Math Help - Prove Theorem

1. Prove Theorem

Does anyone know how to write a FORMAL PROOF for (For all real numbers x)(-(-x)=x)
using the definition (For all real numbers x)(-x = 0-x)
So i need to obviously use the axioms of arithmitec etc... any1 can help me do it? it should take about 20 steps not much less but shouldntbe greater than 20
http://www.math.rutgers.edu/~sussmann/papers/hmw10.pdf is where the axiom rules can be foudn ont he last part of the page

2. Originally Posted by ruprotein
Does anyone know how to write a FORMAL PROOF for (For all real numbers x)(-(-x)=x)
using the definition (For all real numbers x)(-x = 0-x)
So i need to obviously use the axioms of arithmitec etc... any1 can help me do it? it should take about 20 steps not much less but shouldntbe greater than 20
http://www.math.rutgers.edu/~sussmann/papers/hmw10.pdf is where the axiom rules can be foudn ont he last part of the page
Code:
def: for all real x: -x=0-x

1. -x=0-x    ..(def)
2. x+(-x)=0  ..(1 and Sub2)
3. (-x)+x=0  ..(2 and Add2)
4. x=0-(-x)  ..(3 and Sub2)
5. x=-(-x)   ..(4 and def)
RonL

3. Originally Posted by ruprotein
Does anyone know how to write a FORMAL PROOF for (For all real numbers x)(-(-x)=x)
using the definition (For all real numbers x)(-x = 0-x)
So i need to obviously use the axioms of arithmitec etc... any1 can help me do it? it should take about 20 steps not much less but shouldntbe greater than 20
http://www.math.rutgers.edu/~sussmann/papers/hmw10.pdf is where the axiom rules can be foudn ont he last part of the page
This is a consequence of the following propery.

In any group $G$ we have,
$(a^{-1})^{-1}=a, \, \, \, \, \forall \, \, a\in G$
The reals under addition $<\mathbb{R},+>$ for a group. Hence, $-(-x)=x$.

Why is this true?
Because,
$a^{-1}a=aa^{-1}=e$
Thus, the inverse of $a^{-1}$ is $a$.
Thus,
$(a^{-1})^{-1}=a$

4. Originally Posted by ruprotein
using the definition (For all real numbers x)(-x = 0-x)
I just realized this part.
This is not a definition for the negative of a real number. That is the property of zero.

5. Originally Posted by ThePerfectHacker
I just realized this part.
This is not a definition for the negative of a real number. That is the property of zero.
Is it? You can define the negative of a real number as "0 - x", but what do you call it? We are using the definition that 0 - x is the symbol "-x." (Typically at this point if I am remembering correctly we haven't yet defined (-1)x = -x.)

-Dan

6. Originally Posted by topsquark
Is it? You can define the negative of a real number as "0 - x", but what do you call it? We are using the definition that 0 - x is the symbol "-x." (Typically at this point if I am remembering correctly we haven't yet defined (-1)x = -x.)
It is a mistake.

The actual way $-x$ is well-defined as the additive inverse of $x$ in any ring.

$0$ is our identity element.

And the fact that,*)
$(-a)(-b)=ab$
$(-a)b=a(-b)=-ab$
Is a theorem (called "rule of signs") hold true in any ring.
Like the ring of reals.

These are not definitions.

*)I can prove it, if you thus wish.

7. does anyone kno how to write the formnal proof of it using the axioms of arithmetic, it cant be 5 steps, it has to be around 20 i was starting out like

s1. Defintion given
s2. Let a be a real number that is arbitrary
s3. then -a = 0-a
...
...
....
....
...
s17. 0-a=-a
s18. -a = (-a) because for all x = x equality axiom
s19. -(-a) = a
s20. then for all x belonging to real numbers such that (-(-x
)=x)

it would be sumthin lioke that i figure, its all about detail and showing eacxh step, any1 can finish the puzzle?

8. Originally Posted by ruprotein
does anyone kno how to write the formnal proof of it using the axioms of arithmetic, it cant be 5 steps, it has to be around 20 i was starting out like
The proof I wrote is a formal proof.

1)Let $x\in \mathbb{R}$
2)Then there exists $-x \in \mathbb{R}$ such that, $x+(-x)=(-x)+x=0$
3)Thus, $x$ is inverse for $-x$.
4)Thus, $-(-x)=x$.

Notes
------
1)The condition of the problem.
2)Reals form a group.
3)Because $x+(-x)=(-x)+x=0$
4)Because $-(-x)$ is the inverse of $x$.

9. Originally Posted by ruprotein
does anyone kno how to write the formnal proof of it using the axioms of arithmetic, it cant be 5 steps, it has to be around 20 i was starting out like

s1. Defintion given
s2. Let a be a real number that is arbitrary
s3. then -a = 0-a
...
...
....
....
...
s17. 0-a=-a
s18. -a = (-a) because for all x = x equality axiom
s19. -(-a) = a
s20. then for all x belonging to real numbers such that (-(-x
)=x)

it would be sumthin lioke that i figure, its all about detail and showing eacxh step, any1 can finish the puzzle?
If this were so aren't your steps s4 to s16 redundant? Or are you not allowed
to assume that "=" is symmetric (a=b iff b=a)?

RonL

10. im just using the axioms of arithmetic, i understand what you mnean with invrerse, but it ewould e wrong to provee it like that because it isnt done ussing the axioms of arithmetic, ive attached the pdf file if you need to see the axioms of arithmetic, ONLY that can be used, otherewise this would have been an easy problem

11. Frankly, I see nothing in the set of axioms you posted that allows this to be proved. As the Captain has already told you, this is usually done using the eight ring axioms. This is one in a sequence of theorems leading to the proof that (-a)(-b)=ab. Therefore, it is more than the inverse of an inverse is a group. It is about both multiplication and addition.

12. captain black wouldnt step 2 be
x-(-x)=0?

imn trying to also make this a 20 step process, i think im suppopsew to pick another letter such as a belonging to real numbers such that -a=0-a etc...

13. What!
$\boxed {x-y\in \mathbb{R} }$
Is not an axiom.

As Plato said it satisfies the ring axioms...

1)The set $\mathbb{R}$ is an abelian group under addition (5 axioms: closure, associate, identity, inverse, commutativity).

2)The set $\mathbb{R}$ is closed under multication (1 axiom) and left and right distributive laws hold:
$a(x+y)=ax+ay$
$(x+y)a=xa+ya$.

A total of 8 axioms.
(5+1+1+1=8)

That list you gave is faulty.
Too much stuff.
-----
My own question. Why are these called "axioms"? They certainly can be proved, all of them. Is it because they are independant?

14. Originally Posted by ruprotein
captain black wouldnt step 2 be
x-(-x)=0?

imn trying to also make this a 20 step process, i think im suppopsew to pick another letter such as a belonging to real numbers such that -a=0-a etc...
Direct application of Sub2 and the def gives 0=x+(-x), and I'm afraid I have
assumed symmetry of "=" again.

You should be uncomfortable with something like "x-(-x)=0" because we
have in mind this system representing arithmetic with reals (even though
they appear undefined here), so we do not want for every real x: x+x=0 which
contradicts a property of our model that we want this system of axioms to
represent at some level.

I must say that the order axioms look as though they may permit the
proof of the symmetry of "=", but it does not look nice (it looks as
though the burden is shifted to the commutativity of the logical conectives
"and" and "or".) Also it looks to me as though "=" has the same status here
as the other "logical" symbols which seem to be a bit short on defined properties,
which presumably are already defined.

RonL

15. i tihnk i figured out how it's suppose to be done: IM going to restate the professors question so i can show you whyi cant use anything else
WRITE A FORMAL PROOF OF THE STATEMENT (FOR ALL x BELONGING TO REAL NUMBERS)(-(-x=x), using the definition (FOR ALL x BELONGING TO REAL Numbers)(-x=0-x), the axioms of arithmetic, the equality axion (For all x)(x=x), the 17 rules of logic such as (substitution, special case rule, iff-use, witnessrule, etc..), and NOTHING ELSE.

This is how i approached it, tell me what you guys think, i know its a stupid way the professor wants it done but thats his rules lol well here i go:

Proof
S1. (FOR ALL X BELONGING TO REAL Numbers)(-x=0-x) [Definition given]
2. (For all x)(x=x)
3. Let a be an arbitrary real number
4. a=a using step 2
5. 0 belongs to the set of integers NZ3 Axiom
6. then a - a = 0 via Axion SUB2
7. a+(0-a)=0 Substitution of definition in step 1
8. a = -(0-a)
9. -(0-a)-a=0
10. -(-a) - a = 0
11. from step 6 we substitute 0 ith a-a which gives us -(-a)-a=a-a
12. -(-a)=(a-a)+a
13. -(-a)=0+a
14. (For all x)(For y)(x+y=y+x)
15. So from step 14 0+a=a+0
16. (For all x belonging to real numbers)x+0=x
17. Then -(-a)=a+0
18. from step 16 -(-a)=a
19. Hence (for all x belonging to real numbers)(-(-x)=x