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

• Math100
In summary: 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.In summary, if ##a## is a positive integer and ##\sqrt[n]{a}## is rational, then ##\sqrt[n]{a}## must be an integer.
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.

There's a lot of maths there but you've effectively assumed the key point:
Math100 said:
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.

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.

nuuskur said:
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.

nuuskur said:
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##?

## 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##.

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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?

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.

Math100 said:

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:

Math100 said:
## 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.

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 ##.

Math100 said:
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##.

Math100 said:
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

PeroK
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.

PeroK
Math100 said:
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?

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nuuskur said:
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 and nuuskur
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.

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nuuskur
@Math100
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.

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## 1. What is a positive integer?

A positive integer is a whole number that is greater than zero. It can be written without any fractions or decimals.

## 2. What does it mean for a number to be rational?

A rational number is a number that can be expressed as a ratio of two integers, where the denominator is not equal to zero. In other words, it can be written as a simple fraction.

## 3. How do you find the nth root of a number?

To find the nth root of a number, you can use a calculator or you can use the exponentiation operator. For example, to find the square root of a number, you can use the square root symbol (√) or raise the number to the power of 1/2.

## 4. Can a positive integer have a rational nth root?

Yes, a positive integer can have a rational nth root if the exponent (n) is a factor of the number. For example, the 4th root of 16 is 2, which is a rational number.

## 5. Are there any exceptions to the rule that a rational nth root of a positive integer is a factor of the number?

Yes, there are a few exceptions. One exception is when the exponent (n) is 1, in which case the nth root is equal to the original number. Another exception is when the exponent is a prime number, in which case the nth root will not be a factor of the number.

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