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Math and science::Analysis::Tao::05. The real numbers

nth root of a real

The definition of the nth root of a real utilizes the supremum, hence why the nth doesn't exist for rationals.

Let 0x be a non-negative real. Let E be the set E={yR:y0 and yn<x}. We define the nth root of x, denoted as x1n, to be x1n:=supE. Thus,
x1n:=sup{yR:y0 and yn<x}

The existence of the nth root, proposition
Let x0 be a positive real and let n1 be a positive integer. Then x1n is a real number (it exists).

An easy proof
Outline: we must show that the set {yR:y0 and yn<x} has an upper bound; if it has an upper bound, it follows that it has a least upper bound and this least upper bound is the nth root of x, by definition.

Let E be the set {yR:y0 and yn<x}. 0 is an element of E ( 00 and 0k=0 for any natural number k). Thus E is non-empty. E also has an upper bound. To show this consider the two possible cases, x<1 and x1

If x<1, then 1 is an upper bound of E. If it isn't then 1<y for some yE. y>1 implies that yn>1 so that y doesn't meet the criteria for being an element of E, a contradiction.

If x1, then x is an upper bound of E. If it isn't then y>x for some yE. This implies that yn>x, so again, y cannot be an element of E, a contradiction.

So E must have an upper bound. As it is a non-empty set of reals with an upper bound, it must have a least upper bound. Thus x1n is a real number.