square of a real number is always positive I was working on a problem and had to prove firsthand that $a^2>0$ if $a$ is not $0$ and I did not find anything on the site that gave a proof using the properties I'm using so I was wondering if mine was good:
$P$ is the set of all positive numbers
Case 1
If $a>0$ then $(a-0) = a\in P$.
Then $(a-0)a\in P$ (closure under multiplication)
$\Rightarrow (a^2 -0)\in P \Rightarrow a^2>0$
Case 2
If $a<0$ then $(0-a) = -a\in P$.
Then $-a(0-a)\in P$ (closure under multiplication)
$\Rightarrow (0 + a^2)\in P$ considering $((-a)(-b) = ab)$
$\Rightarrow (a^2 -(-0)) = (a^2 - 0)\in P \Rightarrow a^2>0$
I feel like $(-a)(-b) = ab$ kind of implies what I'm trying to prove...
Thank you for your help!!
 A: Short answer:
Let $a$ be positive and note that $-a=(-1)a$. Then by associativity of the multiplication,
$$(-a)^2=(-1)^2a^2=a^2.$$
A: What axioms are you using?  How was $P$ closed under multiplication proven?  I think your answers are probably correct if those indeed are the axioms and properties actually available to you (they are, of course correct) however I think the use of $a-0$ and $0-a$ was clunky and unnecessary.
That standard method is that you are given axioms
$a < b \implies a+x < b+x$
and $x > 0; a < b \implies ax < bx$
And you have that additive and multiplitive identities inverses exist and distribution property applies.
From that you can prove:
i) $-(-a) = a$ Pf:  $-a + a = 0;  - a + -(-a) = 0$.  (It's an axiom that additive inverses always exist.)  So $a + (-a) + a = a + (-a) + -(-a)\implies 0 + a = 0 + -(-a) \implies a = -(-a)$.
ii) If $a > 0$ then $-a < 0$ (and vice versa).  Pf: $a > 0 \implies a+(-a) > 0 - a \implies 0 > -a$.
iii) $x*(-y) = -xy$.  Pf:  Let $x*(-y)+ xy = x(-y + y) = x*0 = 0$ (Oops, we must also prove $a*0 = 0$.  See addendum). And there for $-xy = x*(-y)$.  
iv)  $x^2 \ge 0$ with equalition holding only if $x = 0$.
Pf:  Case 1:  $x > 0$ then $x*x > x*0 =0$.   
Case 2: $x = 0$ then $x^2 = 0$.
Case 3: $x < 0$ then $-x > 0$ and $x*(-x) < 0$ so $-x^2 < 0$ and $x^2 > 0$.
v)  $1 > 0$ Pf:  $1 \ne 0$ and $1^2 = 1$.
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Addendum.  $a*(0) = a*(0 + 0) = a*(0) + a*(0)$.  So $a*(0) + (-a*(0)) = a*(0) + a*(0) + (-a*(0)) \implies 0 = a*0 + 0 = a*0$.
