# Integer power of fraction never is two

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.

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

• Pytagoras proof uses the fact that if a and b are coprime so are a^2 and b^2 – user3105485 Sep 25 '15 at 16:23
• @user3105485 exactly. But by the same reasoning, if $a,b$ are coprime, so are $a^n$ and $b^n$ for any $n$. – Mathmo123 Sep 25 '15 at 16:24
• this fact seems obvious, shall I prove it or just assume it is true? – user3105485 Sep 25 '15 at 16:29
• 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. – 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.