# $(\text{even}/\text{even})\times \text{even} = \text{odd}$? [closed]

Do even natural numbers $$e_1,e_2,e_3$$ exist with $$\frac{e_1}{e_2}\cdot e_3=o$$, such that $$o$$ is an odd natural number?

If exist do they have any relation for this even tuple$$(e_1,e_2,e_3)$$?

## closed as off-topic by Thomas, Morgan Rodgers, Watson, Davide Giraudo, colormegoneJul 5 '16 at 18:10

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• (even / even) * even is even if and only if (even / even) is integer. – barak manos Jul 5 '16 at 11:53
• @barak it can be true even if it isn't an integer. – Matt Samuel Jul 5 '16 at 11:54
• @MattSamuel: Correct!!! – barak manos Jul 5 '16 at 11:55
• @kvk30: What I said is not true (in particular, the "if and only if" part). I'm leaving it here for people to read and learn (and also for the comment-thread to make sense). – barak manos Jul 5 '16 at 11:56
• Write $e_1=2^{k_1}q_1;\ e_2=2^{k_2}q_2;\ e_3=2^{k_3}q_3$ with $q_1,q_2,q_3$ odd numbers. The condition is: if and only if $$\begin{cases}k_1+k_3=k_2\\ q_2\mid q_1q_3\end{cases}$$ – user228113 Jul 9 '16 at 20:02

If $v(n)$ denotes the exponent in the power of $2$ in the prime factorization of the even number $n$ ,

then $\frac{e_1}{e_2}\cdot e_3$, if it is an integer, is odd if and only if $v(e_1)+v(e_3)=v(e_2)$.

The easiest example is $\frac{2}{4}\cdot 2$

If $e_1$, $e_2$ and $e_3$ are even and $o$ is odd, then:$$(e_1/ e_2)e_3=o\iff e_1\cdot e_3=o\cdot e_2\implies e_2\equiv0\bmod 4.$$

On the other hand, if $o$ is odd and $e_2\equiv 0\bmod 4$, then: $$o\cdot e_2\equiv 0\bmod 4\implies \exists e_1,e_3\text{ even, such that } o\cdot e_2=e_1\cdot e_3\iff (e_1/e_2)e_3=o.$$

Here's a simple counterexample: (4/8)*6=3.

• "Counterexample" is for disproving an assertion. Not for exploring a question, unless you yourself make an assertion in the process. – Joffan Jul 5 '16 at 12:18
• Moreover, the OP explicitely asks for a general answer and not an example. – gebruiker Jul 5 '16 at 12:51