O6:    a·a   is greater than or equal to zero.

Note first that  [Graphics:../Images/index_gr_118.gif]  is read as 'a squared' and that  [Graphics:../Images/index_gr_119.gif]  means that 'a squared' is defined to be  a·a.  Also note that  [Graphics:../Images/index_gr_120.gif]  means that  a  is not equal  to  b.

"For any  a  in  R  we have  [Graphics:../Images/index_gr_121.gif],  with equality holding only if  a=0;"  (see ita-2-2-12-pg20)

"[...]more generally the sum of the squares of several elements of  R  is always greater than or equal to zero, with equality only if all the elements in question are zero."  (see ita-2-2-12-pg20)

"For by O5, the statement  [Graphics:../Images/index_gr_122.gif]  implies [Graphics:../Images/index_gr_123.gif], and a sum of positive elements is positive."  (see ita-2-2-12-pg20)

"Note the special consequence that  [Graphics:../Images/index_gr_124.gif]."  (see ita-2-2-12-pg20)

It seems to me that the  1>0  is another one of those deceptively profound results, and therefore I will discuss it in more detail.  Recall from the preliminary section that the terms positive and negative were defined as follows:  Numbers that are in  R+  are called positive.  Numbers,  a,  whose additive inverse,  -a,  is in  R+  are called negative.  Also recall that the greater than symbol,  >,  was defined in the preliminary section as follows:  a>b  means that  a-b  is in  R+.  Consequently,  1>0  means that  1-0=1+(-0)=1+0=1  is in  R+.  But this is an assertion, and not a proof, that  1  is in  R+.  A more detailed proof follows.


Also note, in the assertion above that  1  is in  R+,  it was assumed that  -0=0,  and this is so because  -0=-1·0=0  by F10 and F5.


Converted [ in part ] by Mathematica      December 4, 2007