Ineed asap ! what is a counterexample for the conjecture? conjecture: any number that is divisible by 2 is also divisible by 4.a: 32b: 12c: 40d: 18

Let's go through the list of values and test the claim

Is 32 divisible by 2? Yes because 16*2 = 32. Is it also divisible by 4? Yes because 8*4 = 32. So we can cross choice A off the list

Is 12 divisible by 2? Yes because 6*2 = 12. Is it also divisible by 4? Yes because 3*4 = 12. So we can cross choice B off the list.

Is 40 divisible by 2? Yes because 2*20 = 40. Is it also divisible by 4? Yes because 4*10 = 40. Choice C is also crossed off the list.

Is 18 divisible by 2? Yes because 2*9 = 18. Is it also divisible by 4? No. We can see that by dividing 18/4 = 4.5 which is not a whole number result. Or you can list out the multiples of 4 which are 4, 8, 12, 16, 20 and we see that 18 is not on the list.

So choice D) 18 is the answer. This is a counter example to show that the claim "if number is divisible by 2, then it is also divisible by 4" is a false statement.

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.