If ## a ## is a positive integer and ## \sqrt[n]{a} ## is rational?

[Math100]

Homework Statement

If $a$ is a positive integer and $\sqrt[n]{a}$ is rational, then $\sqrt[n]{a}$ must be an integer.

Relevant Equations

None.

Proof:

Suppose $a$ is a positive integer and $\sqrt[n]{a}$ is rational.
Then we have $\sqrt[n]{a}=\frac{b}{c}$ for some $b,c\in\mathbb{Z}$
such that $gcd(b, c)=1$ where $c\neq 0$.
Thus $\sqrt[n]{a}=\frac{b}{c}$
$(\sqrt[n]{a})^{n}=(\frac{b}{c})^n$
$a=\frac{b^n}{c^n}$,
which implies that $ac^n=b^n$.
This means $c^{n}\mid b^{n}$.
Since $gcd(b, c)=1$, it follows that $c=1$.
Thus $\sqrt[n]{a}=\frac{b}{c}$
$=\frac{b}{1}$
$=b$,
which implies that $b$ is an integer.
Therefore, if $a$ is a positive integer and $\sqrt[n]{a}$ is rational,
then $\sqrt[n]{a}$ must be an integer.

 

[PeroK]

There's a lot of maths there but you've effectively assumed the key point:

which implies that $ac^n=b^n$.
This means $c^{n}\mid b^{n}$.
Since $gcd(b, c)=1$, it follows that $c=1$.

if you assume that, then your proof should be 3 lines long.

 

[nuuskur]

And now the post exam conversation.

1. Why is $a$ positive important here?
2. Lines 4 and 10 are redundant.
3. Why does $c=1$ follow from $c^n\mid b^n$? That is the main (and the only interesting!) part of the argument, but you move past it quickly.

 

[Math100]

And now the post exam conversation.

1. Why is $a$ positive important here?
2. Lines 4 and 10 are redundant.
3. Why does $c=1$ follow from $c^n\mid b^n$? That is the main (and the only interesting!) part of the argument, but you move past it quickly.

1) Because if $a$ is a negative integer, then $\sqrt[n]{a}$ would be irrational.
3) Because the integers $b$ and $c$ are relatively prime, this means there are no common factors except $1$.

 

[nuuskur]

So it follows from $c\mid b$. Why does it follow from $c^n\mid b^n$? Do you claim $c^n\mid b^n$ implies $c\mid b$ if $b,c$ are coprime?

As for 1), I don't agree. Give it some thought.

 

[Math100]

So it follows from $c\mid b$. Why does it follow from $c^n\mid b^n$?

As for 1), I don't agree. Give it some thought.

Suppose $c\mid b$.
Then we have $cm=b$ for some $m\in\mathbb{Z}$.
Note that $(cm)^{n}=b^{n}$
$c^{n} m^{n}=b^{n}$
$b^{n}\equiv 0$ mod $c^{n}$.
Thus $c^{n}\mid b^{n}$.

 

[nuuskur]

Yes, $c\mid b$ implies $c^n\mid b^n$ without any additional assumptions. You did not answer my question, however. Is the converse true? How do you get from $c^n\mid b^n$ to $c=1$?

 

[Math100]

$c^{n}\mid b^{n}$ does not imply $c\mid b$ if $b, c$ are coprime.

 

[nuuskur]

Suppose $b,c$ are coprime. It suffices to show $c\mid b^n$ implies $c\mid 1$ for every $n\in\mathbb N$. The base case holds true. Now assume $c\mid b^n$ implies $c\mid 1$ for some $n$. For the case $n+1$ assume $c\mid b^{n+1}$. Then by Euclid's lemma follows $c\mid b^n$ and therefore, $c\mid 1$ by induction hypothesis.

Now conclude that $c^n\mid b^n$ implies $c\mid b$.

 

[PeroK]

I must admit I would have seen this as the perfect proof for the fundamental theorem. Perhaps it's been used so often recently that it's been worn out?

 

[Math100]

How about this revised proof?

Suppose $a$ is a positive integer and $\sqrt[n]{a}$ is rational.
Then we have $\sqrt[n]{a}=\frac{b}{c}$ for some $b,c\in\mathbb{Z}$ where $c\neq 0$.
Thus $(\sqrt[n]{a})^{n}=(\frac{b}{c})^{n}$
$a=\frac{b^{n}}{c^{n}}$.
Since $a$ is a positive integer,
it follows that $c^{n}=1$.
This means $c=1$.
Thus $\sqrt[n]{a}=\frac{b}{1}$
$=b$,
which implies that $b$ is an integer.
Therefore, if $a$ is a positive integer and $\sqrt[n]{a}$ is rational,
then $\sqrt[n]{a}$ must be an integer.

 

[nuuskur]

Take $a=1$, then $1 = \frac{2^2}{2^2}$, does it follow that $2=1$?! You still haven't answered how $c\mid b$ follows from $c^n\mid b^n$. See #9, for instance, there's a small argument left to complete the proof.

 

[PeroK]

How about this revised proof?

Suppose $a$ is a positive integer and $\sqrt[n]{a}$ is rational.
Then we have $\sqrt[n]{a}=\frac{b}{c}$ for some $b,c\in\mathbb{Z}$ where $c\neq 0$.
Thus $(\sqrt[n]{a})^{n}=(\frac{b}{c})^{n}$
$a=\frac{b^{n}}{c^{n}}$.
Since $a$ is a positive integer,
it follows that $c^{n}=1$.
This means $c=1$.
Thus $\sqrt[n]{a}=\frac{b}{1}$
$=b$,
which implies that $b$ is an integer.
Therefore, if $a$ is a positive integer and $\sqrt[n]{a}$ is rational,
then $\sqrt[n]{a}$ must be an integer.

Most of that proof is unnecessary and yet the key step is just assumed without any justification. In other words, you go into huge detail on steps that are elementary and but skip the step that is hard:

The key step is:

$a=\frac{b^{n}}{c^{n}}$.
Since $a$ is a positive integer,
it follows that $c^{n}=1$.

That is essentially the result assumed without justification. The rest of your proof is largely unnecessary padding.

 

[Math100]

From:

This means $c^{n}\mid b^{n}$.
Since $b\land c$ are relatively prime,
it follows that $b^{n}\land c^{n}$ are also relatively prime.
The fact that $a$ is a positive integer implies $c^{n}=1$.
Thus $c=1$.

 

[PeroK]

From:

This means $c^{n}\mid b^{n}$.
Since $b\land c$ are relatively prime,
it follows that $b^{n}\land c^{n}$ are also relatively prime.
The fact that $a$ is a positive integer implies $c^{n}=1$.
Thus $c=1$.

The key point is that the prime factorisation of $c^n$ has the same primes as $c$, but raised to higher powers. So, if $b$ and $c$ have no common factors, hence no primes in common, then neither do $b^n$ and $c^n$.

 

[nuuskur]

From:

This means $c^{n}\mid b^{n}$.
Since $b\land c$ are relatively prime,
it follows that $b^{n}\land c^{n}$ are also relatively prime.
The fact that $a$ is a positive integer implies $c^{n}=1$.
Thus $c=1$.

We've already established that $a=b^n /c^n$, don't go to back to $a$. Your task is to conclude $c\mid b$ from $c^n\mid b^n$.

Don't write $b\land c$, this is incredibly confusing. Say $b$ and $c$, nothing wrong with that. The symbol $\wedge$ could be also read as "meet" which is the gcd in this context. So $b\wedge c=1$.

You don't get points for rephrasing what you need to prove without proving it. So I'll ask again. Why does $c\mid b$ follow from $c^n\mid b^n$, if $b,c$ are coprime? There are (at least) two options.

1. Use prime factorisation.
2. Complete the argument in #9

 

[Math100]

Here's my newly revised proof:

Suppose $a$ is a positive integer and $\sqrt[n]{a}$ is rational.
Then we have $\sqrt[n]{a}=\frac{b}{c}$ for some $b,c\in\mathbb{Z}$ such that
$gcd(b,c)=1$ where $c\neq 0$.
Let $b=p_{1 }p_{2}\dotsb p_{X}$ and $c=q_{1} q_{2}\dotsb q_{Y}$ such that
$p_{i}\neq q_{j}$.
Now we have $(\sqrt[n]{a})^{n}=(\frac{b}{c})^{n}$
$a=\frac{b^{n}}{c^{n}}$,
and so $ac^{n}=b^{n}$.
This means $(q_{1} q_{2}\dotsb q_{Y})^{n} a=(p_{1} p_{2}\dotsb p_{X})^{n}$,
which implies that $(p_{1} p_{2}\dotsb p_{X})^{n}\mid a$.
Let $a=(p_{1} p_{2}\dotsb p_{X})^{n} k$ for some $k\in\mathbb{Z}$.
Since $(q_{1} q_{2}\dotsb q_{Y})^{n} (p_{1} p_{2}\dotsb p_{X})^{n} k=(p_{1} p_{2}\dotsb p_{X})^{n}$,
it follows that $q_{j}=1$ for every $j$.
Thus $c=1$ and $\sqrt[n]{a}=\frac{b}{c}=\frac{b}{1}=b$,
where $b$ is an integer.
Therefore, if $a$ is a positive integer and $\sqrt[n]{a}$ is rational,
then $\sqrt[n]{a}$ must be an integer.

 

[nuuskur]

Since $(q_{1} q_{2}\dotsb q_{Y})^{n} (p_{1} p_{2}\dotsb p_{X})^{n} k=(p_{1} p_{2}\dotsb p_{X})^{n}$,
it follows that $q_{j}=1$ for every $j$.

You keep repeating the same thing - gliding over the main argument of the proof as if it was as transparent as the rest of the proof. What happens if some $q_j\neq 1$? Why would that necessarily be a contradiction?

 

[Math100]

You keep repeating the same thing - gliding over the main argument of the proof as if it was as transparent as the rest of the proof. What happens if some $q_j\neq 1$? Why would that necessarily be a contradiction?

If some $q_{j}\neq 1$, then $a$ cannot be an integer.

 

[PeroK]

This thread is going round in circles. If $\sqrt[n]{a}$ is rational, then$$ac^n = b^n$$where $gcd(b, c) = 1$.

The key is that if $c \ne 1$, then $c$ has some prime factor $p$, which divides $ac^n$. Hence p divides $b^n$ hence $p$ divides $b$. That contradicts $gcd(b,c) = 1$. Hence $c = 1$.

Alternatively, it's clear that $gcd(b, c) = 1$ iff $gcd(b^n, c^n) = 1$. This is a direct consequence of the fundamental theorem, as $b$ and $b^n$ have the same prime factors. And we see that $\frac b c$ is an integer iff $\frac{b^n}{c^n}$ is an integer.

 

[nuuskur]

@Math100
I should have been reading your post more carefully.

This means $(q_{1} q_{2}\dotsb q_{Y})^{n} a=(p_{1} p_{2}\dotsb p_{X})^{n}$,
which implies that $(p_{1} p_{2}\dotsb p_{X})^{n}\mid a$.

This is rotten. $ka =b$ does not imply $b\mid a$. We are only allowed to conclude $a\mid b$, which is of no help.

Suppose $a$ is a positive integer and take $n\in\mathbb N$. Assume $\sqrt[n]{a} = \frac{p}{q}$ is rational. Then $q^n\mid p^n$. Without loss of generality, we may assume $\mathrm{gcd}(p,q)=1$. Since $q\mid p^n$, it follows that $q\mid 1 \mid p$ (see #9). Therefore $\sqrt[n]{a}$ is an integer.