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I'm reading through Serge Lang's Basic Mathematics and I've fallen into trouble with a particular proof exercise:

Let $a = \frac m n$ be a rational number expressed as a quotient of integers m, n with $m\neq0$ and $n\neq0$. Show that there is a rational number $b$ such that $ab = ba = 1$.

I was able to solve the problem in my own way by assuming $b = \frac r s$ and proving the exercise via the cross-multiplication rule, however Lang instead proves the exercise by giving b the value of $\frac n m$ and simply multiplying both $a$ and $b$. How is this valid? Wouldn't the wording of the problem assume that rational number $b$ has its own unique numerator and denominator values?

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    $\begingroup$ In the last sentence of the statement shouldn't it be "number $b $ such that $ab = ba = 1$"? $\endgroup$
    – RGS
    Commented Jan 23, 2017 at 13:23
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    $\begingroup$ All you have to show is a $b$ exists. It is perfectly valid to just give the value of $b$ and check that it works (you don't even need to show how you came up with $b$, only that it works). $\endgroup$ Commented Jan 23, 2017 at 13:23

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I assume you mean that $ab=ba=1$.

The statement claims that, given $a=\frac{m}{n}$, there exists some $b=\frac{r}{s}$ such that $ab=ba=1$.

The statements says nothing about $r$ and $s$ being different from $m$ and $n$. In fact, the statement is true if you set $r=n$ and $s=m$. The statement can be proven by setting $b=\frac{n}{m}$ and showing that $ab=ba=1$.


Remember, in mathematical statements, nothing is said "implicitly". Everything that is said, is said outright. So, for example, the statement

If $x=2$, there exists some $y$ such that $x+y=4$

is true, because I can set $y=2$ and demonstrate that $x+y=4$. I don't have to worry that I picked $y$ as the same number as $x$, because, if the statement doesn't explicitly demand that $x$ and $y$ are different, I don't have to set them to different values.


So, to sum up, you ask

Wouldn't the wording of the problem assume that rational number $b$ has its own unique numerator and denominator values?

and the answer is no. The wording of the question says that $b$ is a rational number, nothing more. The problem doesn't explicitly say that $b$ has its own unique numerator and denominator, so it doesn't assume that.

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Not sure what your 'unique numerator and denominator values' mean exactly, but would $b=\frac{2n}{2m}$ work?

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