Prove the following consequences of the field axioms using only the six field axioms and the first four theorems of the chapter (the axioms and theorems can be found on p. 18 of Apostol, or anywhere else that lists axioms for a field should give roughly the same information).

- Prove that .
- Prove that .
- Prove that if and , then . Show also that the multiplicative identity 1 is unique.
- Prove that given with there is exactly one such that .
- Prove that if , then .
- Prove that if , then .
- Prove that if then or .
- Prove that and .
- Prove that if and , then .
- Prove that if and , then .
- Prove that if , then .

* Note: For the first few of these I was super explicit, listing each axiom and theorem. After maybe part (c) I was a bit more terse. Each step is still included, and hopefully it is clear which axiom is being used. If not, leave a comment and I can add in a few more details.*
* Proof. * We compute, with justifications on the right,

* Proof. * First, by Axiom 1 (commutativity). Then,

* Proof. * Let with and . Since , we know (Axiom 6) it has an inverse , i.e., there is a such that . Then,

This shows that the multiplicative identity is unique since if is also a multiplicative identity, then

* Proof. * Since and , we know there exists a such that (Axiom 4). Let . Then,

Further, this solution is unique since if is another solution, then and implies implies by part (c)

* Proof. * By definition of , we have

* Proof. * By definition of we know . By the same reasoning, . Hence, by part (c) (Theorem I.7) we have
* Proof. * Let with .

Then, if , we know there exists such that . Thus,

But,

* Proof. * Using the axioms and theorems we have,

And then, from this result (with in place of ) we have , and so,

* Proof. * We know from (d) that for and we have,

* Proof. * Again, since and we know by part (g) that , so exists and we have,

* Proof. * Since we know so exists. Similarly, we know exists. So,

*Related*

$in k everything is right, is just you forgot the double inverse of (c^-1d)$

line 6

The last proof is way more easy:

https://imgur.com/CqqnoRp

Thank you for posting this. This solution is elegant and correct.

I’ve also noticed that in part c), you need to state that b ≠ 0 in order to conclude that 1=1′.

There are some typo’s:

In the beginning of part i), “we know from (d)” should be “we know from (e) (Theorem I.9);

In the beginning of part k), it is (a/b)/(c/d) and not “(a/c)/(b/d)”.

Good job on the solutions, the proofs are clear!

I believe that are also some typo’s in the middle of part k). If you have the time, take a look at it.

I believe that in the middle of part k), you are using the fact that if x ≠ 0 and y ≠ 0, then (xy)^(-1) = x^(-1) ⋅ y^(-1): fact that has not been proved yet (it is exercise 8)

In proof e it is written By definition of b/a,

Can u please explain where is this written in the book?

for part (i) I had a different solution:

Let bx = a and dy = c. Multiply the first equation by d and the second equation by b. This yields: bdx = ad and bdy = bc. Adding up the first and second equations yields: bdx + bdy = bd(x + y) = ad + bc. This implies a/b + c/d = (ad + bc)/(bd)

For a follow up and in a similar vein, for part (j):

Let bx = a and dy = c. If we multiply the two equations together we get: bdxy = bd(xy) = ac. This implies (a/b) (c/d) = (ac/bd).

as a follow up, for part (j):

Let bx = a and dy = c. Multiply the two equations together and this yields bdxy = bd(xy) = ac. This implies (a/b)(c/d) = ac/bd.

You could try something similar for division, as in part k, but admittedly this way of proving it may not be as intuitive.

Hello,

On my Apostol Copy (which is admittedly old) it also asks to prove the first 4 theorems, including the cancellation law for addition. Do you have that anywhere?

Hi, thanks for the solutions.

For part (b), how did you deduce that aa + a(-a) = aa – aa from part (a) ?