Give a recursive definition of each of these sets of ordered pairs of positive integers. Use str

Give a recursive definition of each of these sets of ordered...

Key Concepts

Recursive Definitions

A recursive definition specifies a set by giving one or more initial elements (the base cases) and then providing one or more rules for generating additional elements from those already produced. This approach is effective for systematically defining complex sets by building them step-by-step from simpler elements.

Structural Induction

Structural induction is a method of proof used alongside recursive definitions. It involves showing that the property holds for the base case(s) and then proving that if the property holds for elements constructed through the recursive rules, it also holds for the new elements. This ensures the validity of the property across the entire recursively defined set.

Ordered Pairs of Positive Integers

This concept refers to pairs of numbers where each element is a positive integer. Such ordered pairs are often used in problems where relationships between two quantities are explored, and conditions (such as sums or divisibility) are imposed on these pairs.

Parity Conditions

Parity conditions distinguish between even and odd integers. In the context of ordered pairs, conditions like 'the sum is even' or 'the sum is odd' rely on these parity definitions. Essentially, they dictate how the components of the pair interact under addition to produce an even or odd result.

Divisibility Rules

Divisibility rules are criteria that determine whether one integer divides another without leaving a remainder. When recursive definitions involve specific divisibility conditions (for example, '3 divides b'), these rules become crucial in ensuring that every newly generated pair meets the given constraint.

Pattern Recognition

Pattern recognition involves analyzing or visualizing a set of points (or pairs) to identify underlying structures or regularities. This process can reveal natural groupings or sequences that suggest how to define the recursive steps, making it easier to construct a complete and consistent recursive definition.

Transcript

00:01 We are given sets of ordered pairs of positive integers, and we are asked to give recursive definition of each of these sets, and then to use structural induction to prove that the recursive definition is correct.

00:16 So in part a, we're given the set s is the set of all ordered pairs, ab, such that a and b are positive integers, and we have that a plus b is even.

00:44 So first let's find the point nearest the origin as the basis step.

00:51 To do this, you may want to graph it.

00:53 Otherwise you can sort of just do it by thinking about possibilities.

00:58 So notice that the smallest possible pair, which would have one in one, we have one plus one is two, which is even.

01:08 So it follows that 1 -1 is an s.

01:11 We can't get any closer to the origin than this.

01:14 So this is our base case.

01:16 And as per curses definition, notice that we can add all the other points by adding a multiple of 2 to either coordinate or a multiple of 1 to both coordinates.

01:31 So it follows that if ab lies in s, then we have that a plus 2b lies an s, and a .b.

01:47 Plus 2 lies an s.

01:49 An s and a plus 1, b plus 1, lies in s.

01:57 This is our recursive definition.

02:08 Now we'll use structural induction to prove that this definition is correct.

02:15 So the basis step, now we want to prove that a plus b is even whenever ab, ab lies in s.

02:35 So first of all, for the basis case, we have that 1 -1 lies in s and 1 plus 1 equals 2 of course is even so this checks out.

02:52 Now for the inductive step, assume that ab is an s with ab even, if a plus b is even.

03:19 Then by our definition we have that a plus 2b is an s, a plus 1 b plus 1 is an s, and we have that a, b, plus 2 lies in s.

03:44 We have that a plus 2 plus b is equal to a plus b plus 2, which is going to be even since a plus b is even.

03:58 We have that a plus 1 plus b plus 1 is equal to a plus b plus 2, which of course is even.

04:11 Since a plus b is even.

04:16 And we have the sum of two even numbers is another even number.

04:19 And finally, a plus b plus 2 is the same as a plus b, plus 2, which is even.

04:28 Again, since a plus b is even.

04:35 So it follows that property is true for the recursive step.

04:46 And so by the principle of structural induction, it follows that a plus b is going to be even if ab is an s.

05:19 So we've proven a definition is correct.

05:23 In part b, we're given that s is the set of all pairs ab, such as a and b are positive integers, and a or b is odd.

05:54 So to find a recursive definition, first let's find the point that's closest to the origin.

06:00 This will be our base case.

06:02 You can do this either by graphing or just by thinking about it.

06:06 So consider the smallest possible point.

06:08 Point 1 .1, we have that 1 is odd.

06:16 And so it follows that 1 -1 lies in s.

06:22 And we also have that 1 -2 lies in s.

06:27 And we have that 2 -1 lies in s.

06:32 And this is as close as we're gonna get to the origin.

06:35 So these three points are our base case.

06:42 And you know that all the other points can be obtained by adding two to one of the coordinates.

06:48 Or both because then we preserve the oddness of the coordinate.

06:58 So it follows that if a, b, lie in s, then we have that a plus 2, b lies in s, and a, b plus 2, lies in s.

07:23 And we could technically add two to both, but this is really the same as adding 2 to one coordinate and then adding two to the other coordinate for the new pair.

07:33 So this is a good enough definition...

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