Question concerning rigor of proofs

[Mr.Rockwater]

I've just started Spivak's Calculus and I'm having a few questions concerning the validity of certain of my proofs since some of mine are not the same as the ones in the answer book.

Homework Statement

Here is one of the proof:

I need to prove that $(ab)^{-1}$ = $(a)^{-1}$$(b)^{-1}$

Homework Equations

I must only use the basic laws of numbers :

(PI) (Associative law for addition) a + (b + c) = (a + b) + c.
(P2) (Existence of an additive identity) a + 0 = 0 + a = a.
(P3) (Existence of additive inverses) a + (—a) = (—a) + a = 0.
(P4) (Commutative law for addition) a + b = b + a.
(P5) (Associative law for multiplication) a • (b • c) = (a • b) • c.
(P6) (Existence of a multiplicative a identity) a • 1 = 1 • a = a; 1 ≠ 0
(P7) (Existence of multiplicative inverses) a • $a^{-1}$ = $a^{-1}$• a = 1, for a ≠ 0
(P8) (Commutative law for multiplication) a • b = b • a.
(P9) (Distributive law) a • (b + c) = a • b + a • c.

The Attempt at a Solution

My solution is :

$(ab)^{-1}$ = $(a)^{-1}$$(b)^{-1}$
$\Rightarrow$$(ab)^{-1} ab$ = $(a)^{-1}$$(b)^{-1} ab$ (Multiply both sides by ab)
$\Rightarrow$$(ab)^{-1} ab$ = $(a)^{-1}a$$(b)^{-1}b$ (Commutative law for multiplication)
$\Rightarrow$$1 = 1$ (Existence of a multiplicative inverse)

Would this be regarded as a rigorous proof? I want to get good at this!

 

[Ray Vickson]

I've just started Spivak's Calculus and I'm having a few questions concerning the validity of certain of my proofs since some of mine are not the same as the ones in the answer book.

Homework Statement

Here is one of the proof:

I need to prove that $(ab)^{-1}$ = $(a)^{-1}$$(b)^{-1}$

Homework Equations

I must only use the basic laws of numbers :

(PI) (Associative law for addition) a + (b + c) = (a + b) + c.
(P2) (Existence of an additive identity) a + 0 = 0 + a = a.
(P3) (Existence of additive inverses) a + (—a) = (—a) + a = 0.
(P4) (Commutative law for addition) a + b = b + a.
(P5) (Associative law for multiplication) a • (b • c) = (a • b) • c.
(P6) (Existence of a multiplicative a identity) a • 1 = 1 • a = a; 1 ≠ 0
(P7) (Existence of multiplicative inverses) a • $a^{-1}$ = $a^{-1}$• a = 1, for a ≠ 0
(P8) (Commutative law for multiplication) a • b = b • a.
(P9) (Distributive law) a • (b + c) = a • b + a • c.

The Attempt at a Solution

My solution is :

$(ab)^{-1}$ = $(a)^{-1}$$(b)^{-1}$
$\Rightarrow$$(ab)^{-1} ab$ = $(a)^{-1}$$(b)^{-1} ab$ (Multiply both sides by ab)
$\Rightarrow$$(ab)^{-1} ab$ = $(a)^{-1}a$$(b)^{-1}b$ (Commutative law for multiplication)
$\Rightarrow$$1 = 1$ (Existence of a multiplicative inverse)

Would this be regarded as a rigorous proof? I want to get good at this!

You have more-or-less the right idea, but you can clean it up: using a^(-1)*b^(-1) = b^(-1)*a^(-1), you can see (using associativity) that [a^(-1)*b^(-1)]*(ab) = 1, so a^(-1)*b^(-1) is an inverse of ab. However, this leaves one question: how do you know that a nonzero number has just ONE inverse? (If a number had several inverses, we might not have (ab)^(-1) = a^(-1)*b^(-1).)

RGV

 

[Mr.Rockwater]

Hmm.. I guess I just went with the property that says that (ab)^-1 * (ab) = 1 and since they were equal, I assumed that proved it.

I get your point that my proof doesn't necessarily prove that a^-1 * b^-1 = (ab)^-1 and nothing else (and vice-versa). So basically I should be taking one of the members and playing with it until it gives me 2nd member instead of getting them to equal the same value?

 

[Mark44]

$(ab)^{-1}$ = $(a)^{-1}$$(b)^{-1}$
$\Rightarrow$$(ab)^{-1} ab$ = $(a)^{-1}$$(b)^{-1} ab$ (Multiply both sides by ab)
$\Rightarrow$$(ab)^{-1} ab$ = $(a)^{-1}a$$(b)^{-1}b$ (Commutative law for multiplication)
$\Rightarrow$$1 = 1$ (Existence of a multiplicative inverse)

Would this be regarded as a rigorous proof? I want to get good at this!

The first line above is what you are trying to prove, so don't start off by assuming the two quantities are equal.

Your last line is trivially true, but that doesn't necessarily mean that where you started from must be true.

Here's a somewhat exaggerated example:

Prove: 2 = 3
==> 0*2 = 0*3
==> 0 = 0

Does the fact that 0 is obviously equal to itself then somehow imply that the first equation above is also true?

Your work should conclude that (ab)^-1 = a^-1b^-1, not start off assuming this is so.

 

[Mr.Rockwater]

Thank you for your precious advice! Thanks to the school system, I've never learned to actually prove anything in mathematics. That's exactly the type of advice I was looking for