Can I deduce from $ \frac{a}{b} = \frac{x}{y} $ that a = x and b = y where $ a, x \in Z $ and $ b, y \in Z^{+} $?

Question

Can I deduce from $ \frac{a}{b} = \frac{x}{y} $ that a = x and b = y where $ a, x \in Z $ and $ b, y \in Z^{+} $?

I'm wanting to prove that the function $ f(x, y): \frac{x}{y} $ where $(x , y) \in \mathbb{Z} \times \mathbb{Z} ^{+} $ is injective.

I'm using the definition of injectivity: $ \forall a, b \in X, f(a) = f(b) \Rightarrow a = b $ but I can't seem to find a way to proof that $f(x, y) = f(x', y') \rightarrow (x = x' \wedge y = y') $

Answer 1

Nope:

$\frac ab = \frac {ak}{bk}$ but we can not conclude that $ak =a$ nor that $bk=b$.

For example $\frac 34 = \frac 68$ but $3\ne 6$ and $4\ne 8$.

But if $\frac ab = \frac xy$ and both are "in lowest terms" you can conclude that $a=x$ and $b=y$.

And if $\frac ab$ is in lowest terms but we don't know if $\frac xy$ is (or if we know it is not) the we know thre is an $m\in\mathbb Z$ so that $am = x$ and $bm = y$. (And we know that $m = \frac xa = \frac yb$).

And if neither are necessarily in lowest terms you can conclude that there is a value $k\in \mathbb Q$ so that $ak = x$ and $bk = y$. (we can figure that $k = \frac xa =\frac yb$. And we can figure that if $\frac ab$ is in lowest terms then $k$ is an integer.)

....

A case where we might fall into trouble might be a problem: Suppose $x = n + 3$ and $y = n + 6$ and $\frac xy = .75$ what is $n$.

If you figure $\frac xy = .75 = \frac 34$ so $x=3$ and $y=4$ so $n+3 =3$ and $n+6=4$ so $n=0$ and $n=-2$ then... well, you just stapled yourself six ways to sundays.

But instead you $\frac xy = \frac 34$ so $x = 3m$ and $y = 4m$ for some integer $m$ ($\frac 34$ is in lowest terms so we can assume $m$ is an integer).

So $n+3 = 3m$ and $n+6 = 4m$ so $n = 3m -3$ and $n = 4m -6$ so $3m-3 = 4m-6$ so $6-3 = 4m -3m$ or $3 = m$ so $m=3$ and $n+3 = 3*3=9$ so $n =6$ and $n+6=4*3 =12$ so $n = 6$.

That's the correct answer.

But, of course, the rule $\frac xy = \frac 34 \implies 4x = 3y$ so $4(n+3)=3(n+6)$ so $4n + 12 = 3n + 18$ so $4n -3n = 18-12$ so $n = 6$ is a much better, and more recommended way to do it.

Answer 2

Let $(a,b,x,y)\in\mathbb{Z^*}^4$.

Suppose that $\dfrac{a}{b} = \dfrac{x}{y}$, then we have : $$\dfrac{\frac{a}{\gcd(a,b)}}{\frac{b}{\gcd(a,b)}} = \dfrac{\frac{x}{\gcd(x,y)}}{\frac{y}{\gcd(x,y)}}$$ Then we have the result : $$\left( \frac{a}{\gcd(a,b)}=\frac{x}{\gcd(x,y)} \right) \wedge \left( \frac{b}{\gcd(a,b)}=\frac{y}{\gcd(x,y)} \right)$$