\int _{0}^{1}{\frac {x^{4}-4x^{5}+6x^{6}-4x^{7}+x^{8}}{1+x^{2}}}\,dx}
= ∫ 0 1 ( x 6 − 4 x 5 + 5 x 4 − 4 x 2 + 4 − 4 1 + x 2 ) d x {\displaystyle =\int _{0}^{1}\left(x^{6}-4x^{5}+5x^{4}-4x^{2}+4-{\frac {4}{1+x^{2}}}\right)\,dx}
= x 7 7 − 2 x 6 3 + x 5 − 4 x 3 3 + 4 x − 4 arctan ⁡ x | 0 1 {\displaystyle =\left.{\frac {x^{7}}{7}}-{\frac {2x^{6}}{3}}+x^{5}-{\frac {4x^{3}}{3}}+4x-4\arctan {x}\,\right|_{0}^{1}}
= 1 7 − 2 3 + 1 − 4 3 + 4 − π   {\displaystyle ={\frac {1}{7}}-{\frac {2}{3}}+1-{\frac {4}{3}}+4-\pi \ } (recall that arctan(1) = π/4)
= 22 7 − π . {\displaystyle ={\frac {22}{7}}-\pi .}

Now we see the difference between 22/7 and π is greater than zero, so 22/7 > π.

Appearance in the Putnam Competition

The evaluation of this integral was once one of the problems set in the Putnam Competition. If it seems trivially routine for a Putnam Competition problem, one may perhaps surmise that its inclusion was motivated by the conjunction of the punch line (summarized by the title of this article) with the fairly nice pattern in the integral itself.

See also