Give me a 5-element language set , A , over the alphabet {a,b} whose square is equal to 11.

Question

The alphabet {a,b} should make the up the 5-element formal language. |A| = 5 and |$A^2$| = 11

For example: A could equal {a, b, bb}

So the square of the A, denoted $A^2$, would be {a, b, bb} * {a, b, bb} =

{aa, ab, abb, ba, bb, bbb, bba, bbb, bbbb} =

{aa, ab, abb, ba, bb, bbb, bba, bbbb} because there were two bbb's, which isn't needed. In this case, |A| = 3 and |$A^2$| = 8.

Is there any easier way to do this without simple trial and error? Any thoughts would be appreciated.

Answer 1

How about $A=\{a,aa,aaa,aaaa,aaaaaaa\}$?