Adding really simple fractions

[bonanova]

Johnny was asked to write the sum of the two fractions p/n and q/n where p, q and n are single-digit numbers.

He remembered that the numerators were to be added, but he thought maybe the denominators should be added as well.

Also, he decided that putting digits next to each other could suggest addition.

So his first try was to say that

p/n + q/n = pq/nn. (where pq and nn are two-digit numbers.)

Then he remembered the result should not depend on the order of addition,

but he saw that p/n + q/n = qp/nn was just as bad. So he decided to use them both.

His final representation of the sum was

p/n + q/n = (pq + qp)/nn.

The puzzle asks: Let p, q and n be chosen at random from {1, 2, 3, 4, 5, 6, 7, 8, 9}.

What is the expected error in Johnny's formula?

[Rob_G]

There is no error.

We know p/n+q/n = (p+q)/n

In his formula pq can be written as 10p+q, qp as 10q+p and nn as 10n+n.

So (pq+qp)/nn = ([10p+q]+[10q+p])/(10n+n) = (11p+11q)\11n = (11[p+q])/11n = (p+q)/n = p/n+q/n.