# Irrational roots don't exist [closed]

I'm going through Apostol's Calculus vol. 1. It is excellent. But I was surprised to see him "prove" each number has a square root.

Inverse functions like divide usually introduce some invalid operations like divide by 0.

SQRT(x) (obviously an inverse function) where x is not a "square" number is just such an "invalid operation." It is impossible because there is no such number which satisfies this operation, and that is why it is irrational, you are trying to approximate something that doesn't exist.

I was really surprised such a clear, logical thinker as Apostol had this blind spot.

-

## closed as not constructive by DonAntonio, Stahl, Michael Albanese, Paul, Guess who it is.May 15 '13 at 8:29

As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance. If this question can be reworded to fit the rules in the help center, please edit the question.

Where on earth did you get the idea that irrational numbers don’t exist?! Every non-negative real number does have a non-negative square root; this is one of the things distinguishing the real numbers from the rational numbers. –  Brian M. Scott May 6 '13 at 1:58
Yes there is. Do you know of any definitions of the real numbers? –  anon May 6 '13 at 2:08
There isn't a problem if you know about limits and have a definition of the real numbers at hand. Which is why I asked you if you have any working definition of the real numbers. –  anon May 6 '13 at 2:19
Is matt channeling the ghost of Kronecker? –  Kris Williams May 6 '13 at 2:25
I realize that I had voted to close this question, but I am also now voting to re-open. Reasoning: Although I believe the question text needs work, the OP's position has now been clarified in the comments, and I think it is now possible to provide an answer to this question. Sorry for the flip/flop of votes, but just wanted to state why I did so. –  apnorton May 6 '13 at 2:31

Ted Shifrin's answer suggests that the OP actually carefully read what Apostol wrote, and I think that is the best answer here. This answer is a little more meta.

To the OP: part of being a student of mathematics is fully engaging with what is presented in your courses and texts. You need to have a bit of something -- somewhere between "respect" and "faith" -- that what the instructors and texts are telling you is worth your careful attention. Your last statement:

I was really surprised such a clear, logical thinker as Apostol had this blind spot.

shows of your lack of this quality. Apostol is not just a "clear, logical thinker" who wrote down his own ideas about real numbers. He is presenting an extremely well established mathematical theory, which he has mastered and which you must be trying to learn if you are reading his book. His book was first published in 1957 and has been read by thousands of students and mathematicians since then. Is it really plausible to you that none of these thousands of people have seen this "blind spot" that you, who are just learning this new subject, can perceive? Are you in possession of the truth about real numbers that the worldwide mathematical community does not know? Please try to have more respect/faith. That's not a reprimand: it's sincere advice for how to continue your studies.

In particular, notice that you are not engaging the material. You haven't mentioned what you perceive to be a flaw in his argument, only that it seems to contradict some conclusion that you already "know". You keep talking about dividing by zero, but that's clearly not the issue here: you're reasoning by analogy only. What you need to do is learn more about what the real numbers are and why they are larger, more complicated and more useful than the rational numbers. Apostol's text will teach you these things if you are willing to read it carefully.

-
Yes you are right, that my words may be to harsh... but on the other hand you are wrong to dismiss me as just a student. The big geniuses (Im not saying Im one) of their times thought very differently to their contemporaries and its precisely these people that advanced math by challenging ideas. –  matt May 6 '13 at 4:58
@matt: I'm neither dismissing you nor saying that you're "just" a student of mathematics. Being a student of mathematics is very difficult and demanding. In order to challenge an idea you need to first properly engage with it and understand it. Big geniuses have new ideas, which emerge, in part, from exceptionally deep understanding of the old ideas. Have you read and understood Apostol's proof of the existence of square roots? If the conclusion is false, there must be a flaw in the argument: where is it? –  Pete L. Clark May 6 '13 at 5:14
By the way, denying the existence of irrational numbers is a very ancient idea, not a new one. Greek mathematics was fundamentally based on the notion of commensurable quantities until the Pythagorean school showed that the diagonal of a square is incommensurable with its side. Many intermediate sophisticated ideas lie between this discovery and the modern (say, by 1900) take on real numbers that Apostol provides. –  Pete L. Clark May 6 '13 at 5:23

Apostol, like Spivak, is a sophisticated calculus text. There are no Dedekind cuts, thank goodness. @matt, you need to reread pp. 24–25 very carefully. Here he discusses the completeness axiom (least upper bound property), which is precisely what you're missing. In Exercise I 3.12 #11, he has you prove that there is no rational number whose square is $2$, but in the next section he proves using the completeness axiom that $\sqrt2$ exists as a real number.

-
Thanks for this.. I will revisit these pages. I must admit I did rush over that area hastily because it was getting a little tedious with all the definitions and introducing latin names for things which already had names etc... I thought Id come back to it when Id get in trouble in the calculus part. –  matt May 6 '13 at 4:38
Ive worked thru "apostols" proof in lots of detail (page 29): that all non-negative numbers have a SQRT... and Im not convinced at all. Heres why: 1) He assumes such a number exists for all numbers. What an assumption! 2) Then he says: ONLY 3 cases are possible: b^2>a, b^2<a, b^2=a... what if there is no a ?? What if a is an asimptote where b will come close to but never touch! THen he proves by contradiction that case 1 and 2 cant be true so it must be 3! So he seems to setup 3 cases (on false assumptions) and disprove 2 of them and then say so it must be the 3rd. –  matt May 7 '13 at 15:10
I meant: "what if there is no x? (in the first line x^2<=a) ? (not "what if there is no a") –  matt May 7 '13 at 15:30
He selects $a$, not $b$. Step 1: "select an arbitrary $a$". You can select an arbitrary real number. Step 2: create the set of all positive reals $x$ such that $x^2 < a$. You are allowed to make such a set, correct? Step 3: show that it's non-empty and bounded above. That's busywork but important! Step 4: this has a least upper bound, which we call $b$. This bound MUST exist! That is the key property of the real numbers; all sets bounded above have a least upper bound. Now, since we've established $b$ exists, there are three possibilities for it, as you noted above. Which step don't you like? –  Henry Swanson May 8 '13 at 2:22
Look what is the problem here. You do not try to learn, you try to find something to justify your conception. It has been explained to you in every possible way, but you still refuse to see. To your comment. If you try the same construction for -2, then you get the empty set, so there isn't a supremum. On the other hand, for any positive number $r$ there is a set of numbers whose square is less than $r$. Then this set is bounded, so it has a supremum. Nothing cyclic here. Then he proves that this supremum is the square root. –  tst May 10 '13 at 20:37

What holds true for "numbers" depends on which number system one is using. For example:

• Among the integers, every number has a negative, but among the natural numbers this is not the case.

• Among the rational numbers, every nonzero number has a reciprocal, but among the integers this is not the case.

• Among the real numbers, every positive number has a square root, but among the rational numbers this is not the case.

• Among the complex numbers, every number has a square root, but among the real numbers this is not the case.

One can argue that negative numbers are meaningless because if I have five apples there is no number of apples you can give me such that I have zero apples. One could argue that fractions are meaningless because there is no part of a dog such that if I have three of them, I have a dog. Similarly one can argue that irrational numbers are meaningless, or that imaginary numbers are meaningless, all on philosophical grounds.

However, it doesn't make sense to implicitly accept the notion of real numbers (on which the book you are referring to is based) and then claim that none of them squares to 2. I don't know whether Aposotol defines real numbers via Cauchy sequences or Dedekind cuts, but if we choose the latter (which I think is simpler) then the partition of the set of rational numbers into those whose square is less than two and those whose square is greater than two is a real number, and following the definitions it is easy to see that its square is the (Dedekind cut corresponding to the) rational number 2.

-
yes... but under no set is division of 0 allowed... just like you can say x=SQRT(2) approaches x*x=2, but it never hits 2 so its not the inverse of a square operation. –  matt May 6 '13 at 3:05
@matt Not quite true; see en.wikipedia.org/wiki/Wheel_theory. (Of course you don't have to accept elements of a wheel as numbers, any more than you have to accept real numbers as numbers.) –  Trevor Wilson May 6 '13 at 3:09
@matt Also, what do you mean by "approaches" or "hits"? It seems like you are thinking of SQRT(2) as a sequence of rational approximations to something. That's good, now you can take a certain equivalence class of such sequences, and obtain the object commonly known as the square root of 2. This is the Cauchy sequence formulation of real numbers. –  Trevor Wilson May 6 '13 at 3:13
...about which you can read more at en.wikipedia.org/wiki/Cauchy_sequences –  Trevor Wilson May 6 '13 at 3:19

Matt, the way you seem to be thinking about numbers is that rational numbers exist, but nothing else.

It has been known for a long time that one can construct the number systems logically. If you want to think in terms of inverse functions, you find that within the natural numbers, addition by a fixed number is not invertible. Then one figures a way to construct the integers, where addition is invertible.

Then you find that within the integers, multiplication by a non-zero number other than $1,-,1$ is not invertible. Then one constructs the rationals so that such inverse functions exist.

But then one can see that the rationals have "gaps", i.e. there are Cauchy sequences that are not convergent. Mathematicians have known how to "fill those gaps" for a long time, and by filling those gaps you get the real numbers. And working in the real numbers one can prove that there is a real number (two of them, actually) such that its square is $2$.

Each construction I mention above is not just wishful thinking, but constructions that can be made logically sound.

-
@TrevorWilson This construction seems really silly, what is non-computable dedekind-cuts? This is like assuming the existence of some other uncountable set just so you can prove that there is one you call the real numbers. How is this supposed to be better then whishful thinking? –  I am very happy May 6 '13 at 4:14
So those non-computable dedekind cuts 'exist' by 'whishful thinking' then? And hence so does most real numbers. –  I am very happy May 6 '13 at 4:19
@MartinArgerami So show me one of those uncomputable dedekind cuts that can be constructed to be logically sound? Where is it, I cant see it? It must exist by whishful thinking then –  I am very happy May 6 '13 at 4:25
I can't see any prime number with more than a trillion of trillions of digits, but that doesn't mean that they don't exist. Have you ever read (carefully) any formal construction of the real numbers? –  Martin Argerami May 6 '13 at 4:27
Do we believe there are an infinite number of natural numbers? After all, nobody has the lifespan to write them all down. Nobody can show them all. Yet believing in them is not a mere act of faith: it's clear by our understanding of numbers that given any finite set of them, there will be a bigger one, so their totality cannot be finite. Moral of the story, believing in infinities is not presumptuous. (Not that IAM is necessarily arguing in good faith; this user is likely a prankster that designs their statements according to the potential reactions of others for entertainment purposes.) –  anon May 6 '13 at 5:22

You seem to be confusing the concept of "not existing" with "non-terminating or non-repeating decimal expansion." Just because we cannot write something down in its entirety does not mean that it doesn't exist.

For example, we use the symbol $\sqrt{2}$ to denote the number such that $\sqrt{2}\cdot\sqrt{2} = 2$. This number most certainly exists.

Numbers that "do not exist" are fairly difficult (if not impossible) to come by--systems can be constructed to account for "impossibilities." Irrational numbers take care of those with non-terminating, non-repeating decimal expansions. Complex numbers take care of the "square root" of $-1$. It isn't a matter of existing, so much, but rather a matter of being able to construct a manner with which to work with the numbers.

-
@matt If you could find or construct a system with which division by zero was rigorously defined, certainly. –  apnorton May 6 '13 at 2:29
matt, while one can create algebraic systems containing the rational numbers that have square roots of two, it is not possible for an algebraic system containing the rationals to have division by zero (in fact, this is only possible in the trivial ring). It's not clear why you believe division by zero follows from the logic of this answer. Sometimes "things that don't exist" (square roots of two in the rationals) can exist in some larger system (e.g. the set of real numbers), whereas other times they simply cannot exist while retaining all the other properties of a system. –  anon May 6 '13 at 2:32
Why do you think 1/3 makes sense? (It does, but the why is important.) More pointedly, can you give me a number that when multiplied by 3 yields 1? If so, do. If you believe writing "1/3" is an acceptable response, why do you think writing "$\sqrt{2}$" is not an acceptable response to a request for the solution to $x^2=2$? –  anon May 6 '13 at 2:49
@matt But there is no $x$ such that $x + x + x = 1$. You can pretend the sum is approaching a limit of $1$, but it never touches $1$. (based on your use of the word limit) –  apnorton May 6 '13 at 3:05
How long is the hypotenuse of a right triangle with legs 1 and 1? –  Henry Swanson May 6 '13 at 3:38

So, let's assume that you have already defined the rational numbers, and you would agree that they exist. If not, I can edit this with a definition.

We can construct the real numbers by the use of "Dedekind cuts". Let $A$ be a subset of $\mathbb{Q}$. Let $B$ be all the elements of $\mathbb{Q}$ not in $A$. Our set must satisfy this condition: $\forall x \in A \ \forall y \in B \ x < y$. In other words, every element of $A$ is less than every element in $B$. This set, $A$ we will call a "real number", for reasons which hopefully become clear.

Let's do an example. Let $A = \{ x \in \mathbb{Q} : x < 0 \textrm{ or } x^2 < 2 \}$. This satisfies our criteria above. $B$ would then be $\{ x \in \mathbb{Q} : x > 0 \textrm{ and } x^2 > 2 \}$. But, since every a is less than every b, what lies at their border? Clearly it's not a rational number, because there is no rational number $q$ such that $q^2 = 2$. But, we can create a set that is bounded above by it. So we suspect that there is some use for this mystery value. We define $\sqrt{2}$ to be this $A$!

Then we define operations on it so that when $A$ corresponds to a rational, we get the same results as we would by working directly with the rational numbers.

EDIT: @Ted Shifrin got to what I was trying to say much more clearly. Essentially, we make a set containing the rationals, and give it the least-upper-bound property. Then our set $A$ has upper bound $\sqrt{2}$.

-
You cant construct all real numbers this way. Most of them you cant. –  I am very happy May 6 '13 at 4:08
Are you referring to non-computable reals? Yeah, I don't quite know how those are ever constructed. The sets that correspond to them exist, but we can't write them out. And same thing with Cauchy sequences. There are plenty of sequences that converge to one, but I couldn't write one out. –  Henry Swanson May 6 '13 at 4:14