Multiplication of two 16-bit numbers - Why is the result 32-bit long? [closed]

Answer 1

(2ⁿ - 1)*(2^m - 1) = 2^n+m - 2ⁿ - 2^m + 1

-(2ⁿ + 2^m) is like clearing the bits at index n and m, which does not affect much the result compared to 2^n+m, so you need n+m bits to represent the result.

For example 1111₂*1111₂ = 11100001₂ (15*15 = 225)

In general, (bⁿ - 1)*(b^m - 1) = b^n+m - bⁿ - b^m + 1, so multiply an n-digit by an m-digit number in an arbitrary base b results in a number at most n+m digits

You can see that easily in base 10: 9*9 = 81 (1 digit * 1 digit = 2 digit) or 99*99 = 9801 (2 digit * 2 digit = 4 digit)

Answer 2

For simplicity, let's take unsigned binary numbers. With n binary digits, you can represent 2^n different numbers, ranging from zero upto 2^n-1. When multiplying an n-digit binary number with an m-digit binary number, the result will be a number between zero and (2^n-1) * (2^m-1) = 2^(n+m) - 2^n - 2^m + 1, which typically is only marginally less than 2^(n+m)-1. To represent such a range of results, you need n+m binary digits.