# Question abou Patterson Algorithm

1. Sep 8, 2011

### juaninf

Hi every one

In the preliminaries section, the item c), there a proposition that say: "So by our choice of g we get "$$\theta/p | \psi/p$$" whence "$$\theta | \psi$$" ". I am not understanding this propositión, Please help me

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2. Sep 17, 2011

### micromass

Staff Emeritus
That

$$\theta/p~\vert~\psi/p$$

means that there exists a c such that

$$c\theta/p=\psi/p$$

Now multiply both sides by p...

3. Sep 18, 2011

### juaninf

thanks by your attention, but, a important part is: In the preliminaries section, the item c), there a proposition that say: "So by our choice of g we get "
[tex]\theta/p | \psi/p[\tex], how i use "So by our choice of g we get", Why g havenot any degree?, why choice small as posible that contravening c)?

Last edited: Sep 18, 2011
4. Sep 18, 2011

### HallsofIvy

Staff Emeritus
juaninf, if you keep "bumping" your threads every few minutes, you are going to get banned.