Minus square root of 2, minus square root of 3, minus pi, minus e, etc. Or, to be a bit more fancy, 1 - pi, 2 - pi, 3 - pi, pi - 4, pi - 5, 2 - e, e - 3, square root of 2 minus 2, etc.
8.53973 (rounded)
Because Euler proved it! (No, I can't!)
'pi' and 'e' both fit that description.
E=mc2 =0.111x 300,000,000x 300,000,000= 10,000,000,000,000,000 (10 quadrillon) joules.
About 20.29791
e^pi ~ 23.14069.............., not rational
It is NOT rational, but it IS real.Start with Euler's formula: e^ix = cos(x) + i*sin(x) for all x.When x = pi/2,e^(i*pi/2) = cos(pi/2) + i*sin(pi/2) = 0 + i*1 = ior i = e^(i*pi/2)Raising both sides to the power i givesi^i = e^[i*(i*pi/2)] = e^[i*i*pi/2]and since i*i = -1,i^i = e^(-pi/2) = 0.20788, approx.
i (taken to be sqrt(-1) for this question) requires that you know a bit about writing complex numbers. i = e^(i*pi/2) so i^i = (e^(i*pi/2))^i which equals e^(i*i*pi/2) since i*i = -1 we get e^(-pi/2) so i^i = e^(-pi/2) which is roughly .207879576
%g is more compact. Do some tests, for example:double pi= 3.1415926535897932384626433;printf ("%%f gives %f %f %f %f %f\n", pi, 100*pi, 10000*pi, 1000000*pi, 100000000*pi);printf ("%%e gives %e %e %e %e %e\n", pi, 100*pi, 10000*pi, 1000000*pi, 100000000*pi);printf ("%%g gives %g %g %g %g %g\n", pi, 100*pi, 10000*pi, 1000000*pi, 100000000*pi);%f gives 3.141593 314.159265 31415.926536 3141592.653590 314159265.358979%e gives 3.141593e+00 3.141593e+02 3.141593e+04 3.141593e+06 3.141593e+08%g gives 3.14159 314.159 31415.9 3.14159e+06 3.14159e+08
epi = 23.140692632779. pie = 22.459157718361. Thus, epi is greater.
Minus square root of 2, minus square root of 3, minus pi, minus e, etc. Or, to be a bit more fancy, 1 - pi, 2 - pi, 3 - pi, pi - 4, pi - 5, 2 - e, e - 3, square root of 2 minus 2, etc.
8.53973 (rounded)
9
sqr(e/pi)
e. g. 103993/33102
by euler: i=ei(pi)/2 therifore ii = (ei(pi)/2)i=ei^2(pi)/2=e-(pi)/2 ~0.208