They told me that $\sqrt{2}$ is real.But what about square root of $\sqrt{2}$? [closed]

My question was; is the square root(principal) of non-terminating repeating and non-terminating non-repeating real numbers; $REAL$? What you guys did not understand(the reason you put it on hold)was how I got to this question.
Consider the function $f(x)=2x$.
The domain is really simple;it is $R$.(Because the twice of every real no exists.)
But for the range of this function;before I can say that that is also $R$,I need to confirm it.And there is a very simple way of doing that.In your mind speak any real number(positive,negative,terminating,non-terminating;all kinds of).
Now suppose I speak $-11$ , well , it belongs in the range.$(-5.5,-11)$ is an element present there(in the function $f$).$32.444.....$,again $(16.222....,32.4444....)$ is an element of the set(of the function $f$).
So,from, $-\infty$ to $\infty$,every real number that I can think of belongs in the range.
So,the range is $R$.
And I was trying to do this same thing with $f(x)=x^2$.
The domain is clearly $R$.(Every real number can be squared.)
But for the range there was some confusion.One can intuitively guess that it is the set of non-negative real numbers.
I did that test on this one also;speak any non-negative real number in your mind.
$1$, it belongs in the range; $(1,1)$.
$2$, belongs in the range; $($$\sqrt{2}$$,2)$.
But what about the non-negative real number $1.3333....$ ?
What about the non-negative real number $\sqrt{2}$$=1.41421356....? Do they belong in the range?That's why I asked , is the square root of \sqrt{2} and \frac{4}{3} Real? Because then I could safely say that the range is the set of non-negative real numbers. (I hope I've made it clear this time.) closed as unclear what you're asking by Lord Shark the Unknown, Hans Lundmark, Rohan, eranreches, José Carlos SantosDec 24 '17 at 18:33 Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question. • Every positive real number has a positive real square root. – Lord Shark the Unknown Dec 24 '17 at 9:38 • yes, these are all real numbers. The square root of \sqrt{2} is a real number \omega such that \omega ^2 = \sqrt{2}. – Alvin Lepik Dec 24 '17 at 9:40 • i also used a certain logic and felt like that;a number whose decimal expansion does not end is a real no.and the decimal expansion of \sqrt{2}$$=1.414....$ does not end; so if you start calculating square root of $1.414......$ using the long division method, how will that end?it will not end,the process will go on.so it is real.i just needed a little bit of confirmation and i have got that. – Durgeshwar Ojha Dec 24 '17 at 9:57
• What is $(-5.5, -11)$? And $-11$ does not look like to be in set. – user99914 Dec 26 '17 at 0:13
• @JohnMa Well,$-11$ is in the RANGE of the function $f(x)=2x$.because $( (-5.5),-11)$ is an element (an ordered pair) that belongs in the function $f(x)=2x$.Functions are sets of ordered pairs,isn't it so? – Durgeshwar Ojha Dec 26 '17 at 0:23

That's why we do analysis. We have an intermediate value theorem (IVT) there. Consider the function $$f(x):=x^2-\sqrt{2}\quad(0\leq x\leq2)\ .$$ Then $f(0)<0$ and $f(2)>0$. The IVT then tells us that there is a $\xi\in\>]0,2[\>$ such that $f(\xi)=0$.

This $\xi$ satisfies the equation $$(\xi^2-\sqrt{2})(\xi^2+\sqrt{2})=\xi^4-2=0\ .$$ Newton's method then allows us to find as many decimal places of $\xi$ as we want. We don't even need the decimal expansion of $\sqrt{2}$ for that.

• Wait, IVT only tells us something if $f$ is continuous. How can we know that if we don't already know that non-negative reals are squares? – Dan Brumleve Dec 27 '17 at 8:32
• polynomials are continuous functions – Alvin Lepik Jan 1 '18 at 9:44

We'll use the concept of a Dedekind cut: every real number $x$ is defined by two sets, such that every rational number less than $x$ is in one set, and every rational number greater than or equal to $x$ is in the other.

Since we have an infinite set, we have to do this via a rule: we need to find a way to describe every member of each set. Let's see if we can build that.

For a number $y$ to be in the greater set, it must satisfy $y \ge \sqrt {\sqrt{2}}$. Now we must get rid of the square roots: $y^2 \ge \sqrt{2} \wedge y\ge 0$, then $y^4 \ge 2 \wedge y\ge 0$. Similarly, for y to be in the lesser set, it must satisfy $y^4 < 2 \vee y< 0$.

To complete the proof we must show that adding a small amount to an element in the greater set gives another element in the greater set, and that subtracting a small amount from an element in the lesser set gives another element in the lesser set - thus that all elements in the lesser set are less than all elements in the greater set.

For this, we can rely upon the fact that, for non-negative $y$, $f(y)=y^4$ is strictly increasing: positive elements in the greater set grow as you add more, and positive elements in the lesser set shrink as more is subtracted. In addition, $f(y)=y$ is strictly increasing everywhere, so non-positive elements in the lesser group remain non-positive, and all elements in the greater group remain positive.

Since we can divide the rationals into two piles with our target between them, our target must be real.

This particular number is the positive real solution to $x^4 - 2=0$, so it is also algebraic.

The completeness of real numbers guarantees that every positive number $a$ has unique positive square(or, more generally, $n$th) root. Therefore, $\sqrt{2}$ is definitely in the range of $f(x)=x^2$

Intuitively, the completeness of real numbers means that real numbers have no gaps(or holes). The formal statement of completeness is as follows:

Let $A \subset \mathbb{R}$. If there is a real number $x$ such that $y \le x$ for all $y\in A$, then there is the smallest such $x$.

The sketch of proof is as follows. We can do the similar thing for $n$th root, too.

Let $A$ be the set of all positive rational numbers whose square is less than $a$. Then $a$ is nonempty since $\frac{a}{1+a} \in A$. Also, if $x\in A$, then $(1+a)^2-x^2 >(1+a)^2-a = 1+a+a^2 >0$, so $x<1+a$. Hence, by completeness of real numbers, there is the smallest positive real number $l$ satisfying $x \le l$ for all $x\in A$. Such $l$ can be proved to be $l^2=a$.

Uniqueness can be proved as follows: If $l_1>0$ and $l_2>0$ are two real numbers such that $l_1^2=l_2^2=a$, then clearly $l_1=l_2$.