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Bashyboy
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Homework Statement
Suppose that ##x \in \mathbb{N}## is such that ##n < x < 2n##, where ##n## is another natural number. I want to conclude that it is impossible for ##x## to divide ##2n##. Also, is there a way to naturally generalize this to all integers?
Homework Equations
The Attempt at a Solution
Since ##x \in (n,2n)##< then ##x = n + \ell## for some natural number ##\ell \in (0,n)##. Now, suppose the contrary, that ##x## divides ##2n##. Then ##2n = k(n + \ell)## for some ##k \in \mathbb{Z}##. Immediately we can say that ##k## must be a positive integer. Rearranging the equation gives
##n(2-k) = k(n+\ell)##
If ##k## is either ##1## or ##2##, then we get ##n = \ell## or ##2 \ell = 0##, both of which are contradictions. If ##k > 2##, then the LHS will be negative but the RHS will be positive, a contradiction.
How do I make this a little more rigorous?
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