MHB Question about problem involving gcd

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HelloI am studying the problem given in the attachement. In the solution given, it says "Similarly \( d|\gcd(a,-b) \) ". I could not understand why this is so.thanks
 

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IssacNewton said:
HelloI am studying the problem given in the attachement. In the solution given, it says "Similarly \( d|\gcd(a,-b) \) ". I could not understand why this is so.thanks
$d|\gcd(a,-b)$ follows from the fact that $d=\gcd(a,b)$.
 
thanks...I should have realized that...
 
Thread 'How to define a vector field?'

Thread 'How to define a vector field?'

Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...
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