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

[Mr Davis 97]

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.

 

[Mr Davis 97]

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.

 

[PAllen]

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.

 

[fresh_42]

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.

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?

 

[Mark44]

I deleted my earlier post, and PAllen's reply to it.

 

[PAllen]

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.

Or even simpler, along this line, it implies 1 is even; but 1 is odd.

 

[mfb]

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, $2n \geq 2 > 1$.