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From Algrebra by Gelfand, it says that "When we claim that we have solved the equation $x^2-2=0$ and the answer is $x=\sqrt{2}$ or $x=-\sqrt{2}$, we are in fact cheating. To tell the truth, we have not solved this equation but confessed our inability to solve it; $\sqrt{2}$ means nothing except "the positive solution of the equation $x^2-2=0$

Can someone please explain what he really means? Does he mean every irrational number is meaningless? Thanks in advance.

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  • $\begingroup$ I think he might be saying that irrational numbers could be thought of as solutions to polynomial equations, not the other way around. $\endgroup$ – Airdish May 27 '17 at 16:40
  • $\begingroup$ I believe he is saying the result is circular. $\endgroup$ – Simply Beautiful Art May 27 '17 at 16:44
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    $\begingroup$ The downvote puzzles me. Gelfand is making a very interesting subtle point and the OP is quite right to be puzzled by it, and ask. $\endgroup$ – Ethan Bolker May 27 '17 at 16:55
  • $\begingroup$ @EthanBolker He's a good writer. I just don't understand sometimes and asked. $\endgroup$ – Mathxx May 27 '17 at 16:58
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    $\begingroup$ Indeed he is a good writer. As the comments and answers show, you asked a good question. I hope they help you understand. $\endgroup$ – Ethan Bolker May 27 '17 at 17:12
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He is saying the following.

  1. $\sqrt{2}$ is defined as "the number such that its square is two."
  2. The statement "the solution $x$ to $x^2-2 = 0$ is $\sqrt{2}$" is therefore saying "the solution $x$ to $x^2-2=0$ is the number such that its square is two."

As you can see, this is a rather circular statement. This doesn't mean that irrational numbers are meaningless (indeed, $\sqrt{2}$ does exist -- see the comments and @EthanBolker's answer for more on this), but rather saying that this method of definition limits us to statements like "the number such that its square is two" or "half of the number such that its square is eight."

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    $\begingroup$ Nice answer, except I don't think it is clear that $\sqrt{2}$ exists at all. It is either taken as an axiom, the completeness axiom of the reals, or is to be proved, depending on our construction of the real numbers... $\endgroup$ – SEWillB May 27 '17 at 16:51
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    $\begingroup$ @SEWillB there certainly exist right triangles with legs equal to $1$ unit. Now, does there not exist a hypotenuse to the triangle, since it would be of length $\sqrt 2$ units, so may not exist? $\endgroup$ – Namaste May 27 '17 at 16:56
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    $\begingroup$ SEWillB (continued)... And without a hypotenuse for a right triangle, we cease to have such a triangle? So the only right triangles require legs of lenghts $x, y$ such that the length of the hypotenuse, $\sqrt{x^2 + y^2},$ is rational? $\endgroup$ – Namaste May 27 '17 at 17:04
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    $\begingroup$ @amWhy, I never liked these arguments at all. There is a nice bit in a book I read a long time ago (Liebeck introduction to pure mathematics) showing the 'existence' of $\sqrt{n}$ by geometrical construction like that. But these arguments aren't convincing to me because it's not clear what you're trying to do in my opinion. $\endgroup$ – SEWillB May 27 '17 at 17:08
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    $\begingroup$ So you do not believe the Pythagorean Theorem? Well, then, I think I've wasted my time commenting with you. $\endgroup$ – Namaste May 27 '17 at 17:09
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To elaborate on @fractal1729 's lovely correct answer:

Having defined $\sqrt{2}$ as "the number such that it's square is $2$" it's reasonable to ask whether there is such a number.

The Greeks knew that the answer is "no" if "number" means "rational number". So in order to claim the existence of $\sqrt{2}$ you must enlarge the system of rational numbers. The Greeks essentially sidestepped that question by doing geometry rather than arithmetic, using points on a line instead of numbers. Clearly there's a point on the "number line" with the right length since you can draw the diagonal of a unit square.

In later centuries mathematicians found more formal algebraic ways to enlarge the system of rational numbers to include what we now call all the "real numbers". The new ones are the irrationals. There are several ways to do this. One of the most common is to make precise the notion of an infinite decimal, along with rules for arithmetic with them. Others come with the names "Cauchy sequences" or "Dedekind cuts".

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    $\begingroup$ To add a bit to your answer, note that the very notion of compass and straightedge constructions assumes with no justification that one can construct ideal circles/lines and somehow they always intersect when certain conditions are met (such as closer than the sum of radii). When one thinks about it, that is not at all obvious. Perhaps it is not even physically meaningful, for all we know! If so, it may very well be that $\sqrt{2}$ can't be constructed in any suitable physical sense! Furthermore, the computable reals are elementarily equivalent to the standard reals. Interesting at least. $\endgroup$ – user21820 May 29 '17 at 15:26
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Using definite descriptor notation, we can define:

$$\sqrt{x} = (\iota y \in \mathbb{R})(y \geq 0 \wedge y^2=x)$$

In words: $\sqrt{x}$ is the unique $y \geq 0$ such that $y^2=x$.

From this, you can show that for all $y$ and all $x \geq 0$, we have: $$y^2 = x \iff y \in \pm\sqrt{x}.$$

But in some sense, this is a purely logical construct, at least insofar as we haven't explained how to approximate $\sqrt{x}$ to arbitrary precision. There's ways of doing this, but we haven't given one.

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I think the author is simply wrong. There is another meaningful way of defining $\sqrt{2}$. Consider the following sequence: $$x_{n+1} := \frac{1}{2}\left(x_n + \frac{2}{x_n}\right)$$ with $x_0=1$.

Then $\lim_{n\to\infty} x_n=\sqrt{2}$, even though all $x_n\in\mathbb{Q}$ for arbitrary $n\in\mathbb{N}$. Now instead of defining $\sqrt{2}$ as the solution of $x^2-2=0$ we could have also defined it as the limit on $x_n$.

Proof for the sequence can be found here (German only).

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    $\begingroup$ This is still circular, in a deeper way. How do you define the notion of a limit without some underlying structure of the real numbers? In the rationals that sequence has no limit. What Cauchy showed is that It is possible to define the limit as the sequence itself. It's the necessity for some definition that's the crux of Gelfand's cryptic comment. $\endgroup$ – Ethan Bolker May 27 '17 at 17:10
  • $\begingroup$ There is no need to use $\mathbb{R}$ with the above sequence. For example no transcendental numbers (like $\pi$) are needed. And in some sense $37$ is also just the result of multiple (albeit finite) application of Peano's axioms. The "only" difference with the above definition of $\sqrt{2}$ is that you need to apply the appropriate axioms an infinite number of times. $\endgroup$ – Hannebambel May 27 '17 at 17:18
  • $\begingroup$ No, the author is right. As @EthanBolker says, your sequence has no limit within $\mathbb{Q}$. The key difference vs. the construction of $37$ lies exactly in the finite number of steps: no matter how large the $n$, no number $x_n$ in your sequence will ever equal $\sqrt{2}$. And the fact that $\sqrt{2}$ is not in $\mathbb{Q}$ is equivalent to the notion that $x^2 - 2 = 0$ is unsolvable with the standard algebraic operations. $\endgroup$ – Roland May 28 '17 at 7:55
  • $\begingroup$ I somehow still don't get how the author is correct. The author says "$\sqrt{2}$ means nothing except the positive solution of the equation $x^2−2=0$", whereas I think that $\sqrt{2}$ also means the limit of the aforementioned sequence. $\endgroup$ – Hannebambel May 28 '17 at 18:23
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    $\begingroup$ This answer shows that, in a certain sense, it is possible to "solve" for what $\sqrt{2}$ is. But Gelfand never claimed it wasn't possible to do so--that was not his point at all. His point is that merely naming the answer as $\sqrt{2}$ doesn't solve anything, because all you've done is give a name to what the answer is. $\endgroup$ – Eric Wofsey May 29 '17 at 21:48
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This is a great question (+1) and judging by the quote, Gelfand must be a great writer at least for algebra textbooks (I confess that I have not read any of his books and hence judging only from the quote in question). In fact he has pinned down the essence of solution of polynomial equations via algebra.

If we restrict ourselves to algebra then the solution of polynomial equations via radicals is equivalent to replacing one polynomial equation with a set of binomial equations of type $x^{n} - a=0$. And for binomial equations we don't do anything apart from inventing symbols like $a^{1/n}$.

Next there are some equations which can't be solved via radicals. How does an algebraist deal with the situation now? He smartly handles the situation using field extensions and for any polynomial $f(x) $ over a field $F$ one can create an extension field $E$ which contains all the roots of $f(x) $. The technique of construction of this extension field is so simple that one is tempted to call such mechanism of solution of equations as "cheating". Thus if $f$ is irreducible then one can get a root in quotient $F[x] /(f(x)) $ and guess what the root of $f(x) $ is the coset $x+(f(x)) $.

In effect algebraic methods never try to find the value of root of an equation, but they are rather concerned with the structure of the field extensions which contains the roots. In some very special cases such extensions are radical and these are the familiar ones where we use the quadratic formula, or Cardano's formula (or more complicated stuff of similar type).


Let's now come back to the usual scenario where the equations have coefficients in specific fields $\mathbb{R} $ and $\mathbb{C} $ and lets first handle the case when coefficients are real. Real numbers are the product of a non-algebraic process and their existence is almost on same footing as those of rational numbers. Thus they can and are used to measure magnitude and it is agreed by convention that there are sufficient real numbers to locate every point on a geometric line so that all the geometric measurement is possible via real numbers.

But then the real numbers are a very powerful system and offer us the following justification for the symbol $a^{1/n}$:

Theorem 1: If $a$ is a positive real number and $n$ is a positive integer then there is a unique positive real number $b$ such that $b^{n} =a$ and it is denoted by symbol $a^{1/n}$.

Thus at least some binomial equations of type $x^{n} - a=0$ have a root which has existence in the real number system. By existence we mean that it is possible to give some concrete idea about this root in comparison to the integers and rationals. In other words we can provide some approximation to the value of the root using the rational numbers. And moreover the approximation can be as accurate as we want. It is in this sense of approximation via rationals that every real number exists. And it applies to all sorts of irrational numbers including the famous ones $e, \pi, \sqrt{2}$. Thus irrational numbers are not meaningless, they are as meaningful as the rationals and perhaps mathematically more significant, but unfortunately algebraic techniques cannot help in discovering their true nature.

Let's also note that there are some real numbers which are the roots of certain polynomial equations with rational coefficients and these we call algebraic real numbers. The reason for extending our number system from $\mathbb{Q} $ to $\mathbb{R} $ is not to give some sort of existence to these algebraic numbers (like $\sqrt{2}$). As far roots of polynomials are concerned the algebraic mechanism of field extensions is a sufficient and beautiful approach. Even if we tried to add such numbers to $\mathbb{Q} $ it does not give us $\mathbb{R} $, but it rather gives $\overline{\mathbb{Q}} $ which we call the algebraic closure of $\mathbb{Q} $. It includes all the algebraic numbers both real and complex, but it does not capture all the real numbers or all the complex numbers. In fact one can prove that like $\mathbb{Q} $, the set $\overline{\mathbb{Q}} $ is also countable whereas both $\mathbb{R}, \mathbb{C} $ are uncountable.

Real numbers were invented to deal with the essential/pressing need to arithmetize geometry. The idea was to develop of a system of numbers which could correspond to the points on a line and it was known from the time of Pythagoras (or perhaps even earlier) that there were some points on the line which did not correspond to a rational number. Another need for real numbers was coming from the very powerful techniques of calculus which were based on geometrical intuition, but there was no rigorous justification of these methods using arithmetical means.


To answer your question, we have two ways to look at the symbol $\sqrt{2}$. One is via algebra which says that it is a solution to the equation $x^{2}-2=0$ and in this approach we can't distinguish between $\sqrt{2}$ and $-\sqrt{2}$. Another way is to think of it as a real number represented by its rational approximations. Thus we can say that $\sqrt{2}$ is a positive real number which is greater than all the positive rationals whose square is less than $2$ and it is less than all the positive rationals whose square is greater than $2$. Further in this particular case we are lucky to have a mechanism of locating a point on the number line corresponding to it via geometrical constructions with ruler and compass.


Let's also discuss a bit about the complex number system. But before we do that we need to note that there is another set of equations which the real number system can handle very well:

Theorem 2: If $f(x) $ is polynomial of odd degree with real coefficients then there is at least one real number $c$ such that $f(c) =0$.

But there are many equations with real coefficients for which there is no root in the real number system. The simplest and most famous such equation is $x^{2}+1=0$. Once again an algebraist comes to the rescue and creates a field extension $\mathbb{C} $ as the quotient $\mathbb{R} [x] /(x^{2}+1)$ and surprisingly in one shot the algebraist succeeds in solving all polynomial equations whatsoever :

Fundamental Theorem of Algebra: If $f(x) $ is a polynomial of positive degree with complex coefficients then there is a complex number $c$ such that $f(c) =0$.

But this is all cheating. The power of the above theorem is not due to the algebraic technique of creating field extensions via quotients, but rather due to the properties of real numbers mentioned in theorem 1 and 2 above.

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  • $\begingroup$ Technically, this kind of justification can only reach the computable reals and also the computable complex numbers, which includes the algebraic closure of the rationals. Notice that the computable complex numbers is algebraically closed, so we can't justify going beyond it because we can't actually demonstrate the existence of anything beyond it! Also, you are right that the fact that the complex numbers are algebraically closed is at its crux not really a matter of algebra but of analysis. Even the field-theoretic proof at one point requires IVT for odd-degree polynomials. $\endgroup$ – user21820 May 29 '17 at 16:05
  • $\begingroup$ In other words the only reason we have for believing the existence of all standard real numbers (whether by Cauchy sequences or Dedekind cuts) is that we must believe the power-set axiom or equivalent, which is purely set-theoretic and has no ontological justification. Actually, even function types are not enough to grant the classical power-set, because if we use type theory we can have function types that are compatible with the universe being computable. What breaks is that type membership may not be always true or false. $\endgroup$ – user21820 May 29 '17 at 16:35
  • $\begingroup$ Of course, ZFC set theory is very elegant and convenient for modern mathematics and so, justified or not, it has found acceptance. =) $\endgroup$ – user21820 May 29 '17 at 16:36

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