Show that every positive real number has a unique nth root. That is show that if

Question

Show that every positive real number has a unique nth root. That is show that if p is any positive real number and n is a positive integer, then there is a unique positive real number x such that x^n = p.

Explanation / Answer

We consider a set X = {x in positive reals st x^n = p [its an upper bound] So now if a^n > p then we can consider the number (a-y)^n > p where y>0 and is really small if y < 1 then (a-y)^n > a^n - n* a^(n-1)*y > p hence we can find the required y Hence if a^n > p then we have a smaller bound for X which is a contradiction Hence a^n = p to prove uniqueness just say a^n = b^n the a^n-b^n = (a-b) (a^(n-1) +..b^(n-1)) = 0 but (a^(n-1) +..b^(n-1))>0 hence a-b = 0 hence we have a unique positive soln Hence proved. message me if you have any doubts

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