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How do you prove $+ 0 = - 0$ ?

I have no clue where to start from. (I am a 11th Grader). Can it be done only using concepts I have learned till now or will I need some more concepts?

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    $\begingroup$ I think what you want to prove is that the additive inverse of $0$, denoted by $-0$, equals $0$. Is that correct? $\endgroup$ Jan 31, 2016 at 14:43
  • $\begingroup$ Yes, if I denote $+0$ by $0$, which is correct. (I assume) $\endgroup$
    – SS_C4
    Jan 31, 2016 at 14:45
  • $\begingroup$ I'm jealous that you got to learn these axioms in high school. Didn't see these until I was well into college. $\endgroup$ Jan 31, 2016 at 14:48
  • $\begingroup$ @Clarinetist Where did OP say he learned any axioms? $\endgroup$ Jan 31, 2016 at 14:51
  • $\begingroup$ @AkivaWeinberger Why else would the OP be asking such a question? Furthermore, given the answer by Noble Mushtak below and OP's comment, it's pretty clear that the OP learned the field axioms. $\endgroup$ Jan 31, 2016 at 14:53

2 Answers 2

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$0$ is the unique(!) number with the property $$\tag1x+0=x$$ for all $x$. For any $y$, $-y$ is the unique(!) number with the property $$\tag2y+(-y)=0.$$

From $(1)$ with $x=0$, we get the following: $$0+0=0$$

Also, from $(2)$ with $y=0$, we get the following: $$0+(-0)=0$$

Thus, by the Transitive Property of Equality, we can set the left side of both of the previous equations equal to each other: $$0+0=0+(-0)$$

Hence, by the Subtractive Property of Equality, we subtract $0$ from both sides of this equation to see that $-0=0$.

(NOTE: The $+$ in $+0$ being redundant and/or misleading)

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  • $\begingroup$ Can I say $-0$ is also a number with the property $x + (-0) = x$ $\endgroup$
    – SS_C4
    Jan 31, 2016 at 14:52
  • $\begingroup$ @SS_C4 You'd have to prove that before you can use that. $\endgroup$ Jan 31, 2016 at 14:55
  • $\begingroup$ @SS_C4 How do you prove that without proving $-0=0$ first? $\endgroup$ Jan 31, 2016 at 15:00
  • $\begingroup$ Shouldn't it be the reverse? I should not be able to prove $0 = -0$ without proving that first? $\endgroup$
    – SS_C4
    Jan 31, 2016 at 15:01
  • $\begingroup$ @SS_C4 I clearly proved $0=-0$ without using that property. $\endgroup$ Jan 31, 2016 at 15:07
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From the definition of $0$, $0$ is the additive identity of $\Bbb{R}$, meaning for all $a \in \Bbb{R}$: $$0+a=a$$

Also, from the definition of negative numbers, for all $b \in \Bbb{R}$: $$b+(-b)=0$$

Thus, if we let $b=0$ in the second equation, we get: $$0+(-0)=0$$

Also, if we let $a=-0$ in the first equation, we get: $$0+(-0)=-0$$

Thus, by the Transitive Property of Equality with the last two equations, $-0=0$.

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    $\begingroup$ Thanks! I understand it now. $\endgroup$
    – SS_C4
    Jan 31, 2016 at 14:48
  • $\begingroup$ I accidentally deleted my post when I went to edit it, but I've undeleted it now, so it's back. $\endgroup$ Jan 31, 2016 at 14:54
  • $\begingroup$ It is not "well-known" in the sense that it is some sort of result. It is an axiom. $\endgroup$ Jan 31, 2016 at 14:58
  • $\begingroup$ @MathematicsStudent1122 I've edited my post to clarify that these properties come from the definition of a field and aren't just theorems. $\endgroup$ Jan 31, 2016 at 15:01

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