Proof for $0a = 0$ Is this a valid proof for $0a =0$? I am using only Hilbert's axioms of the real numbers (for simplicity).
$(a+0)(a+0) = a^2 + 0a + 0a + 0^2 = (a)(a) = a^2$
Assume that $0a$ does not equal zero.
Then from $a^2 + 0a + 0a + 0^2$ we get $(a+0)(a+0) > a^2$ or $(a+0)(a+0) < a^2$ which contradicts the proven above statement.
Therefore, $0a = 0$
QED

$0+0=0.\\\text {Multiplying on both sides by a from the right,}\\ [0+0].a=0.a.\\or, 0.a+0.a=0.a.\\\text {Adding -0.a on both sides,}\\-(0.a)+[0.a+0.a]=-(0.a)+0.a\\or, [-(0.a)+-0.a]+0.a=-(0.a)+0.a.\\or, 0+0.a=0.\\or, 0.a=0. $
