Proving that square root of 2 exists

[Mr Davis 97]

I am reading Abbot's "Understanding Analysis," and in this text he assumes that the real numbers are complete, that is, he assumes the least upper bound property, and begins to prove everything from there. Later in the book he proves that the square root of 2 does in fact exist in $\mathbb{R}$. I don't understand completely why this proof is necessary. If we assume that the real numbers are a complete ordered field, doesn't that imply that there are no "gaps" and hence that the $\sqrt{2}$ must exist?

 

[fresh_42]

If you can name a sequence of rationals which converge to $\sqrt{2}$, then completeness guarantees its existence in $\mathbb{R}$. But if you just say, that $\sqrt{2}$ solves $x^2-2=0$, then how do we know it exists in $\mathbb{R}$ and not in an extension of the reals. The solutions of $x^2+1=0$ do not exist in $\mathbb{R}$. What makes $\sqrt{2}$ different from the imaginary unit?

 

[Mr Davis 97]

If you can name a sequence of rationals which converge to $\sqrt{2}$, then completeness guarantees its existence in $\mathbb{R}$. But if you just say, that $\sqrt{2}$ solves $x^2-2=0$, then how do we know it exists in $\mathbb{R}$ and not in an extension of the reals. The solutions of $x^2+1=0$ do not exist in $\mathbb{R}$. What makes $\sqrt{2}$ different from the imaginary unit?

I'm still not completely understanding. If we have already shown that the real numbers are complete, by showing that it satisfies the least upper bound property, doesn't that automatically imply that $\sqrt{2}$ exists, since the real line has no gap? Why, in this specific scenario, do we need to show that it explicitly exists?

 

[fresh_42]

I'm still not completely understanding. If we have already shown that the real numbers are complete, by showing that it satisfies the least upper bound property, doesn't that automatically imply that $\sqrt{2}$ exists, since the real line has no gap? Why, in this specific scenario, do we need to show that it explicitly exists?

If $\sqrt{2}$ is a point on the real number line, then you are right. But this is a tautology. To use completeness, we need a sequence of rationals which converges to $\sqrt{2}$, because $\mathbb{R}$ is the completion of $\mathbb{Q}$, or by the least upper bound property, in which case we must show that $\sqrt{2}$ is such a least upper bound of some set of reals. The problem is, what does $\sqrt{2}$ represent? Could be any number, e.g. one outside the reals as $i$ is. So what makes them different, because one is real whereas the other one is not.

 

[Mr Davis 97]

If $\sqrt{2}$ is a point on the real number line, then you are right. But this is a tautology. To use completeness, we need a sequence of rationals which converges to $\sqrt{2}$, because $\mathbb{R}$ is the completion of $\mathbb{Q}$, or by the least upper bound property, in which case we must show that $\sqrt{2}$ is such a least upper bound of some set of reals. The problem is, what does $\sqrt{2}$ represent? Could be any number, e.g. one outside the reals as $i$ is. So what makes them different, because one is real whereas the other one is not.

What I'm getting from your statement is that, given that the reals satisfy the least upper bound property, we then have the capacity to prove that certain irrational numbers, such as $\sqrt{2}$, exist in $\mathbb{R}$. But we can't possibly prove, on a case by case basis, that every irrational number is in $\mathbb{R}$, so how do or how can we conclude that every irrational is indeed contained in $\mathbb{R}$ as a general rule?

 

[fresh_42]

One can prove that the topological completion of $\mathbb{Q}$ with the standard metric is the real numbers $\mathbb{R}$ defined as the limit points of Cauchy sequences. Then all limits of converging, rational sequences are in $\mathbb{R}$. That the real numbers defined by the supremum property are the same set has to be proven.

Next what is an irrational number? If it is "not rational" then it is wrong, as the example $i$ shows. But usually irrational numbers are per definition $\mathbb{R}-\mathbb{Q}$ and nothing has to be proven. If you mean only square roots of primes, then there are rational sequences which converge to them. It all depends on what is defined and how. The algebraic method I have sketched is best suited for equations like $x^2-2=0$. The method with the least upper bound has the disadvantage, that we do not know a priori whether a given number is real or not and that's why Abbot proved it for $\sqrt{2}$. As this proof likely works for many other numbers in the same way, we get all these for free, but I don't know the connection between the two approaches at the moment, i.e. the proof they are the same set.

 

[Erland]

Mr Davis 97, how do you deduce that the equation $x^2=2$ has a real solution, from the upper bound principle?

 

[lavinia]

I wonder if the old Greek proof would work:

If $R$ is complete then any length determined by a straight-edge connecting two points in the plane will be a real number. The hypotenuse of a right triangle with equal sides is the square root of 2. It exists since it is a length connecting two constructible points.

More generally, any constructible length will be a real number.

 

[lavinia]

What I'm getting from your statement is that, given that the reals satisfy the least upper bound property, we then have the capacity to prove that certain irrational numbers, such as $\sqrt{2}$, exist in $\mathbb{R}$. But we can't possibly prove, on a case by case basis, that every irrational number is in $\mathbb{R}$, so how do or how can we conclude that every irrational is indeed contained in $\mathbb{R}$ as a general rule?

As fresh_42 pointed out the irrational numbers are the reals minus the rationals. So by definition every irrational is real.

You are asking a different question which is: If a number can be defined by some relation - e.g. an algebraic equation - when does such a number exist and when is it a real number? So for instance, are any of the roots of the equation $3x^{52048}-127x^4 + 19 = 0$ real numbers?

Algebraic completion is different than metric completion. The algebraic completion of the reals is the complex numbers.

BTW: I am not sure what you mean by "a case by case basis." If one is given a number by some defining criterion, then why a prioi would there be no way to decide if the number is real or not? This boils down more to what it means to define a specific number.

For instance, there is a subfield of the real numbers called the "computable numbers". These are numbers for which there exists a computer program(recursive algorithm) that generates a Cauchy sequence that converges to the number. Strangely there are only countably many computable numbers since there are only countably many such computer programs. And even more strange, there is no algorithm that can list all of these algorithms. So there is no way even to identify all of the computable numbers.

 

[Mr Davis 97]

Alright, I think it's more clear to me now. However, I am going to refine my question one more time.

If completion doesn't automatically tell us that there exists a real solution to $x^2 = 2$, then what does completion actually tell us? I am told that completion intuitively means that there are no "gaps," but shouldn't that automatically imply that there exists a real solution to $x^2=2$? What I am not understanding is why we have to go out of our way to prove that there exists a real solution to $x^2=2$, since I thought that was why the real numbers existed in the first place, to complete the rationals so that such equations have solutions, and so that numbers such as $\pi$ and $e$ can exist.

 

[lavinia]

Alright, I think it's more clear to me now. However, I am going to refine my question one more time.

If completion doesn't automatically tell us that there exists a real solution to $x^2 = 2$, then what does completion actually tell us? I am told that completion intuitively means that there are no "gaps," but shouldn't that automatically imply that there exists a real solution to $x^2=2$? What I am not understanding is why we have to go out of our way to prove that there exists a real solution to $x^2=2$, since I thought that was why the real numbers existed in the first place, to complete the rationals so that such equations have solutions, and so that numbers such as $\pi$ and $e$ can exist.

Can you rigorously elaborate why it is automatic that the square root of 2 exists once the rationals have been completed? Suppose someone says"I don't see why the square root of 2 is in the completion of the rationals."

 

[fresh_42]

If completion doesn't automatically tell us that there exists a real solution to $x^2 = 2$, then what does completion actually tell us?

Algebraic completion is different than metric [= topological] completion.

I am told that completion ...

... here the topological one ...

... intuitively means that there are no "gaps," ...

Yes.

... but shouldn't that automatically imply that there exists a real solution to $x^2=2$?

Why? The topological completion means, the distances which can be drawn have a length which actually exists, like the diagonal in a unit square, whose length doesn't exist if only rational numbers were allowed. The existence of a solution is an algebraic property. And

The algebraic completion of the reals is the complex.

$\sqrt{2}$ is the diagonal of the unit square (to be proven) which is as such in the topological completion of the rationals (also to be proven, since this was informal and we need a formal proof). So somehow the connection between the topological completion and the algebraic equation has to be made. Just think about the fact, that the solutions of $x^2+1=0$ are not real, so somewhere there has to be a difference.

What I am not understanding is why we have to go out of our way to prove that there exists a real solution to $x^2=2$, since I thought that was why the real numbers existed in the first place, to complete the rationals so that such equations have solutions, and so that numbers such as $\pi$ and $e$ can exist.

No. This would include all complex numbers, too. The reals are only the metric (topological) completion of $\mathbb{Q}$. The connection can be made by Cauchy sequences which would work for $\sqrt{2}$ and would not for $i$.

 

[Mr Davis 97]

... here the topological one ...

Yes.

Why? The topological completion means, the distances which can be drawn have a length which actually exists, like the diagonal in a unit square, whose length doesn't exist if only rational numbers were allowed. The existence of a solution is an algebraic property. And

$\sqrt{2}$ is the diagonal of the unit square (to be proven) which is as such in the topological completion of the rationals (also to be proven, since this was informal and we need a formal proof). So somehow the connection between the topological completion and the algebraic equation has to be made. Just think about the fact, that the solutions of $x^2+1=0$ are not real, so somewhere there has to be a difference.

No. This would include all complex numbers, too. The reals are only the metric (topological) completion of $\mathbb{Q}$. The connection can be made by Cauchy sequences which would work for $\sqrt{2}$ and would not for $i$.

Okay, I kind of see the picture in a better way. The question of whether $x^2 = 2$ as a real solution is an algebraic question, where the answer is yes, just as whether $x^2 = -1$ has a real solution is an algebraic question, in which the answer is no.

So would the more accurate description be that the real numbers are the completion (in the metric sense) of the rational numbers so that we can do calculus rigorously, and it's only incidental that it also can be used to prove that there is a real solution to $x^2 = 2$ and that $e$ and $\pi$ exist?

I am mainly trying to understand the historical intention in introducing all of these analytic concepts. I originally thought they were introduced so that we could solve more equations, such as $x^2=2$, since that is why the integers and the rational numbers were introduced (both for linear equations), but that doesn't seem to be the case.

 

[FactChecker]

Complete on its own does not mean "no gaps". It's easy to define subsets of the reals that are complete and have gaps. [0,1]∪[2,3] is one. So to prove that R contains √2, it is necessary not only to state that the reals are complete, but also to show a sequence of rational numbers in R that, together with complete, proves that √2 is in R.

 

[lavinia]

Okay, I kind of see the picture in a better way. The question of whether $x^2 = 2$ as a real solution is an algebraic question, where the answer is yes, just as whether $x^2 = -1$ has a real solution is an algebraic question, in which the answer is no.

The proof is not algebraic. It is analytic. For instance using the LUB property of the reals one knows that the set of reals whose square is less than two has a least upper bound. One must show that the square of this LUB must equal 2.

So would the more accurate description be that the real numbers are the completion (in the metric sense) of the rational numbers so that we can do calculus rigorously, and it's only incidental that it also can be used to prove that there is a real solution to $x^2 = 2$ and that $e$ and $\pi$ exist?

Many numbers were known to exist geometrically. For instance, the square root of two exists as the diagonal of a square. π exists as the ratio of the circumference of a circle to its radius. These do not require the completion of the rationals under the Euclidean metric (BTW: there are other completions using different metrics). They only require ruler and compass constructions.

I am mainly trying to understand the historical intention in introducing all of these analytic concepts. I originally thought they were introduced so that we could solve more equations, such as $x^2=2$, since that is why the integers and the rational numbers were introduced (both for linear equations), but that doesn't seem to be the case.

Not sure of the history. The problem of taking limits - which is needed in calculus - would on the surface seem important.

 

[hilbert2]

To make a sequence of rationals that converges to $\sqrt{2}$, consider

$a_1 = 1$
$a_n = \frac{a_{n-1}+2/a_{n-1}}{2}$.

A monotone increasing sequence can be made by suitably removing elements from sequence $(a_n )$ - then you can use the least upper bound property of reals.

 

[Erland]

Okay, I kind of see the picture in a better way. The question of whether $x^2 = 2$ as a real solution is an algebraic question, where the answer is yes, just as whether $x^2 = -1$ has a real solution is an algebraic question, in which the answer is no.

So would the more accurate description be that the real numbers are the completion (in the metric sense) of the rational numbers so that we can do calculus rigorously, and it's only incidental that it also can be used to prove that there is a real solution to $x^2 = 2$ and that $e$ and $\pi$ exist?

I am mainly trying to understand the historical intention in introducing all of these analytic concepts. I originally thought they were introduced so that we could solve more equations, such as $x^2=2$, since that is why the integers and the rational numbers were introduced (both for linear equations), but that doesn't seem to be the case.

It would be wrong to say that it is incidental that one can prove that $\sqrt 2$, $\pi$, and $e$ are real numbers (defined by completion). Since these numbers are supposed to correspond to points on a line, and the real numbers, constructed by completion, are designed to correspond to the points on a line, it is certainly expected that these numbers are real numbers.
But the problem is that if we try to use arithmetic to formalize the intuition behind the points on a line, one of its intuitive properties being that it has no "gaps", we end up with Dedekind cuts, Cauchy sequences of rationals, axiomatization based upon the least upper bound property, or something equivalent to these. These are quite technical constructs, but there is probably no simpler way to achieve this, and to use them might be quite elaborate. Therefore, we should not expect it to be trivial to prove that $\sqrt 2$, $\pi$, $e$, etc. are real numbers by this definition.

In Dedekind's original article, where he introduces what is now called "Dedekind cuts", he explains the motivations behind this: http://www.gutenberg.org/files/2101...n_id=725de5f71b604c1df8ac654d9e6b00507f8db932