MathDB

As evidence that the correct answer does not mean the correctness of the proof, the teacher cited next example. Let's take the fraction

\frac{19}{95}

. After crossing out

9

in the numerator and denominator (“reduction” by

9

), we get

\frac{1}{5}

which is the correct answer. In the same way, a fraction

\frac{1999}{9995}

can be “reduced” by three nines (cross out

999

in the numerator and denominator). Is it possible that as a result of such a “reduction” we also get the correct answer, equal to

\frac13

? (We consider fractions of the form

\frac{1a}{a3}

. Here, with the letter

a

we denote several numbers that follow in the same order in the numerator after

1

, and in the denominator before

3

. “Reduce” by

a

Problem Statement