Math and science::Analysis::Tao::05. The real numbers
root of a real
The definition of the nth root of a real utilizes the supremum, hence why the Let be a non-negative real. Let E be the set . We define the root of , denoted as , to be . Thus,
The existence of the nth root, proposition
Let be a positive real and let be a positive integer. Then is a real number (it exists).
An easy proof
Outline: we must show that the set 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 , by definition.
Let E be the set . 0 is an element of E ( for any natural number ). Thus E is non-empty. E also has an upper bound. To show this consider the two possible cases, and .
If , then 1 is an upper bound of E. If it isn't then for some . implies that so that doesn't meet the criteria for being an element of E, a contradiction.
If , then is an upper bound of E. If it isn't then for some . This implies that , so again, 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 is a real number.