Theorem: Let a and b be positive integer, if gcd(a,b)=1, then exist positive integer x and y such that ax+by=c for any integer greater than ab-a-b.Prove the above theorem.
Accepted Solution
A:
Answer with explanation:It is given that , a and b are positive integers.gcd(a,b)=1We have to prove for any positive integer x and y ,a x + by =c, for any integer greater than ab-a-b.Proof:GCD of two numbers is 1, when two numbers are coprime.Consider two numbers , 9 and 7GCD (9,7)=1So, we have to calculate positive integers x and y such that⇒ 9 x +7 y > 9×7-9-7⇒9x +7 y> 47To prove this we will draw the graph of Inequality.So the ordered pair of Integers are x>5 and y>6.So, for any integers , a and b , →ax+ by > a b -a -b, if[tex]\Rightarrow \frac{x}{\frac{ab-a-b}{a}}+ \frac{y}{\frac{ab-a-b}{b}}>1,\frac{ab-a-b}{a},\text{and},\frac{ab-a-b}{b}[/tex]⇒Range of x for which this inequality hold [tex]=[\frac{ab-a-b}{a},\infty)[/tex]if,[tex]\frac{ab-a-b}{a}[/tex] is an Integer ,otherwise range of x [tex]=(\frac{ab-a-b}{a},\infty)[/tex]⇒Range of y for which this inequality hold [tex]=[\frac{ab-a-b}{b},\infty)[/tex]if,[tex]\frac{ab-a-b}{b}[/tex]is an Integer ,otherwise range of y [tex]=(\frac{ab-a-b}{b},\infty)[/tex]