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Topic: Power of 2 divisible by 3
Replies: 19   Last Post: Feb 1, 2006 10:37 PM

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Carlos Moreno

Posts: 51
Registered: 10/31/05
Re: Power of 2 divisible by 3
Posted: Jan 30, 2006 4:02 PM
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Chip Eastham wrote:

> [...]
> One can set aside the theory of primes and give a direct
> proof by induction, based on 1 + 2 = 3. From this clearly
> 3 does not divide 2 = 2^1, so that's the basis step.
> Suppose 3 does not divide 2^k (the induction hypothesis).
> Then from 1 = 3 - 2, we multiply both sides by 2^k:
> 2^k = 3*(2*k) - 2*(k+1)
> Now if (for the sake of contradiction) one assume 3 will
> divide 2^(k+1), then 3 divides both terms on the right
> hand side and must also divide the left hand side 2^k.
> This contradicts the induction hypothesis, so the
> assumption that 3 will divide 2^(k+1) must be false.


It's always refreshing seeing some "unconventional thinking"
that puts aside the obvious ("the obvious" in this case being
the fundamental theorem of arithmetic) to come up with a
simple, nice, refreshing proof like the above.


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