1. Sep 29, 2008

### snipez90

1. The problem statement, all variables and given/known data
Prove that $$(ab)^{-1} = a^{-1}b^{-1}$$, if, $$a,b \neq 0$$

2. Relevant properties
Associative property of multiplication
Existence of multiplicative inverses

3. The attempt at a solution
Since $$a,b \neq 0$$, there exists a number $$(ab)^{-1}$$ such that $$(ab)^{-1}\cdot(ab) = 1$$. Multiplying both sides by $$a^{-1}b^{-1}$$, we have

$$(ab)^{-1}\cdot(ab) \cdot (a^{-1}b^{-1}) = 1\cdot (a^{-1}b^{-1}) \Rightarrow (ab)^{-1}\cdot(a \cdot a^{-1})\cdot(b \cdot b^{-1}) = a^{-1}b^{-1} \Rightarrow (ab)^{-1}\cdot 1 \cdot 1 = a^{-1}b^{-1} \Rightarrow (ab)^{-1} = a^{-1}b^{-1}$$

Any comments on making this shorter or neater or correct would be much appreciated.

2. Sep 29, 2008

### HallsofIvy

Staff Emeritus
You don't want comments pointing out that it is simply wrong?

You whole proof consists of
$$(ab)^{-1}\cdot(ab) \cdot (a^{-1}b^{-1}) = 1\cdot (a^{-1}b^{-1}) \Rightarrow (ab)^{-1}\cdot(a \cdot a^{-1})\cdot(b \cdot b^{-1}) = a^{-1}b^{-1} \Rightarrow (ab)^{-1}\cdot 1 \cdot 1 = a^{-1}b^{-1} \Rightarrow (ab)^{-1} = a^{-1}b^{-1}$$
How does that prove that a-1b-1 is the inverse of ab? Basically you have said
$$(ab)^{-1}\cdot(ab) \cdot (a^{-1}b^{-1}) = a^{-1}b^{-1}[\tex] which is really, trivial because (ab)-1 is, by definition, the inverse of (ab). Suppose I were to assert the "c" is the inverse of (ab) where c is any member of the group, ring, field, or whatever you are working with (you never did say). Following your proof, [tex](ab)^{-1}\cdot(ab)\cdot c= 1\cdot c = c[/itex] No, it does not follow from that that c is the inverse of ab! your real problem is that unless you are working in an Abelian group, and you neither say nor use that, then (ab)-1 is NOT a-1b-1 it is b-1a-1 which, in a non-abelian group is NOT the same. You can prove that by just multiplying (ab)(b-1a-1) and multiplying (b-1a-1)(ab). Your "proof" asserts to prove something that is NOT true. 3. Sep 29, 2008 ### snipez90 Um, what? No, I am pretty sure I am not supposed to use an Abelian group, whatever that is. It sounds like group theory, which is not the point of a first year introduction to theoretical calculus. I am using Calculus by Spivak. He lists properties, most of which are encountered by anyone who has taken an elementary algebra class, that he does not prove for the most part. They are "obvious" algebraic properties of real numbers (some may call them field axioms, whatever). Many of his proofs rely on manipulations or tricks that are based on the fact that these properties are given and true seemingly without the need for proof of these properties. I am not saying that I am not uneasy about the demonstrations but I understand his point of wanting the reader to be familiar with these properties. I did not list many of the properties in my "proof" but I have emulated Spivak's proofs. The point is NOT to examine the core and depth of mathematics such as group theory to understand what an Abelian group is (or perhaps the concept isn't difficult, but it still stands that nothing beyond elementary algebra should be used in these proofs). Maybe it is almost by definition, but it is based on a correct definition. Anyways I am pretty sure I know what I am supposed to do and I made sure everything fit with the properties (or axioms) so until higher math, this problem is done. 4. Sep 30, 2008 ### HallsofIvy Staff Emeritus Properties of WHAT? That's what I am asking. Of the real numbers? Of members of some field, since you mention that. In either case, yes, multiplication is COMMUTATIVE, which is the crucial property here. It that case it is true that (ab)-1= a-1b-1[/sub]. But your proof is much too complicated. Just go ahead and multiply (ab)(a-1b-1) and (a-1b-1)(ab). I would have understood that IF you had said exactly what the problem was! But you just said "prove that (ab)-1= a-1b-1 without saying that a and b are numbers! 5. Sep 30, 2008 ### statdad If [tex] a, b$$ are non-zero numbers, each has a (multiplicative) inverse, and so does $$ab$$ (since the product of non-zero numbers is non-zero)

You need to show that

$$(ab)^{-1} = a^{-1}b^{-1}$$

Do you have an axiom that says inverses are unique? if so, the problem is easy, since all you need to do is show that

$$(ab) \cdot \left(a^{-1}b^{-1}\right) = 1$$

since this is the property the inverse of $$ab$$ must satisfy. A longer, ''chain of equalities" proof, would look like this.

\begin{align*} a^{-1} b^{-1} & = a^{-1} b^{-1} \cdot 1\\ & = \left(a^{-1} b^{-1}\right) \left((ab) \cdot (ab)^{-1} \right) \\ & =\left((a^{-1} b^{-1}) \cdot ab \right) \cdot (ab)^{-1}\\ & = \left(a^{-1} \left(b^{-1} b \right) a\right) \cdot (ab)^{-1}\\ & = \left(a^{-1} \cdot 1 \cdot a\right) \cdot (ab)^{-1} \\ & = \left(a^{-1} \cdot a \right) \cdot (ab)^{-1} \\ & = 1 \cdot (ab)^{-1} = (ab)^{-1} \end{align*}

I tried to put every bit of nitty-gritty into this; clearly one or two steps could be dropped. It is essentially (literally?) the same as the original proof, but I find things on different lines easier to read than all in one list (sue me: I'm older than you, I'm sure)