In the number $15.3\dot{9}$, $9$ is repeated forever. If the number is rational then it can be expressed as a fraction (i suppose it is rational since it's an exercise for me to find it's rational representation?).
This is my attempt to solve it: $$ x=15.3\dot{9} \iff 10x=153.\dot{9} $$ (i want $15.3\dot{9}$ therefore $138 + 15 +0.\dot{9}$) $$10x=138 + 15 +0.\dot{9} $$ (i want $0.4$, since $0.4 +0.\dot{9}$ will give me $1.3\dot{9}$, i basically want the part .$3\dot{9}$) $$10x=137.6 + 0.4 + 15 +0.\dot{9} \iff 10x=137.6 + 14 +1 +1.3\dot{9} \iff 10x = 137.6 + 1 +15.3\dot{9} \iff 10x = 138.6 + x \iff 9x = 138.6$$ $$ \therefore x = \frac{1386}{90} = \frac{154}{10} = \frac{77}{5} $$
Now if i execute $\frac{77}{5}$ with software i get $15.4$ not $15.3\dot{9}$. All other such exercises i've done give me the exact answer (not with infinite accuracy of course, but you understand that it would be correct, for example a software program should have given me $15.39999999999$ in order for me to realize that what i've done is correct).
What is the issue here? Am i wrong? Maybe the number is irrational?
Also is there any other simpler way to get the right answer?
Thanks in advance.
Hint. You may just observe and directly prove that
Have a look at this.