Would two bits be enough to assign a unique binary number to each vowel in the English language?

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.