moeSizlak | more power to you i guess |

centrx | heh |

noway- | TRWBW, and we can assume the bijection because ? |

Tarantulafudge | how would I write x-3y^2+4y+1 in standard form? |

TRWBW | noway-: it's proof by contradiction |

Tarantulafudge | oops x=-3y^2+4y+1 ach x3y^2+4y+1 |

narg | Tarantulafudge: order them by exponent degree |

TRWBW | noway-: just prove this given any map f:A->P(A), there is an element of P(A) that isn't in the image of f. that implies that no bijection can exist. |

noway- | TRWBW, the statement said there is no injection from f: P(A) -> A |

Tarantulafudge | x=3y^2+4y+1 |

noway- | ok |

Tarantulafudge | narg: they are narg: I think |

Olathe | Tarantulafudge: You mean y = ... ? |

Tarantulafudge | Olathe: no this is a horizontal parabola |

narg | Tarantulafudge: is this a quadratic standard form your after? ah |

Tarantulafudge | sorry standard form is x=a(y-k)^2+h |

Olathe | Tarantulafudge: Complete the square, I guess. |

Tarantulafudge | Olathe: how lol Olathe: I mean I know how, but it doesn't look easy on this problem |

Olathe | (y - k)^2 = y^2 - 2ky + k^2, right ? So, you have 3(y^2 + 4/3y + 1). k is 4/6. Well, -2/3 |

Tarantulafudge | hmm I haven't seen this equation before |

Olathe | So, 3(y - 2/3)^2 + h, find h. Bleh. So, 3(y + 2/3)^2 + h, find h. |

geckosen1tor | is a*(b*c) the same as (a*b)*c for matrixes? |

Olathe | % 3(y + 2/3)^2 + 5/9 //Expand |

mbot | Olathe: 17/9 + 4*y + 3*y^2 |

Olathe | % 3(y + 2/3)^2 - 1/3 //Expand |

mbot | Olathe: 1 + 4*y + 3*y^2 |

Olathe | There. geckosen1tor: Yep. |

Tarantulafudge | interesting |

geckosen1tor | Olathe: sweet, so it's associative |