While I was studying Discrete Mathematics, I faced a question that I do not understand how to solve even after looking at the answer. The question asks me to prove that the set of positive rational numbers is countable by showing that:

The function K is a one-to-one correspondence between the set of positive rational numbers and the set of positive integers if $K(m/n)=P_1^{2a_1}P_2^{2a_2}...P_s^{2a_s}Q_1^{2b_1-1}Q_2^{2b_2-1}...Q_t^{2b_t-1}$, where $gcd(m,n)=1$ and the prime-power factorizations of m and n are $m=P_1^{a_1}P_2^{a_2}...P_s^{a_s}$ and $n=Q_1^{b_1}Q_2^{b_2}...Q_t^{b_t}$.

Attempt:

$K(m/n)=P_1^{2a_1}P_2^{2a_2}...P_s^{2a_s}Q_1^{2b_1-1}Q_2^{2b_2-1}...Q_t^{2b_t-1}=(P_1^{a_1}P_2^{a_2}...P_s^{a_s})^2\frac{(Q_1^{b_1}Q_2^{b_2}...Q_t^{b_t})^2}{(Q_1Q_2...Q_t)}=m^2\frac{n^2}{Q_1Q_2...Q_t}$

I am going to prove that the function is a one-to-one correspondence by showing that when $x=y$, then $f(x)=f(y)$, and when $f(x)=f(y)$, then $x=y$.

1. If $x=y$, then $f(x)=f(y)$

Let $x=\frac{a}{b},y=\frac{c}{d}, gcd(a,b)=gcd(c,d)=1$

$b=b_1^{k_1}b_2^{k_2}...b_t^{k_t}$ and $d=d_1^{s_1}d_2^{s_2}...d_i^{s_i}$

Since x=y and gcd(a,b) and gcd(c,d) are 1, a=c and b=d.

Because $b=d, b_1^{k_1}b_2^{k_2}...b_t^{k_t}=d_1^{s_1}d_2^{s_2}...d_i^{s_i}$ and $ b_1b_2...b_t=d_1d_2...d_i$

$K(a/b)=a^2\frac{b^2}{b_1b_2...b_t}=c^2\frac{d^2}{d_1d_2...d_i}=K(c/d)$

Hence, proved.

2. If $f(x)=f(y)$, then $x=y$.

However, I have no idea how to prove this part.

If there are other more formal ways to solve the question, please teach me. I really need your help. Thank you

While I was studying Discrete Mathematics, I faced a question that I do not understand how to solve even after looking at the answer. The question asks me to prove that the set of positive rational numbers is countable by showing that:

The function K is a one-to-one correspondence between the set of positive rational numbers and the set of positive integers if $K(\frac{m}{n})=P_1^{2a_1}P_2^{2a_2}...P_s^{2a_s}Q_1^{2b_1-1}Q_2^{2b_2-1}...Q_t^{2b_t-1}$, where $gcd(m,n)=1$ and the prime-power factorizations of m and n are $m=P_1^{a_1}P_2^{a_2}...P_s^{a_s}$ and $n=Q_1^{b_1}Q_2^{b_2}...Q_t^{b_t}$.

Attempt:

$K(\frac{m}{n})=P_1^{2a_1}P_2^{2a_2}...P_s^{2a_s}Q_1^{2b_1-1}Q_2^{2b_2-1}...Q_t^{2b_t-1}=(P_1^{a_1}P_2^{a_2}...P_s^{a_s})^2\frac{(Q_1^{b_1}Q_2^{b_2}...Q_t^{b_t})^2}{(Q_1Q_2...Q_t)}=m^2\frac{n^2}{Q_1Q_2...Q_t}$

I am going to prove that the function is a one-to-one correspondence by showing that when $x=y$, then $f(x)=f(y)$, and when $f(x)=f(y)$, then $x=y$.

1. If $x=y$, then $f(x)=f(y)$

Let $x=\frac{a}{b},y=\frac{c}{d}, gcd(a,b)=gcd(c,d)=1$

$b=b_1^{k_1}b_2^{k_2}...b_t^{k_t}$ and $d=d_1^{s_1}d_2^{s_2}...d_i^{s_i}$

Since $x=y$ and gcd(a,b) and gcd(c,d) are 1, a=c and b=d.

Because $b=d, b_1^{k_1}b_2^{k_2}...b_t^{k_t}=d_1^{s_1}d_2^{s_2}...d_i^{s_i}$ and $ b_1b_2...b_t=d_1d_2...d_i$

$K(\frac{a}{b})=a^2\frac{b^2}{b_1b_2...b_t}=c^2\frac{d^2}{d_1d_2...d_i}=K(\frac{c}{d})$

Hence, proved.

2. If $f(x)=f(y)$, then $x=y$.

However, I have no idea how to prove this part.

If there are other more formal ways to solve the question, please teach me. I really need your help. Thank you