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jhonyy9

  • 4 years ago

prove that every prim p>2 can be writing in the form of 2n+1 for n>0 ,n number natural,so p-1=2n --- p-1 --- even,because p is odd. so p>2 --- than p>=3 p is prim,p>2 so than p is even so p=2k+1,where k=1,2,3,...,n. so p=2k+1 --- subtract 1 from both sides,than p-1=2k 2n=p-1=2k so 2n=2k --- divide both sides by 2 n=k - so for every p prims there are n>=1,n natural , such that p=2n+1 - is this correct,right ?

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  1. anonymous
    • 4 years ago
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    By division algorithm, every integer can be written in the form 2k or 2k+1.Primes are those which do not have ANY factors except 1 and itself.Thus, the form 2k is out of bounds for k>1. Thus, every prime >2 is of the form 2k+1

  2. jhonyy9
    • 4 years ago
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    ok but how to prove it this ?

  3. anonymous
    • 4 years ago
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    That IS the proof!

  4. jhonyy9
    • 4 years ago
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    what ? can you write this here ?

  5. anonymous
    • 4 years ago
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    1.Every int can be written as 2k or 2k+1. 2.The int 2k is divisible by 2.Thus it is not a prime for k>1 3.The conclusion follows.

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