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 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

⇒Range of x for which this inequality hold

if,

is an Integer ,otherwise range of x

⇒Range of y for which this inequality hold

if,

is an Integer ,otherwise range of y

x> 6

step-by-step explanation:

8x+8> 7x+14

8x-7x> 14-8

x> 6

y=3n + 1

step-by-step explanation:

this is wrong because it says the second number is one less and i added one instead