Currently working through Apostol need a check on my logic

[VectorField]

Homework Statement

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

Homework 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.

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.

 

[SammyS]

Homework Statement

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

Homework 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.

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.

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.