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Famous Theorems of Mathematics/Root of order n
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For all and for all there exists such that .
Proof
Let us define a set .
This set is non-empty (for ) and has an upper-bound (for all we get ).
Therefore, by the completeness axiom of the real numbers it has a supremum . We shall show that .
Suppose that .
It is sufficient to find such that :
hence , but and so . A contradiction.
Suppose that .
As before, it is sufficient to find such that :
hence , but and so . A contradiction.
Therefore .