Keyword | CPC | PCC | Volume | Score | Length of keyword |
---|---|---|---|---|---|

transitive property matrix | 0.36 | 0.3 | 3385 | 82 | 26 |

transitive | 1.67 | 0.3 | 1447 | 37 | 10 |

property | 0.42 | 0.8 | 4636 | 28 | 8 |

matrix | 0.61 | 0.3 | 3411 | 98 | 6 |

Keyword | CPC | PCC | Volume | Score |
---|---|---|---|---|

transitive property math | 1.25 | 0.1 | 8158 | 46 |

transitive property math definition | 0.05 | 0.2 | 6536 | 73 |

transitive property mathematics | 0.07 | 0.7 | 1741 | 70 |

transitive property math meaning | 0.49 | 0.5 | 456 | 34 |

transitive property math is fun | 0.34 | 0.3 | 748 | 5 |

transitive property matrix | 1.27 | 1 | 2461 | 56 |

transitive property definition math geometry | 0.89 | 0.6 | 8981 | 66 |

math definition of transitive property | 1.19 | 0.5 | 7173 | 5 |

definition of transitive property in math | 0.79 | 0.4 | 2737 | 67 |

transitive property definition in math | 1.62 | 0.7 | 6303 | 2 |

transitive property in math | 1.42 | 1 | 5630 | 75 |

math transitive property of equality | 0.55 | 0.8 | 265 | 18 |

Transitive Property. The Transitive Property states that for all real numbers x , y , and z , if x = y and y = z , then x = z . Substitution Property. If x = y , then x may be replaced by y in any equation or expression.

Transitivity on a set of ordered pairs (the matrix you have there) says that if $(a,b)$ is in the set and $(b,c)$ is in the set then $(a,c)$ has to be. So we make a matrix that tells us whether an ordered pair is in the set, let's say the elements are $\{a,b,c\}$ then we'll use a $1$ to mark a pair that is in the set and a $0$ for everything else.

Let us explore some properties of transitive relations: The inverse of a transitive relation is a transitive relation. For example, as we discussed above 'is less than' is a transitive relation, then the converse 'is greater than' is also a transitive relation. The union of two transitive relations need not be transitive.

The relation is transitive if and only if the squared matrix has no nonzero entry where the original had a zero. (If you don't know this fact, it is a useful exercise to show it.)$\endgroup$ – Harald Hanche-Olsen Nov 4 '12 at 14:39 $\[email protected] Hanche-Olsen, I am not sure I would know how to show that fact.$\endgroup$ – Mack