Repeat.

Proving @Munem Sahariar 's solution.

$\begin{aligned} X & = 5.23\overline{457} \\ & = 5.23 + 0.00\overline{457} \\ & = 5.23 + 0.00457\left(1+10^{-3}+10^{-6} + \cdots \right) \\ & = 5.23 + 0.00457 \sum_{n=1}^\infty \left(\frac 1{1000} \right)^n \\ & = 5.23 + 0.00457\left(\frac {1000}{999}\right) \\ & = 5.23 + \frac {457}{99900} \\ & = \frac {5.23(99900) + 457}{99900} \\ & = \frac {5.23(100000-100) + 457}{99900} \\ & = \color{#3D99F6} \frac {523457-523}{99900} & \small \color{#3D99F6} \text{Munem Shariar's solution.} \\ & = \frac {522934}{99900} \\ & = \frac {261467}{49950} \end{aligned}$

Therefore, $a=\boxed{261467}$ .

$5.23\overline{457} = \frac{(523457-523)}{99900} = \dfrac{522934}{99900} = \dfrac{261467}{49950}$

$a= \boxed{261467}$

Let $X=5.23\overline{457}=5.23457457457\dots$ ,

$100X=523.457457457\dots$ ——————————(1)

$100000X=523457.457457\dots$ ————————(2)

(2)-(1), get $99900X=522934\\49950X=261467\\X=\frac{261467}{49950}$

$\boxed{a=261467}$

Let X=5.23457457457.....

Multiple both side by 100

100x=523.457457457457..... (1)

Again multiply both side by 1000

100000x=523457.457457457... (2) Subtract (1) from (2)

We get

X=522934/99900

But here it is not a couple prime numbers so we have to 522934 and 99900 by 2

We get

X=261467/49950=a/b

Also it is cold prime

Hence

a=261467

$\begin{aligned} 5.23\overline{457} \\ &= 5.23 + 0.00\overline{457} \\ &= 5.23 + \frac {1}{100} \times 0.\overline{457} \\ &= \frac {523}{100}+ \frac {1}{100} \times \frac {457}{999} \\ &= \frac {522477}{99900} + \frac {457}{99900} \\ &= \frac {522934}{99900} \\ &= \frac {261467}{49950} \\ a = \boxed{261467} \end{aligned}$