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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

Jimmy360
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    In writing $a^2 + 0a + 0a + 0^2 = (a)(a)$, didn't you make use of the fact that $0a = 0$? – pjs36 Apr 17 '15 at 04:06
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    I thought they were using that $a+0=0$? – Nishant Apr 17 '15 at 04:06
  • @pjs36 Nishant is right. – Jimmy360 Apr 17 '15 at 04:08
  • Ah, I see; I misread the situation. I believe this works for commutative rings (i.e., what if $0a = -a0$?) not of characteristic two. Maybe I'm reading into things too much... – pjs36 Apr 17 '15 at 04:17
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    @pjs36 I think the real problem here is that OP has not listed what s/he is allowed to assume. For example, if you use the fact that $a+0=0$, then how are you allowed to just assume that? Because it's obvious? Well isn't $a\cdot 0=0$ obvious as well? Or should I have written $0\cdot a=0$? Fundamental questions like this, in my opinion, should certainly list what all is allowed to be assumed. Otherwise, almost all answers will be questionable to some degree. – Daniel W. Farlow Apr 17 '15 at 04:30
  • @MagicMan Only Hilbert's axioms of the real numbers. – Jimmy360 Apr 17 '15 at 04:32
  • Is this in context to a ring, or a group, or a field etc.? – 9301293 Apr 17 '15 at 04:32

2 Answers2

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These sorts of questions really do need the surrounding assumptions to be on display, or its hard to know what to assume. That said, proofs that "do the same thing to both sides", or rely on cancelling from both sides or what have you, are a little weak imo.

Consider \begin{align*} 0 & = a + (-a)\\ & = 1\cdot a + (-a)\\ & = (0+1)\cdot a + (-a)\\ & = 0\cdot a + 1\cdot a + (-a)\\ & = 0\cdot a + a + (-a)\\ & = 0\cdot a. \end{align*} It might be a bit stuffier to do proofs like this, but they always feel so much tighter.

Also, the existence of a $1$ isn't needed. You should try to tidy up user223741's proof to see this.

user24142
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$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. $

user223741
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