Number theory help. How did they conclude gh = 1? (at the bottom of page 17)

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- anonymous

Number theory help. How did they conclude gh = 1? (at the bottom of page 17)

- schrodinger

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- anonymous

https://books.google.com/books?id=eVwvvwZeBf4C&lpg=PA3&pg=PA17#v=onepage&q&f=false

- UsukiDoll

Oh I love that book. I've used it last year.

- freckles

how far did you get because right before they say that they have d=ghd
and dividing d on both sides gives 1=gh (Assuming d isn't 0 )

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- anonymous

@UsukiDoll yeah I like it too. Certainly better than the one wrote by Stein

- freckles

also we know d isn't zero since it is given in the theorem that a and b are not 0

- anonymous

O: oh, I see it now.

- anonymous

The whole time i was reading d2 = gh d1 .

- anonymous

This question is a false alarm XD. I was having a brain fart :P sorry people lol

- UsukiDoll

d = ghd ... when d was divided it left
1 = gh
so g = 1 and h = 1
which are the only values that satisfy
1=gh

- UsukiDoll

xD I'm still wondering how people master proofs x.x like I don't mind easy ones, but the hard ones are crazy

- anonymous

@freckles thanks for clarifying :)

- anonymous

You know what. While we're at it, can any one explain how they got 0 <= r_i <= b-i?

- mathstudent55

\(gd_1 = d_2\) and \(hd_2 = d_1\)
\(d_2 = \dfrac{d_1}{h} \)
Equate the two expressions for \(d_2\)
\(gd_1 = \dfrac{d_1}{h} \)
\(ghd_1 = d_1\)
\(gh = 1\)

- anonymous

They said it's by induction though :/
@mathstudent55 thank you. I got that part :)

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