How prove this $|x_{p}-y_{q}|>0$ let $$x_{1}=\dfrac{1}{8},x_{n+1}=x_{n}+x^2_{n},y_{1}=\dfrac{1}{10},y_{n+1}=y_{n}+y^2_{n}$$
show that: for any $p,q\in N^{+}$
we have $$|x_{p}-y_{q}|>0$$
 A: Clearly all terms in both sequences are rational.
We will show by induction that if $y_n = \frac{ a_n}{b_n}$ where $\gcd(a_n,b_n)=1$, then $5 \mid b_n$.
Base case is obvious. Induction step: Let $b_k = 5 c_k$, then $y_{k+1} = \frac{a_k}{5c_k} + \frac{ a_k ^2 } { 25c_k^2} = \frac{ a_k 5 c_k + a_k^2 } { 5 c_k^2}$.
We will show by induction that if $x_n = \frac{d_n}{e_n}$ where $\gcd(d_n, e_n) = 1$, then $5 \not \mid e_n$. This is similar to the above, or almost immediately obvious.
Hence, $x_p$ and $y_q$ cannot represent the same rational number.
A: Note that both $x_n$ and $y_n$ are strictly increasing sequences. Notice that $y_2 = .1221$ and $y_3 > .125 = \frac{1}{8}$. This shows $y_n$ never equals $x_1$.
Solve the recursion to get $x_n$ in terms of $x_{n+1}$:
$$
x_n = \frac{-1 + \sqrt{1 + 4 x_{n+1}}}{2}
$$
If we had $|x_p - y_q| = 0$, so that $x_p = y_q$, then $x_{p-1} = y_{q-1}$, $x_{p-2} = y_{q-2}$, etc. If $p > q$, then get $x_{p-q+1} = y_{q-q+1} = y_{1}$ which is impossible because $x_1 > y_1$ and $x_n$ is strictly increasing. Then we must have $p < $q. But then $x_{p-p+1} = x_1 = y_{q-p+1}$. This is impossible because $y_n$ is increasing and we know the third term is more than $x_1 = .125$. 
