I am wondering how I can change a repeating decimal to a fraction?

Let's take an example.
#r = 1.523523523.....#

This number has three repeating digits. So you can multiply by #1000# and shift it to the left by 3 places, The decimal point moves to the right.

#1000xxr = 1523.523523#

which is #1522 + r#

so you have
#1000r-r = 1522#

or

#999r = 1522#

and thus
#r= 1522/999 = 1.52352352352#

While the answer can be worked out by a full process as explained by another contributor, there is a useful short cut which is quick to use.

If all the decimals recur:

Write the fraction as: #"the recurring digits"/(9 " for each digit"#

eg: #0.676767... = 0.bar(67) = 67/99" "larr# there are 2 recurring digits

eg: #7.394394394... = 7.bar(394) = 7 394/999" "larr# 3 recurring digits

If only some of the decimals recur

Write the fraction as:

#"all the digits - non-recurring digits"/(9 " for each recurring digit and " 0" for each non-recurring digit"#

eg: #0.23555... = 0.23bar5 = (235-23)/900 = 212/900#

eg: #3.4678678.. = 3.4bar(678) =3 (4678-4)/9990 = 3 4674/9990#

eg: #9.461565656.. = 9.461bar(56) = 9 (46156-461)/99000 = 45695/99000#

An alternative method if you have a calculator, but not all of the digits is to use continued fractions.

Find the coefficients of the (terminating) continued fraction by repeatedly separating off the whole number part and taking the reciprocal. Then write down the continued fraction and simplify:

For example, given:

#2.596638655462#

Write down the whole number part #color(red)(2)#, subtract it and take the reciprocal to get (approximately):

#1.67605633803#

Write down the whole number part #color(red)(1)#, subtract it and take the reciprocal to get (approximately):

#1.47916666666#

Write down the whole number part #color(red)(1)#, subtract it and take the reciprocal to get (approximately):

#2.08695652177#

Write down the whole number part #color(red)(2)#, subtract it and take the reciprocal to get (approximately):

#11.4999999959#

There's obviously a truncation error here, so round to:

#11.5#

Write down the whole number part #color(red)(11)#, subtract it and take the reciprocal to get:

#2#

Our final whole number part is #color(red)(2)#.

So:

#2.596638655462 ~~ color(red)(2) + 1/(color(red)(1)+1/(color(red)(1)+1/(color(red)(2)+1/(color(red)(11)+1/(color(red)(2))))))#

#color(white)(2.596638655462) = 2+1/(1+1/(1+1/(2+2/23)))#

#color(white)(2.596638655462) = 2+1/(1+1/(1+23/48))#

#color(white)(2.596638655462) = 2+1/(1+48/71)#

#color(white)(2.596638655462) = 2+71/119#

#color(white)(2.596638655462) = 309/119#