In which of the following base system, 132 is not a valid number ?
Base 2
The rule is simple and absolute: in base r, the only legal digits are 0 to r − 1. A number is valid in a base only if every one of its digits obeys that limit.
Test the digits of 132 — the largest is 3.
| Base | Digits allowed | Is 132 valid? |
|---|---|---|
| 16 | 0–9, A–F | ✓ Yes |
| 10 | 0–9 | ✓ Yes |
| 8 | 0–7 | ✓ Yes |
| 2 | 0 and 1 only | ✗ No — the digit 3 is illegal |
So 132 is invalid only in base 2 — option 4.
What the same string means in each valid base is worth seeing, because it makes the positional principle concrete:
\((132)_{8}=1\times8^{2}+3\times8^{1}+2\times8^{0}=64+24+2=90_{10}\)
\((132)_{16}=1\times256+3\times16+2=306_{10}\)
\((132)_{10}=132\)
The same three symbols denote three different quantities — which is why a base subscript is never optional when more than one base is in play.
The general rule that follows: the larger the base, the more values a given number of digits can represent, since n digits in base r span \(r^{n}\) values. Binary needs the most digits of all, which is exactly why hexadecimal is used as shorthand — one hex digit replaces four binary digits, with no arithmetic required for the conversion.
A practical note on validity checking : the quickest test is to find the largest digit present and add one; that is the smallest base in which the number is legal. Here the largest digit is 3, so 132 is valid in base 4 and every base above it, and invalid in bases 2 and 3.
Hence, 132 is not a valid number in base 2.
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