Question

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.

Answer 1

Answer with explanation:

It is given that , a and b are positive integers.

gcd(a,b)=1

We 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 7

GCD (9,7)=1

So, we have to calculate positive integers x and y such that

⇒ 9 x +7 y > 9×7-9-7

⇒9x +7 y> 47

To 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

$\Rightarrow \frac{x}{\frac{ab-a-b}{a}}+ \frac{y}{\frac{ab-a-b}{b}}>1,\frac{ab-a-b}{a},\text{and},\frac{ab-a-b}{b}$

⇒Range of x for which this inequality hold

      $=[\frac{ab-a-b}{a},\infty)$

if,

$\frac{ab-a-b}{a}$

is an Integer ,otherwise range of x

         $=(\frac{ab-a-b}{a},\infty)$

⇒Range of y for which this inequality hold

      $=[\frac{ab-a-b}{b},\infty)$

if,

$\frac{ab-a-b}{b}$

is an Integer ,otherwise range of y

         $=(\frac{ab-a-b}{b},\infty)$