Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

HARD MATH QUESTION!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!?

HARD MATH HOME WORK! whats 1+1

60 Answers

Relevance
  • 1 decade ago
    Best Answer

    Well you take a 1, then you take another 1, and you got 11.

    But 11 isn't the only answer. There are infinite answers to this problem.

    You can use any variable (any symbol or letter) (s) such as:

    1+1=x

    1+1=y

    1+1=z

    Or, since "1" is a symbol itself, use it as a variable:

    1=[whatever # you want it to be]

    So, 1+1=[twice that #]

    If 1=128,

    1+1=256

    So, 1+1=[the # of your choice]

    ---Ash---

    Source(s): Tested 99th Percentile in Math, but gosh that one is hard for anyone!!!
  • Anonymous
    1 decade ago

    1

  • 1 decade ago

    1+1=2

    Furthermore, somebody with far too much time on their hands decided to prove it...

    "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."

    Wow, he must be a real hit with the ladies...!! :)

    Source(s): Doctor Rob, The Math Forum http://mathforum.org/library/drmath/view/51551.htm...
  • 1 decade ago

    2

  • How do you think about the answers? You can sign in to vote the answer.
  • Anonymous
    1 decade ago

    11

  • 1 decade ago

    The same as 2+0

  • 1 decade ago

    2 or 11

    Source(s): nothing
  • 1 decade ago

    11

  • 1 decade ago

    1+1=2

    1-1=0

    1x1=1

    1/1=1

  • Anonymous
    1 decade ago

    4 - 2

Still have questions? Get your answers by asking now.