Can someone please help me understand an argument in Hardy's A Course of Pure Mathematics

[MaxMath]

Hi there,

I'm not a mathematician, nor did I study in uni with it as my major. I'm an engineer with an interest in mathematics. Recently I came across this book of Hardy and decided to have a crack at it, partly because I always dreamed of having a taste of real or pure mathematics. This recent random find seems like a good fit; it's a classic and is pretty concise. I don't mind its age at all: quite the opposite.

Having only started off, I found myself in difficulty in understanding the argument, supposed to be a proof, that the square root of 2 is not a rational number. You can find it here. It's on page 22, out of 587 of the e-format, or page 6 in the printed form (shown on the top-right corner) of the page. I've also attached a screenshot of the part I'm talking about (with highlight).

The difficulty I have is --- it appears to me that the proof concludes with the sentence "Thus m=p^2, n=q^2, as was to be proved". This does not seem sufficient to me. This only means, starting from the presupposition (p^2/q^2 = m/n), what we have reached is only getting back to square one. It's not helpful, of course. But this, by itself, does not seem to prove that the presupposition is false, hence proving what is to be proved. This does not lead to any contradiction, as what is normally the desired outcome of this technique of proving theorems.

Can someone be so kind and able to enlighten me on this? Many thanks.

 

[The Highlander]

Hi there,

I'm not a mathematician, nor did I study in uni with it as my major. I'm an engineer with an interest in mathematics. Recently I came across this book of Hardy and decided to have a crack at it, partly because I always dreamed of having a taste of real or pure mathematics. This recent random find seems like a good fit; it's a classic and is pretty concise. I don't mind its age at all: quite the opposite.

Having only started off, I found myself in difficulty in understanding the argument, supposed to be a proof, that the square root of 2 is not a rational number. You can find it here. It's on page 22, out of 587 of the e-format, or page 6 in the printed form (shown on the top-right corner) of the page. I've also attached a screenshot of the part I'm talking about (with highlight).

The difficulty I have is --- it appears to me that the proof concludes with the sentence "Thus m=p^2, n=q^2, as was to be proved". This does not seem sufficient to me. This only means, starting from the presupposition (p^2/q^2 = m/n), what we have reached is only getting back to square one. It's not helpful, of course. But this, by itself, does not seem to prove that the presupposition is false, hence proving what is to be proved. This does not lead to any contradiction, as what is normally the desired outcome of this technique of proving theorems.

Can someone be so kind and able to enlighten me on this? Many thanks.

Hi @MaxMath,

It is a bit abstruse but if you follow through the (essentially the same) argument here (& maybe here too) then go back to your book I think you may well then see how it's a valid proof. ?

 

[blamocur]

"Thus m=p^2, n=q^2, as was to be proved". This does not seem sufficient to me.

Your attached page says earlier "unless [imath]m[/imath] and [imath]n[/imath] are both perfect squares.", but then it shows that the existence of a rational square root implies [imath]m[/imath] and [imath]n[/imath] must be perfect squares.
Does it make sense now?

 

[MaxMath]

Hi @MaxMath,

It is a bit abstruse but if you follow through the (essentially the same) argument here (& maybe here too) then go back to your book I think you may well then see how it's a valid proof. ?

Hi!

Thank you very much for taking the time to answer my question, and for the links to two interesting ways of proving this simple but famous proposition. But my difficulty, or my problem, was not about doubting the correctness of the conclusion, but rather the form of the proof.

In other words, I don't think, in the excerpt of the book I mentioned, we can reach the conclusion that the proposition is correct just because, so far, we have only proved what was supposed to be true (i.e. no net work is done).

Am I missing anything here?

 

[Dr.Peterson]

The difficulty I have is --- it appears to me that the proof concludes with the sentence "Thus m=p^2, n=q^2, as was to be proved". This does not seem sufficient to me. This only means, starting from the presupposition (p^2/q^2 = m/n), what we have reached is only getting back to square one. It's not helpful, of course. But this, by itself, does not seem to prove that the presupposition is false, hence proving what is to be proved. This does not lead to any contradiction, as what is normally the desired outcome of this technique of proving theorems.

Can someone be so kind and able to enlighten me on this? Many thanks.

The key part of what you highlighted is "as was to be proved". What was it that he said was to be proved? I've highlighted that in yellow here:

​

His claim is that if the square of a rational number p/q (in lowest terms) is m/n (in lowest terms), then m and n must be perfect squares. That's what he proved: that [imath]m=p^2[/imath] and [imath]n=q^2[/imath], which are both perfect squares.

He didn't just return to what he supposed at the start; the supposition was that [imath]\left(\frac{p}{q}\right)^2=\frac{m}{n}[/imath]. Do you see that that is a different statement? Two fractions being equal does not mean their numerators and their denominators are the same; the fractions could, for example be 2/3 and 10/15. Saying that these are equal is not equivalent to saying that 2 = 10 and 3 = 15.

The final sentence applies this broader fact to the more specific fact, that no integer m = m/1 that is not the square of an integer can be the square of a rational number p/q, and therefore to the even more specific fact that 2 is not the square of a rational number.

Does that help?

 

[MaxMath]

Your attached page says earlier "unless [imath]m[/imath] and [imath]n[/imath] are both perfect squares.", but then it shows that the existence of a rational square root implies [imath]m[/imath] and [imath]n[/imath] must be perfect squares.
Does it make sense now?

Yes, I think you are correct here, though I still need to think about this a little bit more to be very sure.

At this point of the poof, it's been demonstrated that, for any rational number p/q whose square is another rational number m/n, it follows that both m and n must be perfect squares without exceptions.

Yes, this is indeed sufficient as a proof to me now. I think I was confused by the language "as was to be proved".

Thanks!

 

[MaxMath]

The key part of what you highlighted is "as was to be proved". What was it that he said was to be proved? I've highlighted that in yellow here:

View attachment 35294​

His claim is that if the square of a rational number p/q (in lowest terms) is m/n (in lowest terms), then m and n must be perfect squares. That's what he proved: that [imath]m=p^2[/imath] and [imath]n=q^2[/imath], which are both perfect squares.

He didn't just return to what he supposed at the start; the supposition was that [imath]\left(\frac{p}{q}\right)^2=\frac{m}{n}[/imath]. Do you see that that is a different statement? Two fractions being equal does not mean their numerators and their denominators are the same; the fractions could, for example be 2/3 and 10/15. Saying that these are equal is not equivalent to saying that 2 = 10 and 3 = 15.

The final sentence applies this broader fact to the more specific fact, that no integer m = m/1 that is not the square of an integer can be the square of a rational number p/q, and therefore to the even more specific fact that 2 is not the square of a rational number.

Does that help?

Absolutely, this is very clear and very helpful! Thank you.

And as you've already pointed out, what was to be proved is not merely that the square root of 2 is irrational, but a more general statement, as you highlighted.

I see and agree with the rest of what you are saying.

It's now crystal to me now. Perhaps I was just not used to the 'density' of logic in pure mathematics!

 

[MaxMath]

As a new member of this forum, I'm pleasantly surprised by the activeness of this forum and the swift and helpful responses to my question. I posted the same question on another math forum, but received no response whatsoever in one whole day! ?

I also immediately noticed the math/formula-friendly features, which look amazing.

 

[MaxMath]

Due to the great help I get from this community, I now intend to use this (single) thread for any and all of my questions along the way reading this book. Hopefully, I can finish it. Fingers crossed!

Now another question on page 47 (e format) of this book (again a screenshot attached) --

The proposition is --- if class L₁ has a greatest number [imath]\alpha[/imath], then [imath]\alpha[/imath] must also be the greatest number of class L. The argument for this include a proposition that, if this is not true and in that case the greatest number of class L is [imath]\beta[/imath], "there are rational numbers lying between [imath]\alpha[/imath] and [imath]\beta[/imath]". The underlined part is what I don't understand, or don't recall seeing this having been proved before in this book.

Here clearly [imath]\alpha[/imath] is a rational number, but [imath]\beta[/imath] is only an irrational real number. I know it has been demonstrated earlier in this book that we can always construct a chain of rational numbers between any two given rational numbers, where the greatest difference between any two adjacent numbers is however small we please. But I cannot take it for granted that the same holds for a pair of real numbers, the smaller of which is a rational number but the greater is only an irrational number.

Would appreciate if anyone can shed some light here. This might be a dumb question (just like the earlier one), but just trying to think in a mathematician's way.

(By the way, the way of thinking, and the path taken, in exploring the completeness of the "aggregate" (or "set" in modern language) of real numbers, as laid out in the lower half of this page and the top of the following page, is beautiful!)

 

[NotJeffM]

MaxMath

The only Hardy I have read is his ”Apology,” and I am even less of a mathematician than you are (my acdemic training was in European history and languages). But please do not put multiple topics into one thread. Such threads become unwieldy, discourage responses, and are inconsistent with the forum’s guidelines, which encourage a separate thread for each problem.

 

[MaxMath]

MaxMath

The only Hardy I have read is his ”Apology,” and I am even less of a mathematician than you are (my acdemic training was in European history and languages). But please do not put multiple topics into one thread. Such threads become unwieldy, discourage responses, and are inconsistent with the forum’s guidelines, which encourage a separate thread for each problem.

I think that's a good point, though my intention was to contain discussions about one bigger topic (so it's probably tidier). Point taken and will do it next time! Thank you.

By the way, will have a read of his "Apology"!

 

[MaxMath]

A bit off-topic, I couldn't wait but had a read of Hardy's "Apology" in one breath. There is no need for me to say that it is good. I enjoyed it.

 

[Dr.Peterson]

The proposition is --- if class L₁ has a greatest number [imath]\alpha[/imath], then [imath]\alpha[/imath] must also be the greatest number of class L. The argument for this include a proposition that, if this is not true and in that case the greatest number of class L is [imath]\beta[/imath], "there are rational numbers lying between [imath]\alpha[/imath] and [imath]\beta[/imath]". The underlined part is what I don't understand, or don't recall seeing this having been proved before in this book.

Here clearly [imath]\alpha[/imath] is a rational number, but [imath]\beta[/imath] is only an irrational real number. I know it has been demonstrated earlier in this book that we can always construct a chain of rational numbers between any two given rational numbers, where the greatest difference between any two adjacent numbers is however small we please. But I cannot take it for granted that the same holds for a pair of real numbers, the smaller of which is a rational number but the greater is only an irrational number.

I tried looking in the book to see what he does say about rational numbers between other numbers; can you point that out? I lacked the patience.

Also, do you see at least why it does make sense that there are rational numbers between alpha and beta? Then, what theorem might there be to support that?

(If you start a new thread for this question, you might answer me there.)

 

[MaxMath]

I tried looking in the book to see what he does say about rational numbers between other numbers; can you point that out? I lacked the patience.

It's in section 3 (page 19 of the ebook), right under the heading "3. irrational numbers".

Also, do you see at least why it does make sense that there are rational numbers between alpha and beta? Then, what theorem might there be to support that?

After my question, almost immediately I thought along the same line --- just follow the logic of the previous proof, but only substituting an irrational, and see if the conclusion still holds.

This unfortunately leads to more questions:

1. First, I don't know whether the conclusion still holds; and
2. I'm now even not sure why we can say "It is evident that if we choose a positive integer k so that"
[math]k \ldotp BC>1[/math]It's said in the footnote that this is equivalent to the assumption of the Axiom of Archimedes. But I don't understand why.

 

[Dr.Peterson]

It's in section 3 (page 19 of the ebook), right under the heading "3. irrational numbers".

It seems to me that this exactly answers your question:

If the reader will mark off on the line all the points corresponding to the rational numbers whose denominators are 1, 2, 3, . . . in succession, he will readily convince himself that he can cover the line with rational points as closely as he likes. We can state this more precisely as follows: if we take any segment BC on Λ, we can find as many rational points as we please on BC.​

This is not restricted to what you said before,

I know it has been demonstrated earlier in this book that we can always construct a chain of rational numbers between any two given rational numbers, where the greatest difference between any two adjacent numbers is however small we please.

It applies to any interval between any two numbers, which is what you need:

But I cannot take it for granted that the same holds for a pair of real numbers, the smaller of which is a rational number but the greater is only an irrational number.

(It's also interesting that he starts the paragraph with exactly my observation, that it makes sense intuitively.)

 

[MaxMath]

It seems to me that this exactly answers your question:

If the reader will mark off on the line all the points corresponding to the rational numbers whose denominators are 1, 2, 3, . . . in succession, he will readily convince himself that he can cover the line with rational points as closely as he likes. We can state this more precisely as follows: if we take any segment BC on Λ, we can find as many rational points as we please on BC.​

This is true and I understand it for any rational numbers, because however big, or small, it is, its denominator is definite. In the case of irrational number, it involves infinity which I don't want to rely on my intuition, because infinity is clearly not something intuitive. And pure mathematics, as far as I understand, shall not be based on intuitions but rather specific definitions and tight deduction.

This is not restricted to what you said before,

In the book, at this point, it's not explicitly said it applies only to rational numbers. But since the discussion so far has been limited to rational numbers only without introducing the concept of real number or irrational number, I don't think we can automatically extend this argument to irrational numbers without a proper investigation and proof.

Maybe I just need to read this part for more times.

Thank you for taking your time to look at his!

 

[Dr.Peterson]

This is true and I understand it for any rational numbers, because however big, or small, it is, its denominator is definite. In the case of irrational number, it involves infinity which I don't want to rely on my intuition, because infinity is clearly not something intuitive. And pure mathematics, as far as I understand, shall not be based on intuitions but rather specific definitions and tight deduction.

But they don't say it's only an intuition! This statement is followed by what looks like a proof to me:

​

And what's being said here is entirely about rational numbers, except that it doesn't say B and C have to be rational. What am I missing? Is it the fact that this starts with "It is evident", so you don't consider it a proof? In that case, please refer me to the proof you said was given about rationals between rationals, so we can talk about that.

As for that footnote you mentioned ... Have you looked at the Axiom of Archimedes and considered how it is related?

 

[MaxMath]

But they don't say it's only an intuition! This statement is followed by what looks like a proof to me:

View attachment 35317​

And what's being said here is entirely about rational numbers, except that it doesn't say B and C have to be rational. What am I missing?

I may be wrong, but this is how I look at this section. This point is where in the whole book Harday prepares to introduce the concept of irrational number. Before this point, irrational number is not even a thing. So even without an explicit description of its application (as limited only to rational numbers), I must so assume.

In other words, the concept of irrational number is meaningless so far.

Is it the fact that this starts with "It is evident", so you don't consider it a proof? In that case, please refer me to the proof you said was given about rationals between rationals, so we can talk about that.

No. I do consider this as a proof, but with only a supposition that is considered true as a starting point, which must be obvious to Hardy or the intended reader (a mathematician), but unfortunately is too difficult for me to swallow.

This is exactly what I referred to before as an explanation of why we can always find a chain of rational numbers between any two given rational numbers so that the difference between any two adjacent rational numbers in the chain is smaller than whatever we please.

Have you looked at the Axiom of Archimedes and considered how it is related?

Yes, I did. But again this is not very clear and difficult to understand. Some descriptions of this axiom is without much context (or requires extensive further reading to understand the relevant context). Like this one, it's not even clear to me what a magnitude is. Of course I assume it is a number. But is it limited to rational numbers or include irrational numbers? Probably the latter, because according to Google irrational numbers were discovered around the 5th century B.C. before the time of Archimedes (c. 287 -- c. 212 BC)? But I'm not sure. And this page confuses me more --- why is there a "proof" of an axiom? Is axiom meant to be an assumption to start with, which we have to accept as true without needing a proof (and probably there is no way to prove it)?

I also came across this, started reading but it's again quite obscure. Is there a good coverage of this Archimedes Axiom that you consider is relatively self-contained, clear, and easy to understand?

By the way, having looked at section 3 of the book a few more times, it becomes somehow a bit clearer to me, particularly with respect to
[math]k\cdotp BC > 1[/math]I think here BC is the "magnitude", k is the "multiplies" in one description of the Archimedes Property (I assume this is the "Axiom"?) ---
"Magnitudes are said to have a ratio to one another which can, when multiplied, exceed one another." But does this statement apply to --- or does "magnitude" include --- irrational numbers?

Sorry too many dumb questions.

 

[ColabDog]

Hi - I think you have gotten a lot of help but I created this flowchart diagram for you to better understand the argument that is being proposed here. I hope this helps!

 

[Dr.Peterson]

Hi - I think you have gotten a lot of help but I created this flowchart diagram for you to better understand the argument that is being proposed here. I hope this helps!

<link to commercial site removed by moderator>

But that is not the argument that Hardy was giving, nor is this the question currently under discussion.

Are you a spammer?