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 Math 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

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