Answer:
No. Two bits are not enough to assign a unique binary number to each vowel in the English language.
Explanation:
The vowel in the English language is made up of 5 lowercase letters (a, e, i, o, u) and their corresponding 5 uppercase letters (A, E, I, O, U). All together, there are 10 unique vowel letters.
The number, x, of unique characters that can be stored by n bits is given by:
x = 2ⁿ
So if we have 2 bits, it implies that, the number of different characters that can be stored is:
2² = 4
Therefore, with 2 bits, there are 4 unique characters that can be stored.
But then, we need to store a total of 10 characters representing the English Language vowels. Therefore, two bits will not be enough to assign a unique binary number to each vowel.
It can also be seen this way
For two bits, we have the following possibilities:
00
01
10
11
Now, let's assume we have assigned the vowels as follows:
00 = a
01 = e
10 = i
11 = o
Then, only four of the vowel letters can be assigned a unique binary number.
Therefore, two bits are not enough to assign a unique binary number to each vowel of the English language.
We need at least 4 bits which will yield 2⁴ = 16 different combinations. And from that, we are able to assign each vowel a unique binary number like so:
Used
0000 = a
0001 = e
0010 = i
0011 = o
0100 = u
0101 = A
0110 = E
0111 = I
1000 = 0
1001 = U
Remaining:
1010
1011
1100
1101
1110
1111
10 out of 16 different combinations of binary numbers have been used by the 10 vowel characters. We will be left with 6 binary numbers. Better being surplus than being deficit.
