How to show $a+b=b+a$ correctly?

Question

I have to show that $a+b = b+a$ without the use of the first axiom, which states exactly this. I may use commutativity of multiplication, associativity of addition and multiplication, existence of the neutral element in addition and multiplication, and the existence of the (unique) inverse element of an addition or multiplication

$$a+b = b+a |+(-1(b+a))$$ $$a+b+(-1(b+a)) = (b+a)+(-1(b+a))$$
$$a+b+(-b)+(-a) = (b+a)+(-(b+a))$$

with $c = b+a$

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

Would you consider this a solid proof of the Commutative axiom? I translated everything as good as possible.

New attempt:

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

Under condition that I showed $x+(-x)=0$ and $-(a+b)=(-a)+(-b)$

Answer 1

Yes, this is a valid proof, if you state tha each pair of consecutive equations is equivalent. _Then the first statement is equivalrnt to the last one, and the last one is apperently true. But it is simpler to read if you begin with the last ( or better the next to last) statement and write a sequence of statents where a statement is implied by its predecessor end you end with your first statement. But you should check for each step which axioms/rules you use to deduce this step. You must be sure that you do not use the commutative law in one of your steps.

Answer 2

Yes, it is correct, though a little approximate.

$a+b=b+a\\ \equiv\\ a+b+(-1)(b+a)+(b+a)=(b+a)+(-1)(b+a)+(b+a)\\ \equiv\\ a+b-b-a+b+a=(b+a)(1-1)+(b+a)\\ \equiv\\ a-a+b+a=b+a\\ \equiv\\ b+a=b+a$.

You must subtract and add the term $b+a$ (i.e. add $0$), otherwise you need to invoke the rule $a=b\equiv a+c=b+c$, which isn't taken for granted.

For brevity, several intermediate steps have been omitted.

Answer 3

With you idea, you can prove it directly, ie starting with $a+b$ and ending with $b+a$. $$a+b$$ Add $0$ to this equation $$a+b=a+b+(-(b+a))+(b+a)$$ Suppose you previously proved that $-(b+a)=(-b)+(-a)$ $$a+b = a + b +(-b) + (-a) + (b+a)$$ $b + (-b) = 0$ $$a+b = a + (-a) + (b+a)$$ Same for $a + (-a) = 0$ $$a+b = b+a$$

Answer 4

The proof is standard: consider $c=(1+1)(a+b)$. Then, by distributivity on the right, $$ c=(1+1)a+(1+1)b $$ By distributivity on the left, $$ c=(1a+1a)+(1b+1b) $$ By the property of $1$ and associativity, $$ c=a+(a+(b+b))\tag{1} $$ On the other hand, by distributivity on the left, $$ c=1(a+b)+1(a+b) $$ and, by the property of $1$ and associativity, $$ c=a+(b+(a+b))\tag{2} $$ Now we can use $(1)$ and $(2)$ to state $$ a+(a+(b+b))=a+(b+(a+b)) $$ Sum to both sides, on the left, $-a$: $$ -a+(a+(a+(b+b)))=-a+(a+(b+(a+b))) $$ Use associativity: $$ (-a+a)+(a+(b+b))=(-a+a)+(b+(a+b)) $$ which gives $$ 0+(a+(b+b))=0+(b+(a+b)) $$ By the property of $0$, $$ a+(b+b)=b+(a+b) $$ and, by associativity, $$ (a+b)+b=(b+a)+b $$ Add $-b$ to both sides on the right: $$ ((a+b)+b)+(-b)=((b+a)+b)+(-b) $$ Apply associativity: $$ (a+b)+(b+(-b))=(b+a)+(b+(-b)) $$ so we get $$ (a+b)+0=(b+a)+0 $$ Finally, the property of $0$ yields $$ a+b=b+a $$

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Of course the proof can be abbreviated using more properties at each step and avoiding use of parentheses due to associativity.

One can note that associativity of multiplication is not used.

Note also that using $$ -(a+b)=(-a)+(-b) $$ is equivalent to assuming $a+b=b+a$. Indeed, without commutativity, one can only say that $$ -(a+b)=(-b)+(-a) $$ so assuming $-(a+b)=(-a)+(-b)$ is the same as saying $$ (-a)+(-b)=(-b)+(-a) $$ Since the map $a\mapsto -a$ is bijective, this is the same as saying $a+b=b+a$.

This observation can be used to get a perhaps simpler proof.

First step: $0x=0$

Consider $0x=(0+0)x=0x+0x$. Add $-0x$ to both sides and use associativity to conclude $0x=0$

Second step: $(-1)x=-x$

Consider $x+(-1)x=1x+(-1)x=(1+(-1))x=0x=0$. Sum $-x$ to both sides on the left: $$ (-x)+(x+(-1)x)=(-x)+0 $$ Use associativity: $$ ((-x)+x)+(-1)x=-x $$ So $0+(-1)x=-x$ and finally $(-1)x=-x$

Third step $$ -(a+b)=(-1)(a+b)=(-1)a+(-1)b=(-a)+(-b) $$ QED