## All Powers of 2 Are Equal to 1

We are going to prove by induction that,

For all integer n ≥ 0, 2 ^{n} = 1.

The claim is verified for n = 0; for indeed, 2^{0} = 1.

Assume the equation is correct for all n ≤ k, that is

2^{0} = 1, 2^{1} = 1, 2^{2} = 1, ..., 2^{k} = 1.

From these we now derive that also 2^{k+1} = 1:

2^{k+1} = 2^{2k} / 2^{k-1} = 2^{k} × 2^{k} / 2^{k-1} = 1×1/1 = 1.

Induction is complete.

### References

- P. Brown, Proof by Induction Test

|Contact| |Front page| |Contents| |Geometry| |Up|

Copyright © 1996-2018 Alexander Bogomolny

The error in the proof is subtle. The inductive step had to be formulated as

Assume the equation is correct for all 0 ≤ n ≤ k, where k ≥ 0.

The derivation then clearly fails for k = 0; for,

2^{1} = 2^{0} / 2^{-1} = 2^{1},

which is 2^{k+1} = 2^{2k} / 2^{k-1} for k = 0. So ^{1} = 2^{1}^{1} = 1.^{2} = 1,

|Contact| |Front page| |Contents| |Geometry| |Up|

Copyright © 1996-2018 Alexander Bogomolny