Could $(2n-1)\left( \dfrac{1}{a}- \dfrac{1}{b} \right) $ be an integer?

Question

Let $ a,b \in \mathbb{R}_{>0} \backslash \left \{ \tfrac{2N-1}{2M-1} : N,M \in \mathbb{N} \right \} $ and $n \in \mathbb{N}$.

I want to prove that the quantity \begin{equation}\label{key} (2n-1)\left( \dfrac{1}{a}- \dfrac{1}{b} \right) \end{equation} cannot be an integer.

I got the following:

Assume that it is an integer, $N$, say. Then \begin{equation} \dfrac{Nab}{a-b} \end{equation} is odd. However, I get stuck here.

Answer 1

None of the expressions expressed need to be integer.

We can have $(2n-1)(\frac 1a - \frac 1b) = M\in \mathbb Z$

if $\frac 1a - \frac 1b = \frac M{2n-1}$

Nos $a,b$ can be any positive reals that are not a ration of two odd integers.

So we can have $b$ but any irrationa and then $a = \frac 1{\frac M{2n-1}-\frac 1b}$ will not be rational and that will work.

Example $b = \frac 1{\pi}$ and $a = \frac 1{2-\pi}$ and

$(2n-1)(\frac 1a -\frac 1b) = (2n-1)(2-\pi + \pi) = (2n-1)\cdot 2$ is an integer.

If we want $a$ and $b$ to be rational we can have $a$ and $b$ but have an even components. Let $b = \frac 12$ and $a =\frac 14$. Thn

$(2n-1)(\frac 1{\frac 14} - \frac 1{\frac 12})= (2n-1)(4-2) = 2(2n-1)$ is an integer.

Answer 2

If $a$ and $b$ are both odd integers, then $ab$ will be an odd integer. Then

1. $n= \frac{ab+1}{2}$ is an integer.
2. $\frac{2n-1}{a} = b$
3. $\frac{2n-1}{b} = a$
4. $(2n-1)\left( \dfrac{1}{a}- \dfrac{1}{b} \right) = b - a \in \Bbb Z$

Answer 3

Let $a = b$. Then the unnamed quantity is the integer $0$.

Possibly the strongest signal that your argument cannot work is your line "odd plus odd is odd", which is false. The sum of two odd numbers is an even number. Easy example: $1+1$.