A Small Proof in Number Theory

Question

let us suppose that there are values $a_1, a_2, b_1, b_2 \in \mathbb{Z}^+$ such that $gcd(a_1,b_1) = 1$ and $gcd(a_2,b_2) = 1$

I have to prove

$a_1b_2 = a_2b_1 => a_1 = a_2$ and $b_1 = b_2$.

Answer 1

Suppose $a_1 b_2 = a_2 b_1$. Since $a_1 \mid a_1 b_2$, we have $a_1 \mid a_2 b_1$. Since $\gcd (a_1,b_1) = 1$ and $a_1 \mid a_2 b_1$, we must have $a_1 \mid a_2$.

Similarly, $a_2 \mid a_1$. Hence, $a_1 = \pm a_2$. But $a_1 > 0$ and $a_2 > 0$, so $a_1 = a_2$.

Similarly, $b_1 = b_2$.

Answer 2

Notice that if $a_1b_2 = a_2b_1$ then $\frac{a_1}{b_1} = \frac{a_2}{b_2}$. And also notice that $gcd(a_1,b_1) = 1$ implies the fraction $\frac{a_1}{b_1}$ cannot be further simplified and same goes for $\frac{a_2}{b_2}$. So the equality $\frac{a_1}{b_1} = \frac{a_2}{b_2}$ implies $a_1 = a_2$ and $b_1 = b_2$.

Answer 3

If you have Bezout's theorem, then you know that $\gcd(a_1,b_1)=1$ implies $1=a_1m_1+b_1n_1$ for some integers $m_1$ and $n_1$. This implies

$$a_2=a_2a_1m_1+a_2b_1n_1=a_1a_2m_1+a_1b_2n_1=a_1(a_2m_1+b_2n_1)$$

Since $a_1$ and $a_2$ are positive, we must have $a_2m_1+b_2n_1\ge1$, hence $a_2\ge a_1$. But the same argument applied to $\gcd(a_2,b_2)=1$ gives

$$a_1=a_1a_2m_2+a_1b_2n_2=a_2a_1m_2+a_2b_1n_2=a_2(a_1m_2+b_1n_2)$$

which, as before, implies $a_1\ge a_2$. Together these imply $a_1=a_2$, so that $a_1b_2=a_2b_1$ (with $a_1=a_2\not=0$) implies $b_1=b_2$ as well.