Two integers and thus their squares have no common factors

[Happiness]

Btw, "if $\sqrt{m}$ is rational then $\sqrt{p_i}$ is also rational, where $p_i$ is one of the primes in the prime factorisation of $m$" is wrong in general.
$\sqrt 4$ is rational and $2$ is a prime factor of $4$, but $\sqrt 2$ is not rational.
You can probably express what you mean more precisely and get it right, but will you then really get a proof of the proposition in the book which is simpler, or more straightforward, than the proof in the book?

Sorry, I don't follow you.

 

[Erland]

i.e., letting $m$ be a product of primes and then arguing that if $\sqrt{m}$ is rational then $\sqrt{p_i}$ is also rational, where $p_i$ is one of the primes in the prime factorisation of $m$

I just pointed out that this is wrong in general. $m=4$, whose only prime factor is $2$, is a counterexample.

I suggest that you write down your own proof of the proposition in the book in full, to see if it really is simpler or more straightforward than the proof in the book, quoted in your OP.

 

[Happiness]

I just pointed out that this is wrong in general. $m=4$, whose only prime factor is $2$, is a counterexample.

Isn't this expected in a proof by contradiction?

I suggest that you write down your own proof of the proposition in the book in full, and see if it really is simpler or more straightforward than the proof in the book, quoted in your OP.

My proof is more straightforward, but I don't want to convince you that because I understand it is subjective. More crucially, the proof given in the book is not complete. It's deceivingly complete but it's not. The fundamental theorem of arithmetic is not even mentioned in the proof.

As mentioned by PeroK, the book states without further note the most essential part of the proof.

The proof in the book states without further note that "$p$ and $q$, hence $p^2$ and $q^2$ have no common factors". This depends on the FTA. Also, this is essentially the proof in a nutshell, as it shows that proper rationals square to proper rationals, never to whole numbers. QED

 

[Erland]

Isn't this expected in a proof by contradiction?

No, a proof by contradiction should lead to a contradiction in the end, it should not use invalid arguments in its body. But of course, I have not seen your full proof, which might clear this issue up.

More crucially, the proof given in the book is not complete. It's deceivingly complete but it's not. The fundamental theorem of arithmetic is not even stated in the proof.

I agree. Perhaps the author takes the fundamental theorem of arithmetic for granted. Since I am not familiar with the rest of the book, I don't know how appropriate that is.

 

[PeroK]

My proof is more straightforward, but I don't want to convince you that because I understand it is subjective. More crucially, the proof given in the book is not complete. It's deceivingly complete but it's not. The fundamental theorem of arithmetic is not even mentioned in the proof.

As mentioned by PeroK, the book states without further note the most essential part of the proof.

I've seen the proof in the book and similar variations many times. Often it is to show that $\sqrt{2}$ is irrational, or to show that $\sqrt{p}$ is irrational, where $p$ is prime, or (in this case) to show that $\sqrt{m}$ is either a whole number or irrational. Often, they make a big fuss over the proof, with contradiction or reductio ad absurdum. And it all looks quite complicated.

I remember being surprised the first time I noticed that the simple argument involving squares having no new prime factors proves the general case with a minimum of effort.

I don't think this is particularly important, but to see how much fuss can be made of this issue, look at a recent Insight:

https://www.physicsforums.com/insights/irrationality-for-dummies/

I think you should keep the simple proof you've found in your back pocket and move on!

 

[Happiness]

But of course, I have not seen your full proof, which might clear this issue up.

Let $m=p_1^{k_1}p_2^{k_2}...p_n^{k_n}$. Given that $m$ is not a square number, there exist at least one $k_n$ that is odd. Suppose we call them $k_j$'s and denote those even $k_n$'s as $k_i$'s. Then for $\sqrt{m}$ to be rational, $\sqrt{p_1^{k_1}p_2^{k_2}...p_n^{k_n}}$ has to be rational. $\sqrt{p_i^{k_i}}$ are integers and hence rational. So $\sqrt{\Pi_jp_j^{k_j}}$ and hence (we are using $k_j=1$ mod 2) $\sqrt{\Pi_jp_j}$ has to be rational. (But it is not rational, so we [will] have a contradiction.)

In a few more steps, we can show that this leads to a contradiction.

The $p_n$'s are prime numbers while the $k_n$'s are integers, of course.

 

[PeroK]

Let $m=p_1^{k_1}p_2^{k_2}...p_n^{k_n}$. Given that $m$ is not a square number, there exist at least one $k_n$ that is odd. Suppose we call them $k_j$'s and denote those even $k_n$'s as $k_i$'s. Then for $\sqrt{m}$ to be rational, $\sqrt{p_1^{k_1}p_2^{k_2}...p_n^{k_n}}$ has to be rational. $\sqrt{p_i^{k_i}}$ are integers and hence rational. So $\sqrt{\Pi_jp_j^{k_j}}$ and hence $\sqrt{\Pi_jp_j}$ has to be rational. (But it is not rational, so we [will] have a contradiction.)

In a few more steps, we can show that this leads to a contradiction.

My proof would be:

Let $q = \frac{a}{b}$ be a proper rational (where $a,b$ have no common factors and $b \ne 1$). Now $a^2$ and $b^2$ have the same prime factors as $a$ and $b$ respectively, hence have no common factors. And, as $b^2 \ne 1$, $q^2 = \frac{a^2}{b^2}$ is a proper rational.

Hence, a proper rational squares to a proper rational, never to a whole number.

Therefore, the square root of a whole number is either a whole number or irrational (never a rational).

 

[Happiness]

Therefore, the square of a whole number is either a whole number or irrational (never a rational).

I suppose you mean "Therefore, the square root of a whole number is either a whole number or irrational (never a rational)."

 

[PeroK]

I suppose you mean "Therefore, the square root of a whole number is either a whole number or irrational (never a rational)."

Yes, fixed now!

 

[Erland]

PeroK's proof is the best one, in my opinion.

 

[bahamagreen]

not a mathematician (like you couldn't tell...) but here is my attempt...

let p<q, no common factor
if p is even (two is a factor) then q cannot be even
if p is threeven (three is a factor) then q cannot be threeven
if p is fourven then p is also even (an integer sub-multiple of four), so q cannot be either

so generally if p is nven (n is a factor) then q cannot be nven nor any integer sub-multiples of n of nven
q cannot be xp nor p^x where x = int > 1; where x=1 then p is pven (p is a factor) so q cannot be pven

for p and q to share a common factor n entails that n<p
but all nven of p for n on q up to p yield no common factor

so for p^2 and q^2 to be nven (share a common factor n),
then p>n<p^2 is the only range to look for n since all nven of p on q where n</=p are eliminated
but all n>p where n=xp<p^2 are already eliminated as well, including all x from 1 to p (where np=pp=p^2)
therefore p^2 and q^2 have no common factor

 

[Happiness]

not a mathematician (like you couldn't tell...) but here is my attempt...

let p<q, no common factor
if p is even (two is a factor) then q cannot be even
if p is threeven (three is a factor) then q cannot be threeven
if p is fourven then p is also even (an integer sub-multiple of four), so q cannot be either

so generally if p is nven (n is a factor) then q cannot be nven nor any integer sub-multiples of n of nven
q cannot be xp nor p^x where x = int > 1; where x=1 then p is pven (p is a factor) so q cannot be pven

for p and q to share a common factor n entails that n<p
but all nven of p for n on q up to p yield no common factor

so for p^2 and q^2 to be nven (share a common factor n),
then p>n<p^2 is the only range to look for n since all nven of p on q where n</=p are eliminated
but all n>p where n=xp<p^2 are already eliminated as well, including all x from 1 to p (where np=pp=p^2)
therefore p^2 and q^2 have no common factor

What do you mean by "nven of p for n on q"? Could you give an example?

 

[bahamagreen]

"but all nven of p for n on q up to p yield no common factor"

nven is divisibility by n; for p where nven is true, it cannot be true for q

"nven of p" is the case of p divisible by n being true e.g., p is "even" (AKA "twoven") and divisible by 2
"for n" is the factor number for nven e.g., "even" means 2 is a factor
"on q" means q is the object for which this factor cannot apply e.g., therefore q is not even
"up to p" means the n of nven may take integer values from 2 up to but not including p e.g., even (twoven), threeven, fourven... up to but not including pven (because p is a factor)

sorry for the terrible wording and made up terms...

edit - My first intuition about this problem was to assign p and q as y and x intercepts on the axes and wonder if there was anything special that distinguished the lines through p and q (and subsequently the lines through p^2 and q^2) when p and q had vs did not have common factors... but I think that may be similar to what is going on when the various approaches are using p/q?

 

[Happiness]

so for p^2 and q^2 to be nven (share a common factor n),
then p>n<p^2 is the only range to look for n since all nven of p on q where n</=p are eliminated
but all n>p where n=xp<p^2 are already eliminated as well, including all x from 1 to p (where np=pp=p^2)
therefore p^2 and q^2 have no common factor

The third line is not explained clearly. You have only established that $q$ is not divisible by $p$ (since they don't have a common factor). But it has not been established that $q^2$ is not divisible by $p$, which is your deduction in the third line.

For example, $q=30$ is not divisible by $p=12$, but $q^2=900$ is. So it's not clear how you come to the deduction that $q^2$ is not divisible by $p$.

Also, in the second line where you deduce that $q^2$ and $p^2$ have no common factor $n$ for $n\leq p$ because $q$ and $p$ have no common factor $n$ for $n\leq p$, it looks like an inverse error is committed.

It's better to denote the common factor of $p^2$ and $q^2$ as $m$, to distinguish it from $n$, the common factor of $p$ and $q$.