Proof of ax=a: A Simple Proposition in Spivak's Calculus

[chemistry1]

I haven't written a lot of proofs so I need the opinion of the experts on my proof of a simple proposition. Here's the various properties I used: (P10) (Trichotomy law) For every number a, one and only one of the following holds: (i) a = 0, (ii) a is in the collection P, (iii) —a is in the collection P.

(P7) For every number a not equal to 0, there is a number a^-1 such that a • a^-1 = a^-1 • a = 1.

(P6) If a is any number, then a • 1 = 1 • a = a.

If ax=a for some number "a" different from 0, then x=1.(Spivak's calculus.)

I consider two cases: a>0 or a<0. By definition (Given in Spivak's calculus) :

a>b if a-b is in the collection P (P being the collection of all positive numbers.)

a>0 because a-o is in collection P by trichotomy law(P10)

So a*x=a

by P7 a*a^-1*x= a*a^-1

1*x=1

by P6 x=1

Second case :

By definition :

a< to b if b>a

a<0 because 0>a Now, we do the same thing as the previous case.

Proposition proven !

Here's Spivak's answer

1=a^-1*a=a^-1*(a*x)=(a^-1*a)*x=1*x=x

Here's my interpretation line by line (just to be sure I am understanding it)

a^-1*a=1
a^-1*(a*x)=1 (Is my interpretation correct in saying that the number "a" can be factorized in a way that makes a=a*x, "x" being a variable which the value is to be determined)
(a^-1*a)*x=1
1*x=1
x=1

So, basically, Spivak is constructing his proof. Is it correct ?

Any opinions ? Thank you!

 

[mfb]

So a*x=a

by P7 a*a^-1*x= a*a^-1

None of your rules would lead to the left-hand side where a^-1 appears in the middle of the previous multiplication.

Second case :

Second case of what?

a^-1*(a*x)=1 (Is my interpretation correct in saying that the number "a" can be factorized in a way that makes a=a*x, "x" being a variable which the value is to be determined)

That does not make sense. This equation is not proven at this step.

1=a^-1*a=a^-1*(a*x)=(a^-1*a)*x=1*x=x

This is right, but it hides the important step: take the initial equation a*x=x and multiply both sides by a^-1 on the left, afterwards simplify.

 

[chemistry1]

-Wait, what do you mean that my rules won't result in a*a^-1*x= a*a^-1 ? Is it only because I've put a^-1 in the middle that it isn't correct ?

-The way I thought about the proof was to justify that it would work wether we were talking about positive numbers or negative numbers, hence the the second case.

-Then did he just put the the variable "x" just like that ?

-You meant a*x=a. Also, "multiply both sides by a^-1 on the left" you meant to multiply both sides and to simplify on only the left side ? Could you rephrase?

 

[chemistry1]

No need to answer, I understand now. THank you!