Is there any such thing as an absolute?

Your just not doing it right try it with your head first and get some momentum going David copperfild can do it why cant you? Isnt absolute zero absolute or does possibly quantum mechanics allow for something to go below absolute zero?

Um well no its because of mathematical laws we proved that 1+1 = 2 so 2+2 = 4hers the proof i copy pasted it because i know it exists and i googled it but i never tried to grasp what it means or learned it by hart.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 = 2Proof: 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.from http://answers.yahoo.com/question/index?qid=2008062201563...