Okay, so I have a number

*a* and I want to show that I can always define a unique

*x* for every integer

*n* as

which is conversely

for this, I'd want to show that any real number can be divided into

*n* times, right? Also, that for every real number there is only ONE number that can be multiplied by itself

*n* times to produce that original number

*a*? Also given that all components will be positive.

I'm not sure how to apply that theorem, as it says if there is a subset of the reals, that is bounded, then there is a 'least upper bound' and I honestly don't see how that makes sense, seeing as how if there is an upper bound, how can there be a

*least*? Wouldn't all upper bounds be the same, since if there is a an upper bound

**greater** than the least upper bound, wouldn't it nullify the position of said least upper bound?

BUT, going on the idea that there is a subset

*A* of the reals, bounded by

*a* such that

for all numbers

, there exists one

*x* that would be the

*n*th root of

*a*. Would it be sound to state that, in the subset of the reals,

if and only if

, if a=1, x=1=a, because

, no matter how many ones you have. ummmmm..... Stuck, sorry, not sure how to prove it from here. this one seems almost like it should be proved by definition.

the second exercise, prove that, if

, there exists an integer

*n* such that

is equivalent to

or,

if

*y* is the least upper bound of

*S*, then

*x*, by definition, is less than

*y* and, subsequently contained in

*S*.

for every real number n, there exists a number n+1 that is larger than n. So, I can define

*n* **outside** of my subset, that there exists a real number

such that, as long as

since

, since, by definition both x and y are positive. Is that right?