00:01
We are going to prove that every positive real number has a positive nth root.
00:07
That is, for each natural number n and a positive number, a positive real number a, we get to prove that there exists a c positive number such that c raised to the nth power equal a.
00:25
So for that, let n be any natural number, and a real number.
00:34
Number that is positive.
00:46
And now we get to prove that we can find a positive number, a positive real number, such that c raised to the nth power equal a.
00:55
And so let's say first that if n is equal 2, sorry n equal 1, we can take c equal a because c to the nth power will be c to the nth power will be c to the power 1, which is just c, and then if we take c equal a, we fulfill this equation here.
01:42
Okay, so it means that when n equal 1 is very simple to find c, positive c with the property, because that c will be the same number a.
01:57
And with that c is positive.
02:00
So we can assume that n is created down or equal to 2.
02:12
And a is a real number which is positive.
02:16
So now we define the function f of x equal x to the n minus a, where n is a natural number here, which is greater than or equal to 2, and a is a given real number which is positive.
02:38
Okay, so let's define that function there, and let's study two cases.
02:44
If a is less than 1, because a is positive, let's say if a is between 0 and 1, we get to consider f on the interval 0 -1.
03:10
In that case, then f at 0, if we evaluate this function at 0, we get negative a, and because a is positive, this will be a negative number.
03:29
And f at the right end point of this close interval 1 is 1 to the nth power minus a, which is 1 minus a.
03:41
And now because a is less than 1, this number is positive.
03:49
So f at 0 is a negative number and f at 1 is a positive number.
03:54
Since f is a continuous function, because it's a polynomial function, we must have a root of this function that is a number between 0 and 1 for which the function is 0.
04:07
So let's put that here...