Question

Does the check digit of an ISSN detect every error where two consecutive digits are accidentally interchanged? Justify your answer with either a proof or a counterexample.

Short answer

Yes, the check digit of an ISSN detect every every error where two consecutive digits are accidentally interchanged

Step-by-step solution

Step 1

For a International Standard Serial Number (ISSN) \({d_1}{d_2} \ldots {d_8}\), the check digit can be calculated using:

Note: When is used to represents \(d\_\left\{ 8 \right\}.\)

To proof: When two consecutive digits are accidentally interchanged, then this error is always detected.

Step 2

PROOF:

Let \(i\)be an integer from 1 to 6 .

Let \({d_i}\)and \({d_{i + 1}}\)be interchanged.

Note that \({d_i}\)and \({d_{i + 1}}\)need to be different digits, because otherwise the ISSN code doesn't change and thus no error is then made.

When \(d\_\left\{ i \right\} > d\_\left\{ {i + 1} \right\},\)then \(k = {d_i} - {d_{i + 1}}\)is some integer from 1 to 9 (as the difference between two digits is at most 9 ).

However, \(3{d_1} + 4{d_2} + 5{d_3} + 6{d_4} + 7{d_5} + 8{d_6} + 9{d_7}\)then decreases by \(k\), which will result in a different value modulo 11 and thus the check digit will detect the error (as the check digit changes).

When \(d\_\left\{ i \right\} < d\_\left\{ {i + 1} \right\},\)then \(k = {d_{i + 1}} - {d_i}\)is some integer from 1 to 9 (as the difference between two digits is at most 9).

However, \(3{d_1} + 4{d_2} + 5{d_3} + 6{d_4} + 7{d_5} + 8{d_6} + 9{d_7}\) then increases by \(k\), which will result in a different value modulo 11 and thus the check digit will detect the error (as the check digit changes).

Note: I assumed that the check digit was not interchanged. When the check digit can be interchanged as well, then such an error can be detected as we require and thus interchanging \({d_7}and{d_8} = X\)will then be detected.