Well, let's start with a simple case: adding two one-bit numbers. We'll call them X and Y, and since they each only have one

**b**inary dig

**it**they can only have the values 0 or 1. So here are the cases we need to account for: 0+0=0, 0+1=1, 1+0=1, 1+1=10.

Notice the answer for that final case has two digits. A Boolean function can only return one value, so how can get that? By having two functions, one for each digit.

So, the function for the rightmost digit has to satisfy this truth table:

X | Y | Z_{0} | |
---|---|---|---|

0 | 0 | 0 | |

0 | 1 | 1 | |

1 | 0 | 1 | |

1 | 1 | 0 |

The function for the next digit will have to satisfy this truth table:

X | Y | Z_{1} | |
---|---|---|---|

0 | 0 | 0 | |

0 | 1 | 0 | |

1 | 0 | 0 | |

1 | 1 | 1 |

I'm using the subscripts here to indicate which digit in the overall number. So our two bit answer Z has the individual bits Z

_{1}Z

_{0}, where Z

_{1}= X AND Y, and Z

_{0}= X XOR Y;

So, that's how you can add two one-bit numbers. But what if you want to add bigger numbers? Well, the way you add the second bits together is pretty much the same as the first bits, with one major difference: You need to account for the bit that got carried from the sum of the first bits. If that carry bit was 0, then the result of the next bit will be same as the first. But if it's one, then the result is increased by one. So, 0+0+1=1, 0+1+1=10, 1+0+1=10, 1+1+1=11.

So, now we have these two truth tables:

X_{1} | Y_{1} | C | Z_{1} | |
---|---|---|---|---|

0 | 0 | 0 | 0 | |

0 | 1 | 0 | 1 | |

1 | 0 | 0 | 1 | |

1 | 1 | 0 | 0 | |

0 | 0 | 1 | 1 | |

0 | 1 | 1 | 0 | |

1 | 0 | 1 | 0 | |

1 | 1 | 1 | 1 |

X_{1} | Y_{1} | C | Z_{2} | |
---|---|---|---|---|

0 | 0 | 0 | 0 | |

0 | 1 | 0 | 0 | |

1 | 0 | 0 | 0 | |

1 | 1 | 0 | 1 | |

0 | 0 | 1 | 0 | |

0 | 1 | 1 | 1 | |

1 | 0 | 1 | 1 | |

1 | 1 | 1 | 1 |

_{1}= X

_{1}XOR Y

_{1}XOR C and Z

_{2}= (X

_{1}AND Y

_{1}) OR (X

_{1}AND C) OR (Y

_{1}AND C). C, the carry bit, comes from the second bit of the result of the sum of the first two bits, that is C = X

_{0}AND Y

_{0}.

Each subsequent bit works just like the second, with the second bit of the previous result being carried over. Generally, and Z

_{n}= (X

_{n}AND Y

_{n}) OR (X

_{n}AND C

_{n-1}) OR (Y

_{n}AND C

_{n-1}), and C

_{n}= X

_{n}XOR Y

_{n}XOR C

_{n-1}.

Here's another challenge for you: How do you do subtraction in a similar manner?

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