Firstly I would like to apologize for not knowing what forum to post this in. I am taking a calculus class but I am pretty sure this is not calculus.
Also I would like to note that this is my first proof based math class so this is all very new to me.
Here's the problem:
Let 'a' be a member of the real number set. Prove that a has a real square root if and only if a >or= 0.
I don't want the answer. I want help on how to do this problem so that hopefully I can do the entire problem set.
Thanks
Axioms:
1) If a, b, and c are real numbers, then
a + (b + c) = (a + b) + c
2) If a is any number, then
a + 0 = 0 + a = a
3) For every number a, there is a number -a such that
a + (-a) = (-a) + a = 0
4) If a and b are any numbers, then
a + b = b + a
5) If a, b, and c, are any numbers, then
a * (b * c) = (a * b) * c
6) If a is any number, then
a *1 = 1 * a = a
7)For every number a other than 0, there is a number a^-1 such that
a * a^-1 = a^-1 * a = 1
8) If a and b are any numbers, then
a * b = b * a
9) If a, b, and c are any number, then
a * (b + c) = a* b + a * c
*Note: For the final 3 axioms the set of positive numbers will be denoted by P
10) For every number a, one and only one of the following holds:
i) a = 0
ii) a is in P
iii) -a is in P
11) If a and b are in P, then a + b is in P
12) If a and b are in P, then a * b is in P
PS. These are the axioms in the first chapter of my textbook. I also emailed my professor to make sure but for now we can assume that these are the correct axioms.