Solutions To Prime Modulus - Congruence Problems

 1)  Write four additional equivalences for the mod 7 residues (2 already shown)

Solution:

Already given:

1   ÷  2 = 4(mod 7) 

And:

1   ÷   4 = 2 (mod 7) 

So we add:

1   ÷   3 = 5 (mod 7) 

1   ÷   5 = 3 (mod 7) 

1   ÷   6 = 6 (mod 7) 

1   ÷   7 =0 (mod 7)   (Impossible)

2)  Consider the p = 5 prime modulus portrayed as shown below in clock form:

Write four equivalences for the mod 5 residues. Can these be written in the form for the mod 7 cases in Problem 1? Why or why not?

Solution:

Four equivalences from the table:

1   ÷   2 = 1· 3  ≡ 3 (mod 5)

1   ÷   3 = 1 ·2  ≡ 2 (mod 5)

1   ÷   4 = 1 · 4  ≡ 4 (mod 5)   

2   ÷   3 = 2 ·2  ≡ 4 (mod 5) 

 These cannot be written in the same form (as Littlewood's) for the mod 7 cases given there is no counterpart for 4 x 2 = 8 ≡ 1 (mod 7).   I.e. from the mod 5 multiplication table we see no entry: 3 x 2 = 6   ≡ 1 (mod 5)

This is because the commutative and distributive laws are not consistently obeyed by groups.