81
24)
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the sum of x and y is a rational number. which statement is correct?
the number represented by y must be rational.
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the number represented by x must be rational.
the number represented by x may be rational or irrational.
the number represented by x must be rational and the number
represented by y must be rational.

The number represented by x may be rational or irrational

Step-by-step explanation:

Rational and irrational numbers

For this answer, we must remind that a rational number is such that it can be expressed as a fraction. Irrational numbers cannot. Examples of rational numbers are

Examples of irrational numbers are

To help us to better explain this answer, let's suppose

They both are irrational and their sum is rational as shown:

The first option: "The number represented by y must be rational" is false because y is not rational

The second option: "The number represented by X must be rational." and the last option: "The number represented by x must be rational and the number

represented by y must be rational" are equally false.

The only true option is: "The number represented by x may be rational or irrational".  

We can clearly see one member of a sum doesn't necessarily define it as irrational

idk

step-by-step explanation:

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.