Which set of ordered pairs represents y as a function of x? Group of answer choices

{(0, 0), (1, 1), (1, 0), (2, 1)}

{(2, -1), (4, -2), (6, -3), (8, -4)}

{(1, -5), (1, 5), (2, -10), (2, -15)}

{(3, 3), (3, 4), (4, 3), (4, 4)}

{(2, -1), (4, -2), (6, -3), (8, -4)}

In this relation, each point has a unique x coordinate. None of the x coordinates repeat themselves.

For something like choice A, we have x = 1 repeated. The input x = 1 leads to multiple outputs (y = 1 and y = 0 simultaneously).

The same situation happens in choice C also. This time x = 1 and x = 2 repeat themselves.

Choice D has x = 3 and x = 4 repeat themselves.

A function is only possible if each x input leads to exactly one y output.

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.

because that’s us why