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show can find d and b s.t pd-bq=1 , p and q are coprime- GCF =1

(Context probably irrelevant but modular forms- to show that all rational numbers can be mapped to \infty) that is there exists a γ=(abcd) \gamma = ( a b c d) , sorry thats a 2x2 matrix, SL2(Z) \in SL_2(Z) with det(γ)=adbc=1 det (\gamma) = ad-bc=1 s.t γ.t=at+b/ct+d= \gamma . t = at+b/ct+d = \infty , where take t=r t= r , r a rational number. )

So I'm at the stage where I am just stuck on showing that p p and q q co prime implies that b b and d d can be found s.t pdbq=1 pd-bq=1 , b and d integer.

I'm not sure how to do this? I think the argument should be obvious?

Many thanks in advance.
Original post by xfootiecrazeesarax
(Context probably irrelevant but modular forms- to show that all rational numbers can be mapped to \infty) that is there exists a γ=(abcd) \gamma = ( a b c d) , sorry thats a 2x2 matrix, SL2(Z) \in SL_2(Z) with det(γ)=adbc=1 det (\gamma) = ad-bc=1 s.t γ.t=at+b/ct+d= \gamma . t = at+b/ct+d = \infty , where take t=r t= r , r a rational number. )

So I'm at the stage where I am just stuck on showing that p p and q q co prime implies that b b and d d can be found s.t pdbq=1 pd-bq=1 , b and d integer.

I'm not sure how to do this? I think the argument should be obvious?

Many thanks in advance.


Can't say I follow the mapping to infinity stuff, however the bit that you're asking about is a minor variant on Bezout's identity - see here.

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