,.,can anyone please show me the proof of this??i was really confused in my teachers proof,.,thnx
,first,.,he just showed us the properties of P..
1) if a,b belong to P then a+b belongs to P
2) If a,b belong to P, then ab belongs to P
then he proved one theorem " If 0 is not equal to a element R, then a^2 > 0.
the proof goes like this:
either a or -a belongs to P. If a element P, then by a certain theorm, a^2 = aa element of P. If -a element P, then a^2= (-a)(-a) element P. Hence in either case case, a^2 belongs to P.
then he just said that the proof for 1 element of P is the same,.,.but im really confused,.,.
,first,.,he just showed us the properties of P..
1) if a,b belong to P then a+b belongs to P
2) If a,b belong to P, then ab belongs to P
then he proved one theorem " If 0 is not equal to a element R, then a^2 > 0.
the proof goes like this:
either a or -a belongs to P. If a element P, then by a certain theorm, a^2 = aa element of P. If -a element P, then a^2= (-a)(-a) element P. Hence in either case case, a^2 belongs to P.
then he just said that the proof for 1 element of P is the same,.,.but im really confused,.,.
note that 1 = (1)(1) = 1^2.
since 1 is a square, it must be an element of P (by the proof your teacher gave you).
this is rather a high-level proof. in other words, we are defining a positive cone P for the field R.
the definition of such a set P is equivalent to defining the order:
given a subset of P with the given properties, we can define a > b iff a-b is in P,
and given an order >, we can define P as {x in R: x > 0}.
why would one rather define P first, rather than >? well, if one is using Dedekind cuts to define a real number as a certain set of rational numbers,
there is a natural (pre-)order defined by set containment. so we can define P as:
a real number (Dedekind cut) x is in P, if and only if x contains the rational number 0 as an element.
now this might seem a round-about way to get towards something which is pretty simple.
but it turns out to be easier to define (real) addition and multiplication first for positive numbers (elements of P).
negative numbers introduce certain wrinkles, because the negatives of a complement of a Dedekind cut,
may not be a Dedekind cut itself. for an example: suppose we have the real number 4 = {x in Q: x < 4} (the "4" x is less than is the rational number 4).
one might think that -4 might be {x in Q: -x is not in 4}(the real number 4).
however, this makes -4 the set {x in Q: x ≤ -4}, which is not a Dedekind cut (it has a greatest element).
the point being, negative real numbers (defined as Dedekind cuts) are slightly more complicated than positive real numbers.
From your discription, it sounds like you were working with the concept of an "ordered" field. That is, of course, a field with an "order"- a transitive relation, x< y such that:
1) If x< y then, for any z, x+ z< y+ z.
2) If x< y and 0< z then xz< yz.
3) For any two members of the field, x and y, one and only one must be true:
a) x= y
b) x< y
c) y< x
But this is equivalent to saying "there exist a subset of the field, P (called the "positive" members of the field), such that:
1) If x and y are in P then x+ y is in P.
2) If x and y are in P then xy is in P.
3) For any member of the field, x, one and only one of these must be true:
a) x= 0.
b) x is in P.
c) -x is in P.
1 (the multiplicative identity) is not 0, the additive identity so by (3) 1 is in P or -1 is in P. \
Proof by contradiction: Suppose 1 is not in P. Then -1 is in P. By (2) (-1)(-1)= 1 is in P which contradicts the fact that 1 is not in P.
My proof (by contradiction):
Suppose that 1 is not positive
By arithmetic, a negative number multiplied by a negative is a positive
If 1 is negative, then it will be positive when multiplied by -1.
But 1*-1=-1, by arithmetic.
Hence we have reached a contradiction and our supposition is false.
Therefore 1 is positive.
Q.E.D.