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 Skip Navigation LinksMath Help > Math Jokes > Proof that log 2 = 0

Using this proof, you can double any number without changing its value!

Proof that log 2 = 0    *

* note: "log", used here, means natural log, which is sometimes written "ln".

This proof is extremely important, because it can be used to show that 2=1.  It works this way:

log 2 = 0, from the proof, above.

log 1 = 0, because e0=1 

log 2 = log 1, because both logs are equal to 0 

elog 2 = elog 1, taking the exponent of both sides

2 = 1, because elog a = a 

Internet references

I got an email questioning the statement that log 2 = 1-1/2+1/3-1/4+..., so I found these references to support this fact, which, by the way, has nothing to do with the error in the proof, above.

http://numbers.computation.free.fr/Constants/Log2/log2Formulas.html attributes log(2)=1-1/2+1/3-1/4 to Nicolaus Mercator and gives dozens of other formulas for log(2).

http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/harmonic-series gives a sketch of the proof that this series is the integral of dy/y from 1 to 2, which is log(2).

H(n) = 1 + 1/2 + 1/3 + ... + 1/n.
H(2n) - H(n) = 1/(n+1) + 1/(n+2) + ... + 1/(2n).

Prop A: H(2n) - H(n) = 1 - 1/2 + 1/3 - 1/4 ... + 1/(2n-1) - 1/(2n)
Follows from Induction: n=0 : H(0) - H(0) = 0
n−>n+1: H(2(n+1))-H(n+1) =
H(2n) + 1/(2n+1) + 1/(2n+2) - H(n) - 1/(n+1) =
H(2n) - H(n) + 1/(2n+1) - 1/(2n+2) =
1 - 1/2 + 1/3 - 1/4 ... + 1/(2n+1) - 1/(2n+2).

Or Prop A directly:
H(2n)      =  1  + 1/2 + 1/3 + 1/4 + ... + 1/(2n-1) + 1/(2n)
H(n)       = 0/1 + 2/2 + 0/3 + 2/4 + ... + 0/(2n-1) + 2/(2n)
H(2n)-H(n) =  1  - 1/2 + 1/3 - 1/4 + ... + 1/(2n-1) - 1/(2n)

Prop B: H(2n) - H(n) is monotonically increasing and converges to log(2).
The limit of 1 - 1/2 + 1/3 - 1/4 ... was first calculated by Nicolaus Mercator [Logarithmotechnia, 1667]. He found the series
log(1+x) = x - x2/2 + x3/3 - x4/4 ... But you can do that without using the Taylor series.
Adding the estimates 1/(n+1) < Integral[1/x, x, {n,n+1}] < 1/n gives
H(2n)-H(n) < Integral[1/x, x, {n,2n}] < H(2n-1)-H(n-1) = H(2n)-H(n)+1/(2n).
Now Integral[1/x, x, {n,2n}] = Integral[1/y, y, {1,2}] via y = nx. (An interpretation of this transform gives [Yegorov].)
So we get log(2) - 1/(2n) < H(2n) - H(n) < log(2).
As log(t) is defined as the Integral[1/y, y, {1,t}] see [Yegorov].

Regrouping the Harmonic Series, by Dan Teague, 2003, Department of Mathematics and Computer Science, North Carolina School of Science and Mathematics

Related pages in this website

Index of joke proofs

An alternative proof that 2=1

A proof by induction that e = π

 

The webmaster and author of this Math Help site is Graeme McRae.