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 (x)/((ab-a-b)/(a))+ (y)/((ab-a-b)/(b))>1,(ab-a-b)/(a),\text{and},(ab-a-b)/(b)`

⇒Range of x for which this inequality hold

`=[(ab-a-b)/(a),\infty)`

if,

`(ab-a-b)/(a)`

is an Integer ,otherwise range of x

`=((ab-a-b)/(a),\infty)`

⇒Range of y for which this inequality hold

`=[(ab-a-b)/(b),\infty)`

if,

`(ab-a-b)/(b)`

is an Integer ,otherwise range of y

`=((ab-a-b)/(b),\infty)`