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

         

Period is the time it takes for one “cycle” to complete. in the parent sine function, period is 2*pi since sin(0) = 0, sin(pi/2) = 1, sin(pi) = 0, sin(3pi/2) = -1, and sin(2pi) = 0. at that point, the sequence repeats itself. i’m not sure if your graph is similar to the sine function since you didn’t provide a picture, but you can find the period by looking at a graph.

Answer:

13.46

step-by-step explanation:

16.80 x .20 = 13.46