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I want to prove: $a - b = - (b - a)$

I am only allowed to use the following theorems:

“Associativity of +”: (a + b) + c = a + (b + c)

“Associativity of ·”: (a · b) · c = a · (b · c)

“Symmetry of +”: a + b = b + a

“Symmetry of ·”: a · b = b · a

“Additive identity” “Identity of +”: 0 + a = a

“Multiplicative identity” “Identity of ·”: 1 · a = a

“Distributivity of · over +”: a · (b + c) = a · b + a · c

“Zero of ·”: a · 0 = 0

“Unary minus”: a + (- a) = 0

“Subtraction”: a - b = a + (- b)

How would you go about this?

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  • $\begingroup$ Are you learning groups? $\endgroup$ – yuanming luo Sep 21 '19 at 0:31
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By definition, $-(b-a)$ is the unique element (real number?) such that $$-(b-a)+(b-a)=0$$ So, if you are able to prove that $a-b$ holds the same property, we can conclude that $a-b=-(b-a)$ by the uniqueness. Now, observe that $$\begin{align} (a-b)+(b-a)&=\big( a+(-b) \big) + \big( b+(-a) \big) & (\textrm{definition of substraction})\\ &= \big( a+(-b+b) \big) +(-a) & (\textrm{by associativity of +}) \\ &= (a+0)+(-a) & (-b+b=0 \textrm{ for that } b)\\ &= a+(-a) & (a+0=a \textrm{ for that } a) \\ &= 0. \end{align}$$ Thus, $a-b=-(b-a)$.

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I may be misinterpreting the question, but I do believe you apply the Distributive Law. $$-(b-a)=-1\cdot (b+(-a))=-b+-(-a)=\boxed{a-b}.$$ Is this what you want?

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    $\begingroup$ You need before prove that $-a=-1a$ for all $a$. $\endgroup$ – azif00 Sep 21 '19 at 0:52
  • $\begingroup$ You don't have to prove that, right? Isn't it just a notation? $\endgroup$ – guest Sep 21 '19 at 1:13
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    $\begingroup$ Yes, we must do it. The notation is for $a-b$ which means $a+(-b)$. $\endgroup$ – azif00 Sep 21 '19 at 1:20
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$a-b=a+(-b)\\a+(-b)+b-a=0\\a+(-b)=-(b-a)\\a-b=-(b-a)$

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Apply unary minus on both sides:

$$a - b = - (b - a)\\ a - b + b-a= - (b - a)+b-a\\ a-b+b-a=0\\ a-a=0$$

Equality holds, so the proposition holds.

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    $\begingroup$ That could be more persuasive in reverse, starting with $a+(-a)=0$ and ending with the desired result $\endgroup$ – Henry Sep 21 '19 at 9:52
  • $\begingroup$ I could add a few steps and descriptive language, but I think the TR answer does what is required. This is effectively a hint on an alternative approach at this point. $\endgroup$ – abiessu Sep 21 '19 at 20:41
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First, we have following equation:

$$\begin{align}(b - a) +(a - b)&=[b+(-a)]+[a+(-b)]\\&=b+\{(-a)+[a+(-b)]\}\\&=b+\{[a+(-a)]+(-b)\}\\&=b+[0+(-b)]\\&=b+(-b)\\&=0\end{align}$$

Then we can add $-(b-a)$ to the both sides of above equation from the left

$$\begin{align}(-(b-a))+[(b - a) +(a - b)]&=(-(b-a))+0\\ [(-(b-a))+(b - a)] +(a - b)&=-(b-a)\\ 0+(a - b)&=-(b-a)\\ (a - b)&=-(b-a)\end{align}$$

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