proof that square root of a negative does not exist.

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

Re: proof that square root of a negative does not exist.

Quote:

Originally Posted by

**dstathis** Prove that a has a real square root if and only if a >or= 0.

The proof of this theorem really depend on the set of axioms about real numbers you have to use. We do not know what axioms you are using.

In using almost any any system we have $\displaystyle a^2\ge 0\iff a\in\Re~.$

Re: proof that square root of a negative does not exist.

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.