- #1
chollapete
- 4
- 0
Homework Statement
Not a specific exercise per se; just a question from reading the text. In working the proof shown at the bottom of page 4 (2nd edition), I realized that the step "If a + x = a,
then (-a) + (a + x) = (-a) + a" could not be justified by any of Postulates 1, 2, or 3. At this point, that is all we have to work with.
The missing postulate is: If a = b, then a + c = b + c.
Homework Equations
Postulate 1: If a, b, and c are any numbers, then a + (b + c) = (a + b) + c.
Postulate 2: If a is any number, then a + 0 = 0 + a = a.
Postulate 3: For every number a, there is a number -a such that a + (-a) = (-a) + a = 0.
Also, he assumes that given numbers a and b, a+b and ab exist.
The Attempt at a Solution
If a = b,
then a + 0 = b; (P2)
hence a + (c + (-c)) = b; (P3)
hence (a + c) + (-c) = b; (P1)
hence ((a + c) + (-c)) + c = b + c; ? At this point I believe I need the "missing postulate".
hence (a + c) + ((-c) + c) = b + c; (P1)
hence (a + c) + 0 = b + c; (P3)
hence a + c = b + c. (P2)
Any ideas on how to be rigorous here?
P.S. First post on this forum. :-) Can anyone clue me in on how to use TeX code? Thanks.