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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?

gebruiker
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SS_C4
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    I think what you want to prove is that the additive inverse of $0$, denoted by $-0$, equals $0$. Is that correct? – Noble Mushtak Jan 31 '16 at 14:43
  • Yes, if I denote $+0$ by $0$, which is correct. (I assume) – SS_C4 Jan 31 '16 at 14:45
  • I'm jealous that you got to learn these axioms in high school. Didn't see these until I was well into college. – Clarinetist Jan 31 '16 at 14:48
  • @Clarinetist Where did OP say he learned any axioms? – Akiva Weinberger Jan 31 '16 at 14:51
  • @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. – Clarinetist Jan 31 '16 at 14:53
  • I don't know what a field axiom is. But I still can understand NobleMushtak's answer – SS_C4 Jan 31 '16 at 14:55
  • @SS_C4 List of Field Axioms. – Clarinetist Jan 31 '16 at 14:56
  • I learned the zero property of addition ($0+a=a$) in Grade 3 and I think the cancellation property of addition ($a+(-a)=0$) is taught with negative numbers which is introduced in Grade 5. (I skipped Grade 4-6 math, so I'm not sure. – Noble Mushtak Jan 31 '16 at 14:57
  • @SS_C4 Try this one. – Clarinetist Jan 31 '16 at 14:58
  • Ok, yeah. I've learnt them, but not under the name of Field Axioms – SS_C4 Jan 31 '16 at 14:59
  • @SS_C4 Any set satisfying the field axioms is called a field. The set of reals and the set of rationals are both fields, but the set of integers is not since it doesn't have multiplicative inverses. – Akiva Weinberger Jan 31 '16 at 15:45
  • @SS_C4 The set ${0,1}$ is not a field, since the sum $1+1$ is not in this set. However, we can make it a field! Define a new operation $\widetilde+$ by the equations $0\ \widetilde+\ 0=0$, $0\ \widetilde+\ 1=1$, $1\ \widetilde+\ 0=1$, and $1\ \widetilde+\ 1=0$. Now, the set ${0,1}$ is a field where the summation operation is $\widetilde+$ and the multiplication operation is $\times$. (Check the axioms, it works.) – Akiva Weinberger Jan 31 '16 at 15:50
  • Question araises what $-0$ exactly is. Releated: http://math.stackexchange.com/questions/1808347/is-x-x-true-for-x-0 – palsch Jun 18 '16 at 22:22

2 Answers2

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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)

Noble Mushtak
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Hagen von Eitzen
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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, $-0=0$.

Noble Mushtak
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