Solu tion
We know L L
L
L
*
=
∪ ∪
0
1
2 KK
and L
L
L
L
+
= ∪ ∪
1
2
3 KK
Now for given
{
}
L
a b
n
n n
=
≥
+1
0
:
, we have
L
a b
b
L a b
ab
L
a b
a b
0
0 0 1
1
1 1 1
2
2
2 2 1
2 3
=
=
=
=
=
=
+
+
+
.
KKK
KKK
Therefore, we have
L
L
L
L
b ab
a b
*
=
∪ ∪
= ∪
∪
0
1
2
2
2 3
KK
KK
Hence it is true that L = L
* .
Ì Exam ple 0.1.26: Given u a ba b
=
2
3 2 and u bab
=
2 , obtain
(a) uv
(e) ||u||
(i) ||v
2
||
(b) vu
(f) ||v||
(j) ||u
2 ||
(c) v
2
(g) ||uv||
(d) u
2
(h) ||vu||
Solu tion
(a) uv a ba b bab
a ba b ab
=
=
(
)(
)
2
3 2
2
2
3 3
2
(b) vu bab a ba b
bab a ba b
=
=
(
)(
)
2
2
3 2
2 2
3 2
(c) v
vv bab bab
bab ab
2
2
2
3
2
= =
=
(
)(
)
(d) u
uu a ba b bab
a ba b ab
2
2
3
2
2
3 2
2
=
=
=
(
)(
)
(e) ||u|| = 8 (as there are 8 letters in the word u)
(f) ||v|| = 4
(g) ||uv|| = 12
(h) ||vu|| = 12
(i) ||v
2 || = 8
(j) ||u
2 || = 11
Ì Exam ple 0.1.27: For any word u and v, prove that
(a) ||uv|| = ||u|| + ||v||
(b) ||uv|| = ||vu||.
Proof: (a) Let us assume ||u|| = m and ||v|| = n. Therefore uv will have ‘m’
letters of u followed by ‘n’ letters of v.
Introduction
23