Assume both r and s are rational then $\frac{r}{s}\in\mathbb{Q}$

Question

Currently done with showing for both r+s and r $\times s$ rational.

Now stuck with the following part; If we know that r and s are both rationals what could we say about

$\frac{r}{s}$

Assume both r and s are rationals.

Then we would have by definition r = $\frac{a}{b}$ where a and b are integers and b $\neq$ 0

Similarly; s = $\frac{c}{d}$ where c and d are integers and d $\neq$ 0

$\frac{r}{s}$ = ($\frac{a}{b}$) /($\frac{c}{d}$)

Hence $\frac{r}{s}$ = (a $\times d$)/ (b $\times c$)

Since ad and bc are integers, numerator and denumerator are both integers. Integers divided by integers are rational ?

Is this valid proof ?

No, the proof is invalid, because you forgot that $s$ might be $0$. In this case, $c=0$, so $\frac{r}{s}=\frac{ad}{bc}$ is undefined. (Integers divided by non-zero integers are rational!) — TonyK, Jul 05 '20 at 16:03

score 3 · Accepted Answer · answered Jul 05 '20 at 16:08

3

Yes, it's a valid proof.

The only "error" that may be corrected is that you must assume $s\in\mathbb{Q}\backslash\{0\}$, because if $s=0$ then $r/s$ is undefined and consequently we get that $r/s\notin\mathbb{Q}$.

So, as conclussion for your proof: $$r\in\mathbb{Q}\phantom{a}, \phantom{a}s\in\mathbb{Q}\backslash\{0\} \Longrightarrow \frac{r}{s}\in\mathbb{Q}.$$

answered Jul 05 '20 at 16:08

Alejandro Bergasa Alonso

3,534

2

I see, thank you ! – Mark Jul 05 '20 at 16:10

Assume both r and s are rational then $\frac{r}{s}\in\mathbb{Q}$

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