**Theorem:**The digit 4 is a constant number. But sometimes it is not so. It becomes equal to zero.

∴ To prove: 4 = 0

**Proof:**

Statement | Proof |
---|---|

Let we have an equation as follows | |

4² – 4² = 0 | Universal truth as 16 – 16 = 0 |

it can be written | |

( 4 + 4 ) ( 4 - 4 ) = 0 | by the formula |

a² – b² = ( a + b ) ( a – b ) | |

4 – 4 = 0 | dividing both sides by 4 + 4 |

or 2² – 2² = 0 | Given |

( 2 + 2 ) ( 2 – 2 ) = 0 | by the formula |

a² – b² = ( a + b ) (a – b ) | |

2 + 2 = 0 | dividing both sides by 2 – 2 |

4 = 0 | Hence proved. |

Those who'd like to challange above theorem's approach, ever remember solving limit problems at the beginning of calculus?

## 1 comment:

bogus, dividing by 2-2 is dividing by 0, it's undefined.

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