For any

$$\forall a,b,n \in\mathbb{N}\wedge n\ge 2\Rightarrow\Big(\frac{a}{b}\Big)^{n}\ne 2$$

Here is my proof of it :

There are $3$ possibilities:

  • $a = b$
  • $a > b$
  • $b < a$

  • $ a = b $ if two numbers are equal, their ratio is $1$ and $1^n$ is always $1$, and $$1 \neq 2$$

  • $ a < b $ A smaller number divided by a bigger number is less than one, and a number less than one when raised to integer power can only get smaller, so it cannot be $2$.

  • $ a > b $ By taking the $n$-th root on both sides:

$$\frac{a}{b} = \sqrt[n]{2}$$

As the $n$-th sqrt of two can never be expressed as a ratio (Pythagoras proved this I think), this possibility is False too.

Is my proof sensible? Is it rigorous? Is it even fully correct? Is the last item of my last too big / complex to be used in something so simple?

I seek any suggestion, improvement or correction.

  • 7
    $\begingroup$ Your case analysis is not really relevant, and finally your proof is by stating that someone else proved an equivalent theorem :( $\endgroup$ – Yves Daoust Sep 25 '15 at 16:12
  • $\begingroup$ @YvesDaoust I felt something was wrong about it... if I thought it were good I would not have posted it. :) $\endgroup$ – user3105485 Sep 25 '15 at 16:13
  • $\begingroup$ You're basically trying to say that every natural root of $2$ is irrational. $\endgroup$ – barak manos Sep 25 '15 at 16:30
  • $\begingroup$ Fermat's Last Theorem reduces the case to $n=2$... Haha $\endgroup$ – Eoin Sep 26 '15 at 3:29
  • $\begingroup$ See also: math.stackexchange.com/questions/1191176/… $\endgroup$ – Martin Sleziak Sep 26 '15 at 8:03

Let $$\left(\frac ab\right)^n=2$$ or $$a^n=2b^n,$$ where $a,b$ have no common factor (otherwise they can be simplified).

The last equation shows that $a^n$ is even. So is $a$, as odd numbers given only odd powers. Then $a^n$ is a multiple of $2^n$ and $a^n/2$ is a multiple of $2^{n-1}$, i.e. is even ($n>1$). Then $b^n$ is even and so is $b$.

$a$ and $b$ are both even, a contradiction.

Second proof:

Let $a^n=2b^n$, and denote $\alpha$ and $\beta$ the exponents of $2$ in the prime decompositions of $a$ and $b$.

Then as the decomposition is always unique, $$(2^\alpha\cdot p)^n=2\cdot(2^{\beta}\cdot q)^n$$implies $$\alpha n=\beta n+1$$ which is impossible.


Note that your proof of the $a>b$ case is all you need as your reasoning applies to the other two cases, so you can certainly shorten the proof.

The fact that you quote is actually a step in the standard proof that $\sqrt[n]2$ is irrational. As such, you should probably prove it from first principles to avoid any circular logic.

Have you seen the proof that $\sqrt 2$ is irrational? Can you adapt this proof to this more general case?

  • $\begingroup$ Pytagoras proof uses the fact that if a and b are coprime so are a^2 and b^2 $\endgroup$ – user3105485 Sep 25 '15 at 16:23
  • $\begingroup$ @user3105485 exactly. But by the same reasoning, if $a,b$ are coprime, so are $a^n$ and $b^n$ for any $n$. $\endgroup$ – Mathmo123 Sep 25 '15 at 16:24
  • $\begingroup$ this fact seems obvious, shall I prove it or just assume it is true? $\endgroup$ – user3105485 Sep 25 '15 at 16:29
  • 1
    $\begingroup$ You should prove it if you use it. If $p$ is prime and $p\mid a^n$, then can you show that $p\mid a$? In fact you don't need the full strength of your statement - take a look at Yves's answer. $\endgroup$ – Mathmo123 Sep 25 '15 at 16:56

Your proof for the first two cases are perfect.

For the case $a>b$ ; you could do this,in case you do not have the proof that :
$\sqrt[n]{2}$ is irrational for $n\ge 2$ (as Mathmo123 says , is required)

$$(a/b)^n ={a^n}/{b^n}=2$$ $$or, a^n=2.b^n$$

As $2$ is a prime number , this above equation tells that $2|a$. Let $$a=2k$$ Then $$2^n.k^n=2.b^n$$ $$2^{n-1}.k^n=b^n$$

Again this means $$2|b$$ as well so let $$b=2j$$ Then we get $$2^{n-1}.k^n=2^n.j^n$$ $$or\ \ \ , k^n=2.j^n$$

The same procedure applied again shows that $$a=2^s\ \ \ and\ \ \ b=2^l\ \ \ for\ \ \ some\ \ \ s,l\in \mathbb N$$

Then $${(a/b)}^n={({2^s}/{2^l})}^n = (2^{s-l})^n=2^{ns-n}$$

So this can be $2$ iff $$n(s-l)=2$$

Now $2$ being a prime , we have $2$ possibilities :

  • $n=2$,$s-l=1$

  • $n=1$,$s-l=2$ Both of them gives $$(a/b)^n = 4\neq 2$$ So we have it .Proved .


If $(a/b)^n=2$, then $a^n=2b^n$. We claim that the only integer solution to this equation is when $a=b=0$.

We'll prove this by contradiction. Suppose that there was a non-zero solution $(a,b)$. Choose a solution with minimal value of $|a|+|b|$. Since $2b^n$ is even, it follows that $a^n$ must be even, and therefore $a$ is also even. Now we may write $a=2c$ to obtain that $(2c)^n=2b^n$, so canceling a $2$ from both sides yields $b^n=2^{n-1}c^n$. Therefore $b$ is even, so $b=2d$ for some integer $d$, yielding $$ 2^nd^n=2^{n-1}c^n\implies c^n=2d^n. $$ Thus we have obtained a solution $(c,d)$ to the original equation which is strictly smaller, in the sense that $$|c|+|d|=\frac{|a|+|b|}{2}<|a|+|b|.$$

Therefore there is no non-zero integer solution to the equation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.