Give a recursive definition of each of these sets of ordered pairs of positive integers. [Hint: Plot the points in the set in the plane and look for lines containing points in the set.] a) S = {(a, b) | a ∈ Z+ , b ∈ Z+ , and a + b is odd}

(2,1), (1,2) is your base step

if (a,b)   is in the set (a+1,b+1) will be in  the set

if  (a,b) is in the set (a+2,b) will be in the set

if (a,b) is in the set (a,b+2) will be in the set.

Step-by-step explanation:

Think about how to solve this problem in general. How can you assure that the sum a+b is odd ?

Think about this, what happens when you sum two even numbers ? The result is even or odd ?

2+6 = 8 (even  )

10+12 = 22 (even)

And what happens when you sum two odd numbers ? The result will be even or odd ?  Look

3+7 = 10 (even)

5+11 = 16 (even)

Therefore to assure that a+b is odd, one of them has to be odd and one of them has to be even, that is why

(2,1), (1,2) is your base step

if (a,b)   is in the set (a+1,b+1) will be in  the set

if  (a,b) is in the set (a+2,b) will be in the set

if (a,b) is in the set (a,b+2) will be in the set.

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?