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I How to prove that 2n=1 has no integer solutions

  1. Feb 19, 2017 #1
    This might seem like a very simple problem, because we could just say that the only possible solution is n = 1/2, which is not an integer. But I am curious as to how to prove that there is no solution, with no knowledge of rational numbers, just as we can prove that x^2 = 2 as no rational solutions without any knowledge of irrational numbers.
     
  2. jcsd
  3. Feb 19, 2017 #2
    I am confused. 1/2 is never an integer or even number. I am asking how I would prove that 2n=1 has no integer solutions, without just saying that n=1/2 is not an integer.
     
  4. Feb 19, 2017 #3

    PAllen

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    There are different approaches. For example, derive (from a chosen set of integer axioms) that n solving this equation must be both even and odd, a contradiction.
     
  5. Feb 19, 2017 #4

    fresh_42

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    The first thing to do - as always - check the statement: What is an integer? The proof of the irrationality of ##\sqrt{2}## which you mentioned, uses, that rationals can be written as quotients of integers, which are products of primes.

    If you take the definition of primes again, then ##2n=1## implies the prime ##2\,\vert \,1## and is therefore a unit. But primes aren't allowed to be units, so ##2 \nmid 1## and ##n \notin \mathbb{Z}\,.## Or if you like: the only units in ##\mathbb{Z}## are ##\pm 1## so it's impossible because of that.

    I know this is a bit like cheating, because it plays with definitions, so again: what is an integer?
     
  6. Feb 19, 2017 #5

    Mark44

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    I deleted my earlier post, and PAllen's reply to it.
     
  7. Feb 19, 2017 #6

    PAllen

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    Or even simpler, along this line, it implies 1 is even; but 1 is odd.
     
  8. Feb 20, 2017 #7

    mfb

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    With inequalities: 2*0 is not 1 (skip this if you don't include 0 in the integers). For every integer n larger than 0, ##2n \geq 2 > 1##.
     
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