# Currently working through Apostol need a check on my logic

1. Jul 30, 2012

### VectorField

1. The problem statement, all variables and given/known data
Prove the following theorem.
If a≠0, then (a-1)-1=a

Theorems proven before/axioms I am allowed to use:
Existence of reciprocals axiom: there exist real numbers x and y where x≠0 such that xy=1
Possibility of division Thm: basically a-1=1/a

2. Relevant equations
I just need a check on my proof since I am self studying the book. Any logical mistakes on my proof or suggestions etc? This is my first time writing proofs.

3. The attempt at a solution
Assume (a-1)-1=a is true.
Then by the existence of reciprocals axiom there exists an x such that (a-1)-1x=1.
Hence,
(a-1)-1x=ax​
ax=1​
Now by the possibility of division theorem we have,
x=1/a=a-1
Finally since we assumed (a-1)-1=a was true it follows that:
(a-1)-1x=ax=aa-1=1​
Q.E.D​

It looks fine to me, I'm just a noob at proofs. Thanks.

2. Jul 30, 2012

### SammyS

Staff Emeritus
According to the first line of your proof,
Assume (a-1)-1=a is true. ​
you're assuming that the thing you're proving is true.

That's a definite No-No .

I suggest starting with:
a-1 has a reciprocal, (a-1)-1.

Therefore, (a-1)-1 a-1 = 1
...
Now, multiply that by a, with a being on the right.

Last edited: Jul 30, 2012
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