Is there a simpler way to prove that b = 0 if a + b = a?

[Spoo Money]

Homework Statement

I referenced this theorem from the following webpage:

http://mathonline.wikidot.com/theorems-on-the-properties-of-the-real-numbers

Homework Equations

The Attempt at a Solution

The proof makes perfect sense, but why must it be so roundabout? A real number axiom is a + 0 = 0. Can I not just state this, making b = 0 obvious, and be done with the proof? The actual proof does reference other identities, but it seems roundabout in a way. The same can be said for theorem 2 (if a x b = a, can I not simply reference the real number property a x 1 = a ( making b = 1 obvious ) and be done with it? )

To state it more concisely, what would be wrong with the following line of thinking?

Suppose a + b = a

According to Axiom A3, a + 0 = a.

Clearly, since the right side of the equation shows that a is the answer, the only number that a on the left side could have been added to was 0.

Therefore, b must be 0

 

[andrewkirk]

Clearly, since the right side of the equation shows that a is the answer, the only number that a on the left side could have been added to was 0.

It's a good rule of thumb that, if one finds oneself needing to say things like 'clearly', one doesn't have a proper proof.

The right hand side of the equation shows that 'a' is an answer, not the answer. We need to use additional information about the nature of the '+' operation in order to show that it is unique. That's what the linked proof does.

Imagine, for instance that, instead of '+' having its usual meaning, $a+b$ means $a\cdot a^b$ if $a\neq 0$ and is 0 otherwise. Then it will still be the case that, for any $a$ we have $a+0=a$. But if $a=\pm 1$ then 0 is not the only value of $b$ that can satisfy the equation $a+b=a$.

 

[Mark44]

A real number axiom is a + 0 = 0. Can I not just state this, making b = 0 obvious, and be done with the proof?

There is no such axiom that a + 0 = 0. Was this a typo? Did you mean a + 0 = a? The govening axiom here is that 0 is the additive identity element for addition in the field of real numbers.

 

[FactChecker]

They are going through the proof so that every single step can be justified by a very fundamental property of numbers (the Axioms). The explanations on each line below explain how that line was arrived at from the prior line.

a+b=a              given
(a+b)+(-a)=a+(-a)  subtract equal additive inverse from both sides (Axiom A4 and substitution of equals)
(b+a)+(-a)=a+(-a)  commutitive property (Axiom A1)
b+(a+(-a))=a+(-a)  associative property (Axiom A2)
b+0=0              additive inverse (Axiom A4)
b=0                0 is defined as additive identity (Axiom A3)

It's important when starting out learning pure mathematics proofs to get used to having a rock-solid justification of every step of a proof.