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#### CONRADDODD

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- Oct 5, 2013

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**Re: Help solve a problem...**

I have another one....

Prove that \(\displaystyle \sqrt[5]{672}\) is irrational.

Thanks!

- Thread starter CONRADDODD
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- Oct 5, 2013

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I have another one....

Prove that \(\displaystyle \sqrt[5]{672}\) is irrational.

Thanks!

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I also replaced the attached image with $\LaTeX$.

Can you show what you have tried, so that our helpers know where you are stuck?

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- Oct 5, 2013

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Thank you!

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- Oct 5, 2013

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Thank you! Can you give me an example of how you would work this out ?

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Let’s look at Euclid’s proof that the square root of two is irrational.Thank you! Can you give me an example of how you would work this out ?

Let us assume that the root of two is rational and represent it as such:

$\displaystyle \sqrt{2}=\frac{p}{q}$

Solve for $p$ and square both sides:

$\displaystyle p^2=2q^2$

We see that $p$ must be even, since its square is even, making $q$ odd since they are co-prime. We can then let $\displaystyle p=2m$ and $\displaystyle q=2n+1$ to represent an even and odd number respectively.

$\displaystyle (2m)^2=2(2n+1)^2$

Expansion of the squares yields:

$\displaystyle 4m^2=2\left(4n^2+4n+1 \right)$

Divide through by 2 and partially factor the right side:

$\displaystyle 2(m^2)=2\left(2\left(n^2+n \right) \right)+1$

Let $\displaystyle u=m^2$ and $\displaystyle v=2(n^2+n)$ and substitute:

$\displaystyle 2u=2v+1$

Here we absurdly have an even natural number being equal to an odd one. This impossibility means without a doubt that the assumption we made regarding the rationality of the square root of two is false. We can also look at it this way by dividing through by 2:

$\displaystyle u=v+\frac{1}{2}$

This of course is also impossible for natural numbers, to be one half unit apart. This is what's known as proof by

This is essentially the proof given by Euclid in his

- Feb 15, 2012

- 1,967

note that $672 = (2^5)(21)$, so:

$\sqrt[5]{672} = \sqrt[5]{2^5}\sqrt[5]{21} = 2\sqrt[5]{21}$.

It therefore suffices to show that $\sqrt[5]{21}$ is irrational; for if not, then surely

$2\sqrt[5]{21}$ is rational, as well.

So suppose there are two integers $a,b$ with $\text{gcd}(a,b) = 1$ such that:

$\dfrac{a^5}{b^5} = 21$.

This is the same as saying that:

$a^5 = 21b^5$.

Now 3 divides the RHS, so 3 must divide a, and therefore 243 divides $a^5$.

So $a^5 = 243t = 21b^5$.

Thus $81t = 7b^5$.

Since 3 divides 81, 3 divides 81t, so 3 divides $7b^5$. Can you continue?

- Mar 22, 2013

- 573

(It has been quite some time afterwards this has been post, but since this is the first time I had a through look at the NT posts here, I am giving an answer anyways)

The best thing for such large quantities isn't really reduction absurdium.

Note that the desired number is a root of the polynomial

$$x^5 - 672 = x^5 - 3 \cdot 7 \cdot 2^5 \tag{1}$$

Also, keep in mind that $\sqrt[5]{672}$ lies inside the interval $[3, 4]$. The trick here is to apply the rational root theorem to prove that there is no such rational root of $(1)$ in $[3, 4]$. This can easily be accomplished by trying all the factors of $672$ inside $[3, 4]$, namely,

$\{3, -3, 2^2, -2^2\}$

It can easily be seen that none of the above suffices, as the only real root of $(1)$ is positive and both $3$ and $4$ are either under or overestimation of the root by under or overestimating the resultant $0$ by an exponential size.

The best thing for such large quantities isn't really reduction absurdium.

Note that the desired number is a root of the polynomial

$$x^5 - 672 = x^5 - 3 \cdot 7 \cdot 2^5 \tag{1}$$

Also, keep in mind that $\sqrt[5]{672}$ lies inside the interval $[3, 4]$. The trick here is to apply the rational root theorem to prove that there is no such rational root of $(1)$ in $[3, 4]$. This can easily be accomplished by trying all the factors of $672$ inside $[3, 4]$, namely,

$\{3, -3, 2^2, -2^2\}$

It can easily be seen that none of the above suffices, as the only real root of $(1)$ is positive and both $3$ and $4$ are either under or overestimation of the root by under or overestimating the resultant $0$ by an exponential size.

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- Feb 15, 2012

- 1,967

By Gauss' lemma, if $x^2 - 2$ factors over $\Bbb Q$, it factors over $\Bbb Z$, thus any rational square root of 2 must be an integer!

As 2 is prime, the only possibilities are {-2,-1,1,2}, and none of these are viable candidates.

- Mar 22, 2013

- 573

Equivalent to using integer root theorem. I avoid it as much as possible because it usually reminds me of Gauss' lemma on quadratic reciprocity.Deveno said:By Gauss' lemma

- Feb 15, 2012

- 1,967

....which is why i said: ..."in the same vein"....Equivalent to using integer root theorem. I avoid it as much as possible because it usually reminds me of Gauss' lemma on quadratic reciprocity.

But my point was a bit subtler:

When asked to prove something, it is natural to ask: which tools can I use...or, as on these forums, which things has the poster already learned?

An argument can be made to appealing to facts known about polynomials as continuous functions in number theory on the grounds that someone taking a number theory course has likely already taken calculus.

Of course, invoking Euclid's proof begs the question: has the fact that the integers are a UFD been established (being "told" this is true isn't quite the same thing....even if you BELIEVE it....)?

It *is* true, however, that if gcd(a,b) = 1, and:

$\left(\dfrac{a}{b}\right)^n = m$ where $a,b,m \in \Bbb Z^+$

That $b = 1$.

To see this, suppose $p$ is a prime such that $p^k|a$ but $p^{k+1} \not\mid a$.

Since b is co-prime to a, we have that p necessarily divides m. Repeating this process with:

$\dfrac{a^n}{p} = \left(\dfrac{m}{p}\right)b^n$

and so on, for powers of $p$ up to $p^k$, we obtain that $p^k|m$, so that:

$\dfrac{a^n}{p^k} = \left(\dfrac{m}{p^k}\right)b^n$

where this is a relation among 3 integers: $\dfrac{a^n}{p^k},\dfrac{m}{p^k},b^n$.

Now $p$ is not a factor on either side.

Repeating this for every prime factor of a, eventually leads to:

$1 = \left(\dfrac{m}{a^n}\right)b^n$ so that $b^n|1 \implies b^n = 1 \implies b = 1$.

This (perhaps) is unexpected:

every rational $n$-th root of an integer is also an integer.

We can thus conclude:

If $m$ is an $n$-th power-free integer, $\sqrt[n]{m} \not \in \Bbb Q$.

So an even SHORTER proof of the irrationality of $\sqrt[5]{672}$ is:

21 is 5th-power free.

This reduces checking the irrationality of many $n$-th roots of integers to a purely mechanical process of looking at the prime factorization.

We can, without any additional effort, actually go a bit further:

There is a natural isomorphism between $(\Bbb Z^+,\ast)$ and the abelian monoid:

$\displaystyle \bigoplus_{i = 1}^{\infty} \Bbb N$ under addition (this is actually the free abelian monoid on infinitely many generators),

where we can represent any positive integer as a finite $k$-tuple, for example:

$24 \mapsto (3,1)$

So writing $m \mapsto (k_1,\dots,k_r)$ we have:

$\sqrt[n]{m}$ is rational if and only if $n|k_j$ for all $j = 1,\dots,r$.

In other words the "real" reason $\sqrt[5]{672}$ is irrational is because 5 does not divide 1 and:

$672 \mapsto (5,1,0,1)$.

(I'm about 95% sure this is correct, but I never really thought about this problem in this way, so I would appreciate it if someone could confirm I haven't made some egregious error).

(Additonal musing: is the Grothendieck groupification of this monoid just $\Bbb Q^{\ast}$? If so, can we devise a similar mechanical check on the irrationality of $\sqrt[n]{q}$ for any $q \in \Bbb Q$?).

- Jan 29, 2012

- 1,151

Here, [tex]x^5- 672= 0[/tex], the coefficient of [tex]x^5[/tex] is 1 and the constant term is -672 so any rational root would have to be an integer that evenly divides 672.

[tex]672= 2^5(3)(7)[/tex] so any rational root must be one of

1, 2, 3, 6, 7, 8, 12, 14, 16, ...

It easy to see that [tex]1^5= 1[/tex], [tex]2^5= 32[/tex], [tex]3^5= 243[/tex], [tex]4^5= 1024[/tex].

None of those is equal to 672 and the last is already larger than 672.